Chapter 14 Statistics (Additional Questions)

The question asks for the formula used to calculate the mean of grouped data using the direct method.

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Let's examine the given options:

• (A) $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$
  This is the formula for the Assumed Mean Method, where $a$ is the assumed mean and $d_i = x_i - a$ are the deviations of the class marks from the assumed mean.

• (B) $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$
  This is the formula for the Direct Method for grouped data, where $f_i$ is the frequency of the $i$-th class and $x_i$ is the class mark (midpoint) of the $i$-th class. The sum $\sum f_i x_i$ represents the sum of the products of frequencies and their corresponding class marks, and $\sum f_i$ represents the sum of all frequencies (total number of observations).

• (C) $\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h$
  This is the formula for the Step-Deviation Method, where $a$ is the assumed mean, $u_i = \frac{x_i - a}{h}$ are the step-deviations, and $h$ is the class size.

• (D) $\bar{x} = \frac{\sum x_i}{n}$
  This is the formula for the mean of ungrouped data, where $x_i$ are individual observations and $n$ is the total number of observations.

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Comparing the options with the required formula for the direct method for grouped data, we find that option (B) is the correct formula.

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The correct answer is (B).

The question asks for the meaning of $d_i$ in the formula for the mean of grouped data using the Assumed Mean Method: $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$.

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In this formula:

• $\bar{x}$ represents the mean.

• $a$ represents the assumed mean.

• $f_i$ represents the frequency of the $i^{th}$ class.

• $d_i$ represents the deviation of the class mark of the $i^{th}$ class ($x_i$) from the assumed mean ($a$). Mathematically, $d_i = x_i - a$.

• $\sum f_i d_i$ represents the sum of the products of the frequencies and their corresponding deviations.

• $\sum f_i$ represents the sum of all frequencies, which is the total number of observations.

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Based on the definitions, $d_i$ represents the deviation of the class mark from the assumed mean.

Let's check the given options:

• (A) Frequency of the $i^{th}$ class - This is represented by $f_i$.
• (B) Class mark of the $i^{th}$ class - This is represented by $x_i$.
• (C) Assumed mean - This is represented by $a$.
• (D) Deviation of the class mark from the assumed mean ($x_i - a$) - This is the definition of $d_i$.

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Therefore, the correct option is (D).

The question asks how to obtain the class mark of a class interval.

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The class mark (or midpoint) of a class interval is the representative value of that interval. It is calculated as the average of the lower limit and the upper limit of the class interval.

Let's evaluate the given options:

• (A) Subtracting the lower limit from the upper limit. This gives the class size or width of the interval, not the class mark.

• (B) Adding the lower and upper limits and dividing by 2. This is the standard formula for calculating the midpoint or class mark of an interval. If the lower limit is $L$ and the upper limit is $U$, the class mark is $\frac{L+U}{2}$.

• (C) Multiplying the lower and upper limits. This is not how the class mark is calculated.

• (D) Dividing the frequency by the class size. This does not yield the class mark of the interval.

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Therefore, the correct method to obtain the class mark is by adding the lower and upper limits and dividing by 2.

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The correct answer is (B).

The question asks for the formula used to calculate the mode of grouped data.

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The mode for grouped data is calculated using a formula that takes into account the frequencies of the modal class (the class with the highest frequency) and the classes immediately preceding and succeeding it, as well as the lower limit and size of the modal class.

The standard formula for the mode of grouped data is:

Mode $= l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

Where:

• $l$ is the lower limit of the modal class.

• $f_1$ is the frequency of the modal class.

• $f_0$ is the frequency of the class preceding the modal class.

• $f_2$ is the frequency of the class succeeding the modal class.

• $h$ is the class size (assuming equal class sizes).

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Let's compare this formula with the given options:

• (A) Mode $= l + (\frac{f_1 - f_0}{2f_1 - f_0 - f_2}) \times h$ - This matches the standard formula.

• (B) Mode $= l + (\frac{f_0 - f_1}{2f_0 - f_1 - f_2}) \times h$ - The numerator has $f_0 - f_1$ instead of $f_1 - f_0$.

• (C) Mode $= l + (\frac{f_1 - f_0}{f_1 - f_0 + f_2 - f_0}) \times h$ - The denominator is incorrect ($f_1 - f_0 + f_2 - f_0 = f_1 - 2f_0 + f_2$, which is not $2f_1 - f_0 - f_2$).

• (D) Mode $= l + (\frac{f_1 - f_0}{2f_1 - f_0 - f_2}) / h$ - The last term is divided by $h$ instead of multiplied by $h$.

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Thus, option (A) correctly represents the formula for the mode of grouped data.

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The correct answer is (A).

The question asks for the meaning of $f_1$ in the formula for the mode of grouped data: Mode $= l + (\frac{f_1 - f_0}{2f_1 - f_0 - f_2}) \times h$.

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In this formula, which is used to find the mode of grouped data:

• $l$ is the lower limit of the modal class (the class with the highest frequency).

• $f_1$ is the frequency of the modal class.

• $f_0$ is the frequency of the class immediately preceding the modal class.

• $f_2$ is the frequency of the class immediately succeeding the modal class.

• $h$ is the class size of the modal class.

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Based on these definitions, $f_1$ specifically represents the frequency of the modal class.

Let's examine the given options:

• (A) Frequency of the class preceding the modal class. This is $f_0$.
• (B) Frequency of the modal class. This matches the definition of $f_1$.
• (C) Frequency of the class succeeding the modal class. This is $f_2$.
• (D) Cumulative frequency. This term is used in finding the median, not the mode, using this formula.

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Therefore, the term $f_1$ in the given mode formula represents the frequency of the modal class.

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The correct answer is (B).

The question asks about the definition of cumulative frequency.

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In a frequency distribution, the cumulative frequency of a class is the sum of the frequency of that class and the frequencies of all classes that come before it in the distribution.

Let's look at the options:

• (A) Succeeding: Summing frequencies of succeeding classes is not the standard definition of cumulative frequency.

• (B) Preceding: This aligns with the definition where cumulative frequency is the running total of frequencies up to and including the current class. It is the sum of the current class frequency and all frequencies of classes that precede it.

• (C) Having the same frequency as: This is not how cumulative frequency is calculated.

• (D) In the entire distribution: The sum of frequencies in the entire distribution is the total frequency, which is the cumulative frequency of the last class interval, but not necessarily for any intermediate class.

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Therefore, the cumulative frequency of a class is the sum of its frequency and the frequencies of all classes preceding it.

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The correct answer is (B).

The question asks about the starting point for accumulating frequencies in a 'less than' cumulative frequency distribution table.

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In a 'less than' cumulative frequency distribution, the cumulative frequency for a given class interval is the sum of the frequencies of that class and all class intervals preceding it. It represents the total number of observations that are less than the upper limit of the given class interval.

To construct such a table, you start with the frequency of the first class interval. The cumulative frequency for the first class is simply its frequency, as there are no preceding classes.

For the second class interval, the cumulative frequency is the frequency of the first class plus the frequency of the second class.

This process continues, adding the frequency of each class to the cumulative frequency of the preceding class, until the last class interval is reached.

Thus, the frequencies are accumulated sequentially, beginning with the frequency of the first class interval.

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Let's consider the options:

• (A) First: This is correct. The accumulation starts with the frequency of the very first class interval.

• (B) Last: Accumulating from the last class interval is typically done for a 'more than' cumulative frequency distribution.

• (C) Middle: The accumulation process is sequential and does not start arbitrarily from a middle class.

• (D) Modal: The modal class is related to the mode (the most frequent value) and is not relevant to the method of calculating cumulative frequency.

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Therefore, in a 'less than' cumulative frequency distribution table, the frequencies are accumulated starting from the first class interval.

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The correct answer is (A).

The question asks for the formula used to calculate the median of grouped data.

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To find the median of grouped data, we first need to identify the median class. This is the class interval whose cumulative frequency is greater than or equal to $\frac{n}{2}$, where $n$ is the total number of observations.

Once the median class is identified, the median is calculated using the following formula:

Median $= l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h$

Where:

• $l$ is the lower limit of the median class.

• $n$ is the total number of observations (sum of frequencies).

• $cf$ is the cumulative frequency of the class immediately preceding the median class.

• $f$ is the frequency of the median class.

• $h$ is the class size (width) of the median class.

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Let's compare this standard formula with the given options:

• (A) Median $= l + (\frac{\frac{n}{2} - cf}{f}) \times h$ - This matches the standard formula for the median of grouped data.

• (B) Median $= l + (\frac{n - cf}{2f}) \times h$ - The numerator contains $n$ instead of $\frac{n}{2}$.

• (C) Median $= l + (\frac{\frac{n}{2} - cf}{f}) / h$ - The last term is divided by $h$ instead of multiplied by $h$.

• (D) Median $= l - (\frac{\frac{n}{2} - cf}{f}) \times h$ - There is a subtraction sign instead of an addition sign before the fraction term.

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Therefore, option (A) is the correct formula for calculating the median of grouped data.

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The correct answer is (A).

The question asks for the meaning of $cf$ in the formula for the median of grouped data: Median $= l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h$.

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In this formula, which is used to calculate the median of grouped data:

• $l$ is the lower limit of the median class.

• $n$ is the total number of observations (sum of frequencies).

• $cf$ is the cumulative frequency of the class immediately preceding the median class.

• $f$ is the frequency of the median class.

• $h$ is the class size (width) of the median class.

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Based on these definitions, $cf$ represents the cumulative frequency of the class preceding the median class.

Let's examine the given options:

• (A) Frequency of the median class. This is represented by $f$.
• (B) Lower limit of the median class. This is represented by $l$.
• (C) Cumulative frequency of the class preceding the median class. This matches the definition of $cf$.
• (D) Cumulative frequency of the median class. The cumulative frequency of the median class itself is used to identify the median class (it's the first class with a cumulative frequency greater than or equal to $\frac{n}{2}$), but the term $cf$ in the formula refers specifically to the cumulative frequency of the class *before* the median class.

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Therefore, the term $cf$ in the given median formula represents the cumulative frequency of the class preceding the median class.

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The correct answer is (C).

The question asks for the empirical relationship between the three measures of central tendency: Mean, Median, and Mode.

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For a moderately skewed distribution, there is an approximate relationship between the mean, median, and mode. This relationship is known as the empirical formula.

The widely accepted empirical formula is:

Mode $\approx$ 3 Median - 2 Mean

We need to check which of the given options matches this relationship when rearranged.

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Let's examine each option:

• (A) 3 Median = Mode + 2 Mean
  Rearranging this equation to solve for Mode: Mode = 3 Median - 2 Mean
  This matches the standard empirical relationship.

• (B) 2 Median = Mode + 3 Mean
  Rearranging: Mode = 2 Median - 3 Mean. This does not match the standard formula.

• (C) Median = 3 Mode + 2 Mean
  Rearranging: 3 Mode = Median - 2 Mean, so Mode = $\frac{\text{Median} - 2 \text{ Mean}}{3}$. This does not match.

• (D) Mode = Median - Mean
  This is a simple difference and does not represent the empirical relationship.

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Comparing the rearranged options with the empirical formula Mode $\approx$ 3 Median - 2 Mean, we find that option (A) is the correct representation.

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The correct answer is (A).

The question asks what an ogive is a graphical representation of.

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An ogive, also known as a cumulative frequency polygon, is a graph used in statistics. It is specifically designed to represent the cumulative frequency distribution of a dataset.

There are two common types of ogives:

• 'Less than' Ogive: Plots the cumulative frequency against the upper class limits of the class intervals. It starts at 0 on the lower boundary of the first class and ends at the total frequency on the upper boundary of the last class.

• 'More than' or 'Greater than' Ogive: Plots the cumulative frequency (or sometimes the frequency greater than the lower limit) against the lower class limits of the class intervals. It starts at the total frequency on the lower boundary of the first class and ends at 0 on the upper boundary of the last class.

In both cases, the ogive visually shows how the cumulative frequency changes across the class intervals.

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Let's evaluate the given options:

• (A) Frequency distribution: A frequency distribution is typically represented by a histogram or a frequency polygon.

• (B) Cumulative frequency distribution: This is precisely what an ogive represents. It shows the running total of frequencies.

• (C) Class marks: Class marks are the midpoints of the class intervals and are often used in plotting frequency polygons, but an ogive plots cumulative frequency against class boundaries (limits), not just class marks.

• (D) Deviations from mean: Deviations from the mean are related to measures of dispersion and are not directly represented by an ogive.

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Therefore, an ogive is a graphical representation of a cumulative frequency distribution.

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The correct answer is (B).

The question asks about the graphical method to estimate the median of grouped data.

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The median of grouped data is the value that divides the distribution into two equal halves. Graphically, this corresponds to the value on the horizontal axis (the variable) at the point where the cumulative frequency is equal to half of the total number of observations ($\frac{n}{2}$).

Among the standard graphical representations for grouped data:

• A histogram represents the frequency distribution. The mode can be estimated graphically from a histogram.

• A frequency polygon also represents the frequency distribution, connecting the midpoints of the tops of the histogram bars.

• An ogive (or cumulative frequency curve) represents the cumulative frequency distribution. It plots cumulative frequency against the class boundaries.

• A bar graph is typically used for categorical or discrete data, not continuous grouped data distributions for finding measures like the median in this way.

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Since the median is the value corresponding to the cumulative frequency $\frac{n}{2}$, it can be estimated graphically from a graph that displays cumulative frequency. The ogive is precisely this type of graph.

To find the median graphically from an ogive:

• For a 'less than' ogive, locate the point on the cumulative frequency axis corresponding to $\frac{n}{2}$. Draw a horizontal line from this point to intersect the ogive. From the intersection point, drop a vertical line to the horizontal axis. The value on the horizontal axis is the estimated median.

• If both 'less than' and 'more than' ogives are drawn on the same graph, the x-coordinate of their intersection point gives the estimated median.

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Therefore, the median of grouped data can be estimated graphically from an ogive.

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The correct answer is (C).

The question describes the graphical method for estimating the median using a 'less than' ogive and asks what $n$ represents in this process.

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In the context of a frequency distribution, whether grouped or ungrouped, the total number of observations is represented by $n$. When dealing with grouped data, $n$ is the sum of all frequencies, i.e., $n = \sum f_i$.

The median is the middle value of a dataset arranged in order. For a total of $n$ observations, the median is the value corresponding to the $\left(\frac{n}{2}\right)^{th}$ observation (or the average of the $\left(\frac{n}{2}\right)^{th}$ and $\left(\frac{n}{2}+1\right)^{th}$ observations for an even $n$).

On a cumulative frequency distribution graph like an ogive, the cumulative frequency on the y-axis reaches a maximum value equal to the total frequency, $n$.

To find the median graphically from a 'less than' ogive, we locate the cumulative frequency value of $\frac{n}{2}$ on the y-axis. This is because the median is the value below which $\frac{n}{2}$ observations lie.

Therefore, in the expression $\frac{n}{2}$, $n$ represents the total number of observations, which is the sum of all frequencies.

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Let's review the options:

• (A) Class size: This is the width of a class interval ($h$).

• (B) Total frequency: This is the sum of all frequencies ($\sum f_i$) and equals the total number of observations ($n$).

• (C) Number of class intervals: This is the count of the groups in the distribution.

• (D) Median value: This is the value of the variable at the median position, which is what we are trying to find on the x-axis.

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Based on the definition and its use in finding the median position, $n$ represents the total frequency.

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The correct answer is (B).

The question is an Assertion-Reason type question about measures of central tendency.

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Assertion (A): The mean, median, and mode are measures of central tendency.
This statement is true. The mean, median, and mode are the three most common measures used to find a central or typical value of a dataset.

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Reason (R): They all describe the central or typical value of a data set.
This statement is also true. The very definition of a measure of central tendency is a single value that represents the center point or typical value around which other data points cluster. Mean, median, and mode each do this in a different way.

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Now, we need to determine if Reason (R) is the correct explanation for Assertion (A). Assertion (A) states that these three are measures of central tendency. Reason (R) provides the characteristic that defines something as a measure of central tendency (describing the central or typical value). Therefore, the reason directly explains *why* mean, median, and mode are classified as measures of central tendency.

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Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

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The correct answer is (A).

The question presents an Assertion and a Reason regarding measures of central tendency and their relationship in different types of distributions.

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Assertion (A): For a symmetrical distribution, the mean, median, and mode are approximately equal.

This statement is true. In a perfectly symmetrical, unimodal distribution (like the normal distribution), the mean, median, and mode all coincide at the exact center of the distribution. For distributions that are symmetrical but not strictly unimodal, or slightly imperfectly symmetrical, they will be very close or approximately equal.

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Reason (R): The empirical relationship applies mainly to moderately skewed distributions.

This statement is also true. The empirical relationship, typically given as Mode $\approx$ 3 Median - 2 Mean, is an approximate relationship. It is most useful and provides a reasonably good estimate for distributions that are not symmetrical but also not extremely skewed. It describes how the measures of central tendency tend to shift relative to each other as the distribution becomes moderately skewed. For perfectly symmetrical distributions, the relationship holds trivially (since Mean = Median = Mode, say $x$, then $x = 3x - 2x$, which is true), but the relationship itself is not primarily used to describe symmetrical distributions where the equality is inherent to the symmetry. Its practical application is primarily in estimating one measure from the other two when the distribution is moderately skewed.

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Now, let's consider if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states a property of mean, median, and mode in symmetrical distributions (they are approximately equal). Reason (R) states where the empirical relationship is *mainly applied* (moderately skewed distributions).

The fact that mean, median, and mode are equal in a symmetrical distribution is a direct consequence of the symmetry of the distribution itself. The empirical relationship (Mode $\approx$ 3 Median - 2 Mean) is an approximate formula that describes their relationship in skewed distributions, not the reason why they coincide in symmetrical distributions. The empirical relationship is a different concept, primarily useful for understanding or estimating measures in non-symmetrical settings.

Therefore, while both statements are true, the reason provided does not explain why the mean, median, and mode are approximately equal in a symmetrical distribution.

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Both A and R are true, but R is not the correct explanation of A.

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The correct answer is (B).

The question asks to match the statistical measures in Column A with their properties in Column B.

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Let's analyze each statistical measure and its corresponding property:

• (i) Mean: The mean is calculated by summing all values and dividing by the number of values. It is directly influenced by every value in the dataset, including unusually large or small (extreme) values. Therefore, the mean is affected by extreme values.

  Matches with (b) Affected by extreme values.

• (ii) Median: The median is the middle value of a dataset that has been ordered from least to greatest. Its position depends on the number of observations, but its value is not directly calculated from all values, making it less sensitive to extreme values compared to the mean. It represents the central point of the ordered data.

  Matches with (d) Middle value.

• (iii) Mode: The mode is the value that appears most frequently in a dataset. Its determination depends only on the frequency of values, not their magnitude. It is not directly affected by extreme values unless those extreme values happen to be the most frequent ones.

  Matches with (a) Most frequent value.

• (iv) Cumulative Frequency: Cumulative frequency for a given value or class interval is the sum of the frequencies of that value or class interval and all preceding values or class intervals. It represents the total count of observations up to a certain point in the distribution.

  Matches with (c) Sum of frequencies up to a point.

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Based on the matches:

• (i) - (b)
• (ii) - (d)
• (iii) - (a)
• (iv) - (c)

Now, let's check the given options:

• (A) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c) - This matches our findings.
• (B) (i)-(a), (ii)-(d), (iii)-(b), (iv)-(c) - Incorrect match for (i) and (iii).
• (C) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c) - Incorrect match for (ii) and (iii).
• (D) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c) - Incorrect match for (i) and (ii).

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The correct option is the one that lists the correct pairs.

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The correct answer is (A).

The question asks to identify the modal class from the given frequency distribution table of heights of 50 students.

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The modal class is defined as the class interval that has the highest frequency.

Let's look at the provided table:

Height (cm)	Number of Students (Frequency)
150-155	15
155-160	13
160-165	10
165-170	8
170-175	4

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We need to find the highest frequency in the "Number of Students" column:

• Frequency for 150-155 is 15.
• Frequency for 155-160 is 13.
• Frequency for 160-165 is 10.
• Frequency for 165-170 is 8.
• Frequency for 170-175 is 4.

The highest frequency is 15, which corresponds to the class interval 150-155.

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Therefore, the modal class is 150-155.

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Let's check the options:

• (A) 150-155 - This is the class with the highest frequency (15).
• (B) 155-160 - The frequency is 13.
• (C) 160-165 - The frequency is 10.
• (D) 165-170 - The frequency is 8.

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The correct answer matches the class with the highest frequency.

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The correct answer is (A).

The question refers to the height data provided in Question 17 and asks for the value of $\frac{n}{2}$.

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In the context of finding the median of grouped data, $n$ represents the total number of observations.

From the problem description in Question 17, the total number of students is given as 50. This means the total frequency is 50.

So, $n = 50$.

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We need to find the value of $\frac{n}{2}$.

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The value of $\frac{n}{2}$ is 25. This value is used to locate the median class, which is the class interval containing the 25th observation.

Let's check the options:

• (A) 50 - This is $n$.
• (B) 25 - This is $\frac{n}{2}$.
• (C) 15 - This is the frequency of the first class.
• (D) 13 - This is the frequency of the second class.

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The correct option is (B).

The question refers to the height data provided in Question 17 and asks to identify the median class.

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The median class is the class interval that contains the $\left(\frac{n}{2}\right)^{th}$ observation.

From Question 18, we determined that the total number of observations is $n = 50$, and thus $\frac{n}{2} = \frac{50}{2} = 25$.

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To find the class containing the 25th observation, we need to calculate the cumulative frequencies for each class.

Let's construct the cumulative frequency table:

Height (cm)	Frequency ($f_i$)	Cumulative Frequency ($cf$)
150-155	15	15
155-160	13	$15 + 13 = 28$
160-165	10	$28 + 10 = 38$
165-170	8	$38 + 8 = 46$
170-175	4	$46 + 4 = 50$
Total	50

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Now we look for the class whose cumulative frequency is greater than or equal to 25.

• The cumulative frequency for the class 150-155 is 15. This is less than 25.

• The cumulative frequency for the class 155-160 is 28. This is the first cumulative frequency that is greater than or equal to 25.

This means the 25th observation falls within the class interval 155-160.

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Therefore, the median class is 155-160.

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Let's check the options:

• (A) 150-155 - Cumulative frequency is 15.
• (B) 155-160 - Cumulative frequency is 28 (contains the 25th observation).
• (C) 160-165 - Cumulative frequency is 38.
• (D) 165-170 - Cumulative frequency is 46.

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The correct answer is the class interval 155-160.

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The correct answer is (B).

The question asks us to find the approximate mode of a dataset, given its mean and median, using the empirical formula relating the three measures of central tendency.

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The empirical relationship between the mean, median, and mode for a moderately skewed distribution is given by:

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We are given:

• Mean = 20

• Median = 22

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Now, we substitute these values into the empirical formula (1):

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Therefore, the approximate mode of the data set is 26.

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Let's check the given options:

• (A) 26 - This matches our calculated approximate mode.
• (B) 24
• (C) 20
• (D) 18

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The correct answer is (A).

The question asks which of the given options is a correct property of the mode.

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Let's recall the properties of the mode:

• The mode is the value that appears most frequently in a dataset.

• A dataset can have one mode (unimodal), more than one mode (bimodal, multimodal), or no mode at all (if all values have the same frequency).

• The mode is not affected by extreme values, unlike the mean.

• The mode is a measure of central tendency, representing the most typical or common value.

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Now, let's evaluate the given options:

• (A) It is always unique.
  This is incorrect. As mentioned, a dataset can have multiple modes or no mode.

• (B) It is affected by extreme values.
  This is incorrect. The mode is determined by the frequency of occurrence, not the magnitude of extreme values (unless an extreme value happens to be the most frequent).

• (C) It is the value that occurs most frequently.
  This is correct. This is the definition of the mode.

• (D) It is used to find the average of the data.
  This is incorrect. While the mode is a measure of central tendency, the term "average" typically refers to the mean. The mode identifies the most frequent value, not the arithmetic average.

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Based on the properties of the mode, the statement "It is the value that occurs most frequently" is a correct property.

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The correct answer is (C).

The question asks what the cumulative frequencies are plotted against when drawing an ogive.

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An ogive is a graphical representation of a cumulative frequency distribution. There are two main types of ogives:

• 'Less than' Ogive: In this type, the cumulative frequency is plotted against the upper limits of the class intervals.

• 'More than' Ogive: In this type, the cumulative frequency (or the frequency greater than or equal to) is plotted against the lower limits of the class intervals.

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The question refers to drawing "an ogive" without specifying 'less than' or 'more than'. However, the points plotted on the graph relate the cumulative frequency (on the y-axis) to a value from the class interval boundaries (on the x-axis). These values are either the lower limits or the upper limits.

Let's look at the options:

• (A) Lower limits: Used for drawing a 'more than' ogive.

• (B) Upper limits: Used for drawing a 'less than' ogive.

• (C) Class marks: Used for drawing frequency polygons, not ogives.

• (D) Frequencies: Cumulative frequencies are plotted against class limits/boundaries, not against frequencies themselves.

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Both lower and upper limits are used as the values on the x-axis when drawing ogives, depending on the type. However, given that only one option can be selected, and the 'less than' ogive plotting against upper limits is frequently the first introduced or primary example, 'Upper limits' is a common answer in such multiple-choice questions when the type of ogive is not specified. Also, Question 13 specifically refers to the 'less than' ogive, which might suggest the context favors this type.

Therefore, based on common convention and the provided options, the cumulative frequencies are plotted against the upper limits of the class intervals for a 'less than' ogive, which is often referred to simply as an ogive.

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The correct answer is (B).

The question asks about the position of the median in a dataset when the number of observations is odd, after arranging the data in order.

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The median is the middle value of a dataset when it is arranged in either ascending or descending order.

When the number of observations, denoted by $n$, is an odd number, there is a single value that is exactly in the middle of the ordered dataset.

The position of the median observation in an ordered dataset with an odd number of observations ($n$) is given by the formula:

For example, if $n=5$, the median is the $\left(\frac{5+1}{2}\right)^{th} = 3^{rd}$ observation. If $n=7$, the median is the $\left(\frac{7+1}{2}\right)^{th} = 4^{th}$ observation.

This $\left(\frac{n+1}{2}\right)^{th}$ observation is the one located precisely in the center of the ordered list.

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Let's evaluate the given options:

• (A) First: This is the first observation, not the median.

• (B) Last: This is the last observation, not the median.

• (C) Middle: The $\left(\frac{n+1}{2}\right)^{th}$ observation is indeed the middle observation when $n$ is odd.

• (D) Most frequent: This describes the mode, not the median.

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Therefore, if the number of observations is odd, the median is the middle observation after arranging the data in order.

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The correct answer is (C).

The question describes a method for calculating the mean of grouped data where an arbitrary value is chosen from the class marks. We need to identify which method fits this description.

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Let's review the common methods for calculating the mean of grouped data:

• Direct Method: The formula is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$. This method directly uses the class marks ($x_i$) and frequencies ($f_i$) without choosing an arbitrary value as a reference point.

• Assumed Mean Method: The formula is $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$, where $a$ is an assumed mean (an arbitrary value, usually chosen from the class marks to simplify calculations) and $d_i = x_i - a$ are the deviations of the class marks from the assumed mean. This method explicitly involves choosing an arbitrary value ($a$).

• Step-Deviation Method: The formula is $\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h$, where $a$ is an assumed mean, $u_i = \frac{x_i - a}{h}$, and $h$ is the class size. This is a simplified version of the assumed mean method and also involves choosing an arbitrary value ($a$).

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The question specifically mentions "choosing an arbitrary value from the class marks". This is the fundamental step in the Assumed Mean Method. The Step-Deviation method also uses an assumed mean, but the Assumed Mean method is the direct name for the technique based on using an arbitrary mean as a reference.

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Let's look at the options:

• (A) Direct - This method does not involve choosing an arbitrary mean.
• (B) Step-deviation - This method uses an assumed mean, but the Assumed Mean method is the broader term for the principle described.
• (C) Assumed mean - This method is defined by choosing an arbitrary value (assumed mean) to simplify calculations.
• (D) Median - This is a measure of central tendency, not a method for calculating the mean.

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Therefore, the method of calculating mean that involves choosing an arbitrary value from the class marks is called the Assumed Mean method.

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The correct answer is (C).

The question asks which information is required to calculate the median of grouped data using the formula.

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The formula for the median of grouped data is:

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Let's identify the terms in this formula and what they represent:

• $l$: This is the lower limit of the median class. The median class is the class interval where the median lies. To use the formula, we need this value.

• $n$: This is the total number of observations, which is the sum of all frequencies ($\sum f_i$). This value is needed to calculate $\frac{n}{2}$, which helps in identifying the median class and is used in the numerator of the formula.

• $cf$: This is the cumulative frequency of the class immediately preceding the median class. This value is used in the numerator of the formula.

• $f$: This is the frequency of the median class. This value is used in the denominator of the formula.

• $h$: This is the class size (width) of the median class. This value is used to scale the fraction term.

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Based on the formula and the meaning of its components, the required information is:

• The lower limit of the median class ($l$).
• The total frequency ($n$), used to find the position $\frac{n}{2}$.
• The cumulative frequency of the class preceding the median class ($cf$).
• The frequency of the median class ($f$).
• The class size of the median class ($h$).

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Let's examine the given options:

• (A) Lower limit of the median class - Required ($l$).
• (B) Cumulative frequency of the class preceding the median class - Required ($cf$).
• (C) Frequency of the median class - Required ($f$).
• (D) Class size - Required ($h$).
• (E) Total frequency - Required ($n$).

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All the listed options correspond to the components needed to apply the formula for calculating the median of grouped data. Therefore, all of them are required.

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The correct answer is (A), (B), (C), (D), (E).

The question asks what the intersection point of the 'less than' ogive and the 'more than' ogive represents on the x-axis.

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An ogive is a graphical representation of a cumulative frequency distribution.

• The 'less than' ogive plots the cumulative frequency against the upper limits of the class intervals. At any point on this curve, the y-value represents the number of observations less than or equal to the corresponding x-value (which is an upper limit).

• The 'more than' ogive plots the cumulative frequency (or frequency greater than) against the lower limits of the class intervals. At any point on this curve, the y-value represents the number of observations greater than or equal to the corresponding x-value (which is a lower limit).

When both ogives are drawn on the same graph, their intersection point has a significant meaning. At the point of intersection, the number of observations less than the value on the x-axis is equal to the number of observations greater than or equal to that same value on the x-axis.

This value that divides the dataset into two equal halves (where half the observations are below it and half are above it) is the definition of the median.

The y-coordinate of the intersection point corresponds to $\frac{n}{2}$, where $n$ is the total frequency, and the x-coordinate corresponds to the value of the variable at the median.

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Let's evaluate the given options:

• (A) Mean: The mean cannot be directly determined from the intersection of ogives.
• (B) Median: The intersection point's x-coordinate represents the median.
• (C) Mode: The mode can be estimated graphically from a histogram.
• (D) Quartiles: Quartiles can also be estimated from an ogive, but they correspond to different cumulative frequency values ($\frac{n}{4}, \frac{3n}{4}$), not necessarily the intersection point of both ogives (which is specifically at $\frac{n}{2}$).

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Therefore, the intersection point of the 'less than' ogive and 'more than' ogive gives the median on the x-axis.

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The correct answer is (B).

The question asks to identify the modal class from the given frequency distribution table of the daily expenditure of 30 families.

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The modal class is the class interval that has the highest frequency.

Let's look at the provided table:

Daily Expenditure ($\textsf{₹}$)	Number of Families (Frequency)
100-150	6
150-200	7
200-250	12
250-300	3
300-350	2

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We need to find the highest frequency in the "Number of Families" column:

• Frequency for 100-150 is 6.
• Frequency for 150-200 is 7.
• Frequency for 200-250 is 12.
• Frequency for 250-300 is 3.
• Frequency for 300-350 is 2.

The highest frequency is 12, which corresponds to the class interval $\textsf{₹}$ 200-250.

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Therefore, the modal class is 200-250.

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Let's check the options:

• (A) 100-150 - The frequency is 6.
• (B) 150-200 - The frequency is 7.
• (C) 200-250 - This is the class with the highest frequency (12).
• (D) 250-300 - The frequency is 3.

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The correct answer matches the class with the highest frequency.

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The correct answer is (C).

The question asks for the lower limit of the modal class based on the daily expenditure data provided in Question 27.

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From the solution to Question 27, we determined that the modal class is the class interval with the highest frequency. The frequencies are:

• 100-150: 6
• 150-200: 7
• 200-250: 12
• 250-300: 3
• 300-350: 2

The highest frequency is 12, corresponding to the class interval 200-250.

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For a class interval written in the form "Lower Limit - Upper Limit", the lower limit is the smaller value and the upper limit is the larger value.

In the modal class 200-250, the lower limit is 200 and the upper limit is 250.

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Let's check the options:

• (A) 200 - This is the lower limit of the modal class 200-250.
• (B) 250 - This is the upper limit of the modal class 200-250.
• (C) 150 - This is the lower limit of the class 150-200.
• (D) 100 - This is the lower limit of the class 100-150.

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The correct answer is the lower limit of the modal class 200-250, which is 200.

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The correct answer is (A).

The question asks to calculate the mode of the daily expenditure using the data provided in Question 27.

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First, we need to identify the modal class, which is the class with the highest frequency.

From the table in Question 27:

Daily Expenditure ($\textsf{₹}$)	Number of Families (Frequency, $f_i$)
100-150	6 ($f_0$ for next class)
150-200	7 ($f_0$ for modal class)
200-250	12 ($f_1$, Modal Class)
250-300	3 ($f_2$ for modal class)
300-350	2 ($f_2$ for previous class)

The highest frequency is 12, which corresponds to the class interval 200-250.

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Now, we use the formula for the mode of grouped data:

Where:

• $l$ is the lower limit of the modal class = 200.

• $f_1$ is the frequency of the modal class = 12.

• $f_0$ is the frequency of the class preceding the modal class = 7.

• $f_2$ is the frequency of the class succeeding the modal class = 3.

• $h$ is the class size of the modal class = $250 - 200 = 50$.

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Substitute these values into the formula:

Now, calculate the value of $\frac{125}{7}$:

So,

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Comparing our calculated approximate mode ($\textsf{₹} 217.86$) with the given options:

• (A) $\textsf{₹} 216.67$ (approx.) - Difference $|217.86 - 216.67| = 1.19$
• (B) $\textsf{₹} 208.33$ (approx.) - Difference $|217.86 - 208.33| = 9.53$
• (C) $\textsf{₹} 225$ - Difference $|217.86 - 225| = 7.14$
• (D) $\textsf{₹} 233.33$ (approx.) - Difference $|217.86 - 233.33| = 15.47$

The calculated value $\textsf{₹} 217.86$ is closest to option (A) $\textsf{₹} 216.67$. While there is a slight discrepancy, this option is the best approximation among the choices provided, suggesting potential rounding or a minor variation in the source of the question or options.

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The calculated mode is approximately $\textsf{₹} 217.86$. Based on the options, the closest value is $\textsf{₹} 216.67$.

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The correct answer is (A).

The question asks to find the median class from the daily expenditure data provided in Question 27.

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The median class is the class interval that contains the $\left(\frac{n}{2}\right)^{th}$ observation, where $n$ is the total number of observations.

From the table in Question 27, the total number of families (observations) is:

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Now, we calculate $\frac{n}{2}$:

We need to find the class interval that contains the 15th observation. To do this, we calculate the cumulative frequencies.

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Let's add a cumulative frequency column to the table:

Daily Expenditure ($\textsf{₹}$)	Frequency ($f_i$)	Cumulative Frequency ($cf$)
100-150	6	6
150-200	7	$6 + 7 = 13$
200-250	12	$13 + 12 = 25$
250-300	3	$25 + 3 = 28$
300-350	2	$28 + 2 = 30$
Total	30

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We look for the class where the cumulative frequency is greater than or equal to 15.

• For the class 100-150, $cf = 6$, which is less than 15.

• For the class 150-200, $cf = 13$, which is less than 15.

• For the class 200-250, $cf = 25$, which is greater than or equal to 15.

The class interval 200-250 is the first class whose cumulative frequency is 25, which includes the 15th observation.

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Therefore, the median class is 200-250.

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Let's check the options:

• (A) 150-200 - Cumulative frequency up to this class is 13.
• (B) 200-250 - Cumulative frequency up to this class is 25, which contains the 15th observation.
• (C) 100-150 - Cumulative frequency up to this class is 6.
• (D) 250-300 - Cumulative frequency up to this class is 28.

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The correct answer is (B).

The question asks for the correct order of steps involved in calculating the mean of grouped data using the step-deviation method.

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Let's analyze the provided steps and their dependencies:

• S: Determine class marks $x_i$, choose assumed mean $a$, and calculate class size $h$.
  These are initial values needed to perform further calculations ($u_i$). This step must come early.

• P: Calculate $u_i = \frac{x_i - a}{h}$.
  This step requires the class marks ($x_i$), the assumed mean ($a$), and the class size ($h$). These are determined in step S. So, P must come after S.

• Q: Find $\sum f_i$ and $\sum f_i u_i$.
  This step requires the frequencies ($f_i$, which are part of the original data) and the calculated values of $u_i$. So, Q must come after P (and also after the initial data collection/organization implied before these steps, but S is the first listed action). $\sum f_i$ is the total frequency, which is also needed in the final formula (R). $\sum f_i u_i$ requires $f_i$ and $u_i$.

• R: Calculate $\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h$.
  This is the final formula calculation. It requires the assumed mean ($a$), the sum of products of frequencies and step-deviations ($\sum f_i u_i$), the total frequency ($\sum f_i$), and the class size ($h$). These values are obtained from steps S and Q. So, R must come after Q.

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Based on the dependencies, the logical sequence of steps is:

1. Start by preparing the basic necessary information: determine class marks, choose an assumed mean, and find the class size (Step S).
2. Calculate the step-deviation for each class using the values obtained in step S (Step P).
3. Calculate the sum of frequencies ($\sum f_i$) and the sum of the products of frequencies and step-deviations ($\sum f_i u_i$) using the frequencies from the data and $u_i$ from step P (Step Q).
4. Finally, substitute the values from steps S and Q into the formula for the mean (Step R).

The correct order is S $\to$ P $\to$ Q $\to$ R.

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Let's check the options:

• (A) S $\to$ P $\to$ Q $\to$ R - Matches the derived order.
• (B) P $\to$ S $\to$ Q $\to$ R - Incorrect. P requires information from S.
• (C) S $\to$ Q $\to$ P $\to$ R - Incorrect. Q requires $u_i$ from P.
• (D) Q $\to$ S $\to$ P $\to$ R - Incorrect. Q requires $f_i u_i$, which requires $u_i$ from P, which requires information from S.

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The correct answer is (A).

The question asks which of the given options is a measure of central tendency for grouped data.

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Measures of central tendency are values that represent the center or typical value of a dataset. The most common measures of central tendency are the Mean, Median, and Mode.

Measures of dispersion, on the other hand, describe the spread or variability of the data. Examples include Range, Variance, and Standard Deviation.

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Let's analyze each option:

• (A) Range: The range is the difference between the highest and lowest values in a dataset. It is a measure of dispersion, not central tendency.

• (B) Variance: Variance measures the average of the squared differences from the mean. It is a measure of dispersion, not central tendency.

• (C) Median: The median is the middle value of a dataset when ordered. It divides the data into two equal halves. The median is a measure of central tendency, and there is a specific method and formula to calculate it for grouped data.

• (D) Standard Deviation: The standard deviation is the square root of the variance. It measures the typical distance of data points from the mean. It is a measure of dispersion, not central tendency.

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Among the given options, only the Median is a measure of central tendency. All options listed can be calculated for grouped data, but only the median serves to describe the central position of the data.

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The correct answer is (C).

The question asks about the meaning of the y-coordinate of the intersection point of the 'less than' ogive and the 'more than' ogive.

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As established in a previous question (Question 26), the x-coordinate of the intersection point of the 'less than' and 'more than' ogives gives the median of the grouped data.

Let $n$ be the total frequency (total number of observations).

• The 'less than' ogive shows the number of observations less than or equal to a certain value (plotted on the x-axis). It starts at a cumulative frequency of 0 and ends at a cumulative frequency of $n$.

• The 'more than' ogive shows the number of observations greater than or equal to a certain value (plotted on the x-axis). It starts at a cumulative frequency of $n$ and ends at a cumulative frequency of 0.

The intersection point is where the cumulative frequency from the 'less than' perspective is equal to the cumulative frequency from the 'more than' perspective.

The median is the value (on the x-axis) that divides the distribution into two equal parts. This means that half of the total observations are less than the median, and half are greater than the median.

Therefore, at the median value (the x-coordinate of the intersection), the cumulative frequency ('less than' type) is equal to half of the total frequency. The cumulative frequency ('more than' type) is also equal to half of the total frequency at this point.

The y-coordinate of the intersection point represents this cumulative frequency value where the number of observations below the median equals the number of observations above the median. This value is exactly $\frac{n}{2}$.

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Let's check the options:

• (A) The median value: This is the x-coordinate.
• (B) The total frequency ($n$): This is the maximum cumulative frequency, not the value at the intersection.
• (C) Half of the total frequency ($\frac{n}{2}$): This matches the cumulative frequency at the median.
• (D) The mode: The mode is related to peak frequency, not directly to cumulative frequency intersection.

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The y-coordinate of the intersection point of the 'less than' and 'more than' ogives represents half of the total frequency.

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The correct answer is (C).

The question asks us to find the approximate median of a dataset, given its mode and mean, using the empirical formula relating the three measures of central tendency.

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The empirical relationship between the mean, median, and mode for a moderately skewed distribution is approximately:

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We are given:

• Mode = 26

• Mean = 28

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Substitute these values into the empirical formula (1):

Now, we solve for the Median:

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The approximate median of the data set is $27.333...$.

Let's check the given options to find the closest value:

• (A) 26
• (B) 27
• (C) 28
• (D) 29

The value $27.333...$ is closest to 27.

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The correct answer is (B).

The question asks about the most preferred method for calculating the mean of grouped data when dealing with large data sets and large class marks.

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Let's consider the three main methods for calculating the mean of grouped data:

• Direct Method: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$. This method involves multiplying frequencies ($f_i$) by the class marks ($x_i$) and summing these products. If the class marks ($x_i$) are large, the products $f_i x_i$ will also be large, and calculating their sum can involve working with large numbers, which can be cumbersome, especially for large data sets (meaning large frequencies or many class intervals).

• Assumed Mean Method: $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$, where $d_i = x_i - a$. This method involves choosing an assumed mean ($a$) and calculating the deviations ($d_i$) of class marks from the assumed mean. Since $a$ is usually chosen near the center of the distribution, the deviations $d_i$ are typically smaller than the class marks $x_i$. Calculating $f_i d_i$ and their sum involves smaller numbers compared to the direct method, simplifying calculations.

• Step-Deviation Method: $\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h$, where $u_i = \frac{x_i - a}{h}$ and $h$ is the class size. This method further simplifies calculations by dividing the deviations ($d_i$) by the class size ($h$) to get $u_i$. The values of $u_i$ are usually small integers. Calculating $f_i u_i$ and their sum involves even smaller numbers than the assumed mean method. This significantly reduces the computational effort, making it the most efficient method, particularly when class intervals are of equal size and class marks are large.

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When class marks are large, working directly with them in the Direct Method can lead to very large products and sums. The Assumed Mean method simplifies this by using deviations. The Step-Deviation method offers the greatest simplification by working with small integer values ($u_i$), which makes the calculations easier and less prone to errors, especially with large data sets where there are many frequencies and terms to sum.

Therefore, for large data sets and large class marks, the Step-Deviation method is generally preferred for its computational efficiency and simplicity.

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Let's check the options:

• (A) Direct method - Less preferred for large class marks.
• (B) Assumed mean method - Better than the direct method, but less simplified than step-deviation.
• (C) Step-deviation method - Provides the greatest simplification in calculations when class marks are large and class intervals are uniform.
• (D) All methods are equally efficient - Incorrect. The computational complexity differs significantly.

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The correct answer is (C).

The question asks about the process of forming a 'more than' cumulative frequency distribution table.

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In a 'more than' cumulative frequency distribution, the cumulative frequency for a given class interval represents the total number of observations that are greater than or equal to the lower limit of that class interval.

To construct such a table, the frequencies are accumulated starting from the last class interval and moving upwards towards the first class interval.

• The cumulative frequency of the last class is simply its frequency (representing observations greater than or equal to its lower limit).

• The cumulative frequency of the second to last class is its frequency plus the cumulative frequency of the last class (representing observations greater than or equal to the lower limit of the second to last class). This process continues.

• The cumulative frequency of the first class will be the sum of all frequencies, which is the total number of observations, as all observations are greater than or equal to the lower limit of the first class.

Thus, the accumulation proceeds from the end of the distribution towards the beginning.

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Let's evaluate the given options:

• (A) Beginning, end: This describes the accumulation for a 'less than' cumulative frequency distribution.

• (B) End, beginning: This correctly describes the accumulation for a 'more than' cumulative frequency distribution.

• (C) Middle, ends: This is not the standard method.

• (D) Modal class, end: The modal class is not relevant to the cumulative frequency calculation process.

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Therefore, a 'more than' cumulative frequency distribution table is formed by accumulating frequencies from the end of the distribution towards the beginning.

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The correct answer is (B).

The question asks for a TRUE statement about the mode of grouped data.

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Let's analyze each statement:

• (A) It is always one of the class marks.
  This is false. The mode for grouped data is calculated using a formula that interpolates within the modal class. The result is generally a value within the class interval, but not necessarily the class mark (midpoint).

• (B) It is always in the class with the highest frequency.
  This is true. The modal class is defined as the class with the highest frequency. The formula for the mode of grouped data is designed to calculate a value that lies within the boundaries of this modal class.

• (C) It is the exact midpoint of the modal class.
  This is false. The midpoint of the modal class is the class mark. The mode is calculated based on the frequencies of the modal class and its adjacent classes ($f_0, f_1, f_2$) and is only the midpoint in specific symmetrical cases where $f_0 = f_2$.

• (D) It is less affected by extreme values than the mean.
  This statement is also generally true as a property comparing the mode and the mean. For grouped data, the mode calculation focuses on the frequencies of the most common classes, making it less sensitive to the specific values in extreme classes compared to how the mean's calculation uses class marks across the entire distribution. However, option (B) is a more direct and fundamental property related to the definition and calculation of the mode for grouped data. Both B and D are true, but B is a property directly defining the location of the mode in grouped data. Given standard multiple choice question structure, the most specific and fundamental true statement is often expected.

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While statement (D) is a correct general property comparing the mode and mean, statement (B) is a fundamental property directly related to the definition and calculation of the mode within grouped data, stating where the calculated mode value will be located.

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The most accurate statement directly describing a property of the mode as calculated for grouped data is that it lies within the modal class.

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The correct answer is (B).

The question asks for the cumulative frequency for the age group 'Less than 30' based on the provided frequency distribution table.

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The cumulative frequency for 'Less than 30' is the sum of the frequencies of all class intervals whose upper limit is less than or equal to 30.

From the table:

Age (Years)	Number of People (Frequency, $f_i$)
0-10	100
10-20	150
20-30	200
30-40	120
40-50	80

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The class intervals with ages less than 30 are 0-10, 10-20, and 20-30.

The cumulative frequency for 'Less than 30' is the sum of the frequencies of these classes:

Let's perform the addition:

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Alternatively, we can construct the 'less than' cumulative frequency table:

Age (Years)	Number of People ($f_i$)	Cumulative Frequency (Less than upper limit)
0-10	100	Less than 10: 100
10-20	150	Less than 20: $100 + 150 = 250$
20-30	200	Less than 30: $250 + 200 = \textbf{450}$
30-40	120	Less than 40: $450 + 120 = 570$
40-50	80	Less than 50: $570 + 80 = 650$
Total	650

From the cumulative frequency table, the cumulative frequency for 'Less than 30' is 450.

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Let's check the options:

• (A) 200 - This is the frequency of the 20-30 class.
• (B) 350 - This is the cumulative frequency for 'Less than 20'.
• (C) 450 - This is the cumulative frequency for 'Less than 30'.
• (D) 570 - This is the cumulative frequency for 'Less than 40'.

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The correct answer is (C).

The question asks for the cumulative frequency for the age group 'More than or equal to 20' based on the frequency distribution table from Question 38.

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In a 'more than or equal to' cumulative frequency distribution, the cumulative frequency for a given class is the sum of the frequencies of that class and all subsequent classes.

From the table in Question 38:

Age (Years)	Number of People (Frequency, $f_i$)
0-10	100
10-20	150
20-30	200
30-40	120
40-50	80

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The class intervals with age 'More than or equal to 20' are 20-30, 30-40, and 40-50.

The cumulative frequency for 'More than or equal to 20' is the sum of the frequencies of these classes:

Let's perform the addition:

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Alternatively, we can find the total frequency ($n$) and subtract the cumulative frequency 'Less than 20'.

Let's perform the subtraction:

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Both methods yield the same result.

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Let's check the options:

• (A) 200 - This is the frequency of the 20-30 class.
• (B) 400 - This is the cumulative frequency for 'More than or equal to 20'.
• (C) 500 - This is the cumulative frequency for 'More than or equal to 10' ($150+200+120+80 = 550$). Or possibly sum up to 30 (100+150+200 = 450), plus something? Not clear.
• (D) 650 - This is the total frequency.

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The correct answer is (B).

The question asks about the required characteristic of class intervals when calculating the mean using the step-deviation method.

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The formula for the mean using the step-deviation method is:

In this formula, $u_i = \frac{x_i - a}{h}$, where $h$ is the class size. The purpose of the step-deviation method is to simplify calculations by dividing the deviations ($d_i = x_i - a$) by a common factor, which is typically the class size ($h$).

For $h$ to be a common divisor applied uniformly across all class intervals to calculate $u_i$, the class size must be the same for all (or at least most) class intervals. If the class intervals have unequal sizes, a single value of $h$ cannot be used in the $u_i$ calculation for all classes in a way that simplifies the numbers across the entire distribution. While it is theoretically possible to adapt the method, the standard step-deviation formula assumes equal class intervals.

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Let's consider the options:

• (A) Equal: This condition allows a common class size $h$ to be used in the simplification $u_i = \frac{x_i - a}{h}$, which is central to the step-deviation method.

• (B) Unequal: The standard step-deviation method is not applicable in a straightforward manner when class intervals are unequal. The Assumed Mean method would be more appropriate in such cases.

• (C) Any: This is incorrect, as the method relies on uniform class size for simplification.

• (D) Zero: Class size cannot be zero.

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Therefore, to calculate the mean using the step-deviation method, the class intervals must be of equal size.

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The correct answer is (A).

Given:

The frequency distribution is given as:

Class Interval	Frequency
10-20	5
20-30	8
30-40	12
40-50	15
50-60	10

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To Find:

The mean of the given frequency distribution using the Direct Method.

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Solution:

The Direct Method formula for calculating the mean ($\overline{x}$) of a grouped frequency distribution is:

Where:

$f_i$ is the frequency of the $i$-th class interval.

$x_i$ is the class mark (midpoint) of the $i$-th class interval.

$\sum f_i$ is the sum of all frequencies.

$\sum f_i x_i$ is the sum of the products of frequencies and their respective class marks.

We construct a table to calculate the required values:

Class Interval	Frequency ($f_i$)	Class Mark ($x_i$)	$f_i x_i$
10-20	5	$\frac{10+20}{2} = 15$	$5 \times 15 = 75$
20-30	8	$\frac{20+30}{2} = 25$	$8 \times 25 = 200$
30-40	12	$\frac{30+40}{2} = 35$	$12 \times 35 = 420$
40-50	15	$\frac{40+50}{2} = 45$	$15 \times 45 = 675$
50-60	10	$\frac{50+60}{2} = 55$	$10 \times 55 = 550$
Total	$\sum f_i = 50$		$\sum f_i x_i = 1920$

From the table, we have:

Now, substituting these values into the formula (i):

Simplifying equation (ii):

Therefore, the mean of the given distribution is $38.4$.

The final answer is 38.4.

Given:

The frequency distribution is given as:

Marks	Number of Students (Frequency, $f$)
0-10	3
10-20	9
20-30	15
30-40	30
40-50	18
50-60	5

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To Find:

The mode of the given frequency distribution.

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Solution:

To find the mode of a grouped frequency distribution, we first need to identify the modal class. The modal class is the class interval with the highest frequency.

In the given data, the highest frequency is 30, which corresponds to the class interval 30-40.

So, the modal class is 30-40.

The formula for finding the mode of a grouped data is:

Where:

$l$ = lower limit of the modal class

$f_1$ = frequency of the modal class

$f_0$ = frequency of the class preceding the modal class

$f_2$ = frequency of the class succeeding the modal class

$h$ = class size (width of the class interval)

From the given data and the identified modal class (30-40), we have:

Now, substituting these values into the formula (i):

Simplifying equation (ii):

The mode of the given data is approximately $35.56$.

The final answer is approximately 35.56.

Given:

The frequency distribution of weekly wages is given as:

Weekly Wage ($\textsf{₹}$)	Number of Workers (Frequency, $f$)
100-120	12
120-140	14
140-160	8
160-180	6
180-200	10

---

To Find:

The median weekly wage of the workers.

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Solution:

To find the median of a grouped frequency distribution, we need to calculate the cumulative frequencies.

Let's construct a table with cumulative frequencies:

Weekly Wage ($\textsf{₹}$)	Number of Workers (Frequency, $f_i$)	Cumulative Frequency (cf)
100-120	12	12
120-140	14	$12 + 14 = 26$
140-160	8	$26 + 8 = 34$
160-180	6	$34 + 6 = 40$
180-200	10	$40 + 10 = 50$
Total	$N = \sum f_i = 50$

The total number of workers is $N = 50$.

We need to find the class where the $\frac{N}{2}$-th observation lies.

The cumulative frequency just greater than or equal to 25 is 26, which corresponds to the class interval 120-140.

Therefore, the median class is 120-140.

Now, we use the formula for the median of a grouped data:

Where:

$l$ = lower limit of the median class

$N$ = total number of observations

$cf$ = cumulative frequency of the class preceding the median class

$f$ = frequency of the median class

$h$ = class size

From the median class (120-140) and the table, we have:

Now, substituting these values into the formula (i):

Simplifying equation (ii):

The median weekly wage is approximately $\textsf{₹}$ 138.57.

The final answer is approximately $\textsf{₹}$ 138.57.

Given:

The frequency distribution is given as:

Class Interval	Frequency
50-60	4
60-70	7
70-80	12
80-90	9
90-100	5

---

To Convert:

Convert the given distribution into a 'less than' type cumulative frequency distribution.

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Solution:

To convert a frequency distribution into a 'less than' type cumulative frequency distribution, we list the upper limits of the class intervals along with their corresponding cumulative frequencies. The cumulative frequency for each upper limit is the sum of the frequencies of all classes with an upper limit less than or equal to that upper limit.

Let's calculate the cumulative frequencies:

For the class 50-60, the upper limit is 60, and the frequency is 4. The number of observations less than 60 is 4.

For the class 60-70, the upper limit is 70, and the frequency is 7. The number of observations less than 70 is the sum of frequencies up to this class: $4 + 7 = 11$.

For the class 70-80, the upper limit is 80, and the frequency is 12. The number of observations less than 80 is: $11 + 12 = 23$.

For the class 80-90, the upper limit is 90, and the frequency is 9. The number of observations less than 90 is: $23 + 9 = 32$.

For the class 90-100, the upper limit is 100, and the frequency is 5. The number of observations less than 100 is: $32 + 5 = 37$.

The 'less than' type cumulative frequency distribution is as follows:

Marks Less Than	Cumulative Frequency
60	4
70	11
80	23
90	32
100	37

This table represents the required 'less than' type cumulative frequency distribution.

The empirical relationship between the three measures of central tendency (Mean, Median, and Mode) is a relationship that holds true for a moderately skewed distribution.

The relationship is:

Mode = 3 $\times$ Median - 2 $\times$ Mean

This relationship can be written using mathematical notation as:

$\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$

This relationship is useful when one of the measures is unknown, and the other two are known, especially in cases of moderately skewed distributions.

Given:

Mode of the distribution = $35.6$

Median of the distribution = $32.5$

---

To Find:

The mean of the distribution.

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Solution:

We use the empirical relationship between the Mean, Median, and Mode, which is:

Substitute the given values of Mode and Median into equation (i):

Calculate the product of 3 and 32.5:

Substitute this value back into equation (ii):

Rearrange equation (iii) to solve for $2 \times \text{Mean}$:

Calculate the difference on the right side of equation (iv):

So, equation (iv) becomes:

Now, solve for Mean by dividing equation (v) by 2:

Therefore, the mean of the distribution is $30.95$.

The final answer is 30.95.

Given:

The frequency distribution is given as:

Class Interval	Frequency
0-10	4
10-20	7
20-30	10
30-40	8
40-50	6

---

To Find:

The mean of the given frequency distribution using the Assumed Mean Method.

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Solution:

The Assumed Mean Method formula for calculating the mean ($\overline{x}$) of a grouped frequency distribution is:

Where:

$a$ is the assumed mean.

$f_i$ is the frequency of the $i$-th class interval.

$d_i$ is the deviation of the class mark ($x_i$) from the assumed mean ($a$), i.e., $d_i = x_i - a$.

$\sum f_i$ is the sum of all frequencies.

$\sum f_i d_i$ is the sum of the products of frequencies and their respective deviations.

We first find the class mark ($x_i$) for each interval. The class mark is the midpoint of the class interval.

We choose an assumed mean ($a$). Let's choose the class mark of the middle class (20-30), which is 25. So, $a = 25$.

Now, we construct a table including frequency, class mark, deviation, and the product of frequency and deviation:

Class Interval	Frequency ($f_i$)	Class Mark ($x_i$)	Deviation ($d_i = x_i - 25$)	$f_i d_i$
0-10	4	$\frac{0+10}{2} = 5$	$5 - 25 = -20$	$4 \times (-20) = -80$
10-20	7	$\frac{10+20}{2} = 15$	$15 - 25 = -10$	$7 \times (-10) = -70$
20-30	10	$\frac{20+30}{2} = 25$	$25 - 25 = 0$	$10 \times 0 = 0$
30-40	8	$\frac{30+40}{2} = 35$	$35 - 25 = 10$	$8 \times 10 = 80$
40-50	6	$\frac{40+50}{2} = 45$	$45 - 25 = 20$	$6 \times 20 = 120$
Total	$\sum f_i = 35$			$\sum f_i d_i = -80 - 70 + 0 + 80 + 120 = -150 + 200 = 50$

From the table, we have:

Now, substitute these values into the Assumed Mean formula (i):

Simplify the fraction in equation (ii):

So, equation (ii) becomes:

Calculate the value of $\frac{10}{7}$:

Add this to 25:

The mean of the given distribution is approximately $26.43$.

The final answer is approximately 26.43.

Given:

Mode of the data = $18$

Mean of the data = $24$

---

To Find:

The median of the data.

---

Solution:

We use the empirical relationship between the Mean, Median, and Mode for a moderately skewed distribution:

Substitute the given values of Mode and Mean into equation (i):

Calculate the product of 2 and 24:

Substitute this value back into equation (ii):

Rearrange equation (iii) to solve for $3 \times \text{Median}$. Add 48 to both sides:

Calculate the sum on the left side of equation (iv):

So, equation (iv) becomes:

Now, solve for Median by dividing equation (v) by 3:

Therefore, the median of the data is $22$.

The final answer is 22.

Given:

The frequency distribution is given as:

Height (cm)	Number of Students (Frequency, $f$)
145-150	8
150-155	10
155-160	15
160-165	9
165-170	8

---

To Convert:

Convert the given distribution into a 'more than' type cumulative frequency distribution.

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Solution:

To convert a frequency distribution into a 'more than' type cumulative frequency distribution, we list the lower limits of the class intervals along with their corresponding cumulative frequencies. The cumulative frequency for each lower limit is the sum of the frequencies of all classes with a lower limit greater than or equal to that lower limit.

Let's calculate the cumulative frequencies:

The total number of students is $N = 8 + 10 + 15 + 9 + 8 = 50$.

For "Height more than or equal to 145": This includes all students. So, the cumulative frequency is 50.

For "Height more than or equal to 150": This excludes students in the 145-150 class. So, the cumulative frequency is $50 - 8 = 42$.

For "Height more than or equal to 155": This excludes students in the 145-150 and 150-155 classes. So, the cumulative frequency is $42 - 10 = 32$.

For "Height more than or equal to 160": This excludes students in the classes below 160. So, the cumulative frequency is $32 - 15 = 17$.

For "Height more than or equal to 165": This excludes students in the classes below 165. So, the cumulative frequency is $17 - 9 = 8$.

The 'more than' type cumulative frequency distribution is as follows:

Height More Than or Equal To (cm)	Cumulative Frequency
145	50
150	42
155	32
160	17
165	8

This table represents the required 'more than' type cumulative frequency distribution.

Given:

The frequency distribution is given as:

Class Interval	Frequency
0-5	2
5-10	5
10-15	12
15-20	10
20-25	6

---

To Find:

The modal class of the given frequency distribution.

---

Solution:

The modal class is the class interval that has the highest frequency.

Looking at the frequencies in the given table (2, 5, 12, 10, 6), the highest frequency is 12.

The class interval corresponding to the frequency 12 is 10-15.

Therefore, the modal class is 10-15.

The final answer is 10-15.

Given:

The frequency distribution is given as:

Class Interval	Frequency ($f_i$)
10-30	6
30-50	10
50-70	15
70-90	8
90-110	11

---

To Find:

The median class of the given frequency distribution.

---

Solution:

To find the median class, we first need to calculate the cumulative frequencies and the total number of observations ($N$).

Let's construct a table with frequencies and cumulative frequencies:

Class Interval	Frequency ($f_i$)	Cumulative Frequency (cf)
10-30	6	6
30-50	10	$6 + 10 = 16$
50-70	15	$16 + 15 = 31$
70-90	8	$31 + 8 = 39$
90-110	11	$39 + 11 = 50$
Total	$N = \sum f_i = 50$

The total number of observations is $N = 50$.

We need to find the class where the $\frac{N}{2}$-th observation lies.

Now, we look for the cumulative frequency that is just greater than or equal to 25 in the cumulative frequency column.

The cumulative frequency 6 is less than 25.

The cumulative frequency 16 is less than 25.

The cumulative frequency 31 is greater than 25.

The class interval corresponding to the cumulative frequency 31 is 50-70.

Therefore, the median class is 50-70.

The final answer is 50-70.

Given:

Number of observations, $n = 20$

Mean of 20 observations, $\overline{x} = 15$

Each observation is multiplied by 2.

---

To Find:

The new mean after multiplying each observation by 2.

---

Solution:

Let the original observations be $x_1, x_2, ..., x_{20}$.

The mean of these observations is given by:

We are given that $\overline{x} = 15$. Substituting this into equation (i):

From equation (ii), the sum of the original observations is:

Now, each observation is multiplied by 2. Let the new observations be $y_1, y_2, ..., y_{20}$, where $y_i = 2x_i$ for $i = 1, 2, ..., 20$.

The new mean, $\overline{y}$, is given by:

Substitute $y_i = 2x_i$ into equation (iii):

Using the property of summation ($\sum kx_i = k \sum x_i$), we can take the constant 2 out of the summation:

Rearranging equation (v):

From equation (i), we know that $\frac{\sum \limits_{i=1}^{20} x_i}{20} = \overline{x}$. Substitute this into equation (vi):

Substitute the given value of the original mean ($\overline{x} = 15$) into equation (vii):

---

Alternatively, we can use the property that if each observation in a data set is multiplied by a constant $k$, then the new mean is $k$ times the original mean.

Original Mean ($\overline{x}$) = 15

Constant multiplied = 2

New Mean = $2 \times$ Original Mean

The new mean is 30.

The final answer is 30.

Given:

The observations arranged in ascending order are: 10, 15, x, x+5, 35, 40.

The median of the observations is 27.5.

---

To Find:

The value of $x$.

---

Solution:

The given observations are already arranged in ascending order.

The number of observations is $n = 6$.

Since the number of observations is even, the median is the average of the $\left(\frac{n}{2}\right)$-th observation and the $\left(\frac{n}{2} + 1\right)$-th observation.

Here, $\frac{n}{2} = \frac{6}{2} = 3$.

So, the median is the average of the 3rd observation and the $(3+1)=4$-th observation.

From the given data:

3rd observation = $x$

4th observation = $x+5$

The formula for the median when $n$ is even is:

Substitute the values into equation (i):

Simplify the numerator in equation (ii):

Multiply both sides of equation (iii) by 2:

Subtract 5 from both sides of equation (iv):

Divide both sides of equation (v) by 2:

Thus, the value of $x$ is 25.

Let's verify the observations with $x=25$: 10, 15, 25, 25+5=30, 35, 40. The ordered observations are 10, 15, 25, 30, 35, 40. The 3rd observation is 25 and the 4th observation is 30. The median is $\frac{25+30}{2} = \frac{55}{2} = 27.5$, which matches the given median.

The final answer is 25.

Solution:

The formula for calculating the median of grouped data is:

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Explanation of the terms involved:

In the formula for the median of grouped data:

$l$ = lower limit of the median class. The median class is the class interval whose cumulative frequency is just greater than or equal to $\frac{N}{2}$.

$N$ = total number of observations, which is the sum of all frequencies ($\sum f_i$).

$cf$ = cumulative frequency of the class preceding the median class.

$f$ = frequency of the median class.

$h$ = class size (or width) of the median class interval. It is the difference between the upper limit and the lower limit of the median class ($h = \text{Upper Limit} - \text{Lower Limit}$).

The term $\frac{N}{2}$ represents the position of the median observation in the cumulative frequency distribution.

Definition of Ogive:

An ogive (pronounced as 'oj-ive' or 'oh-jive') is a graphical representation of a cumulative frequency distribution. It is a smooth curve that shows the cumulative frequency against the upper or lower boundaries of the class intervals. Ogives are useful for determining the median, quartiles, and other percentiles from grouped data.

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Types of Ogives:

There are two main types of ogives:

1. Less Than Ogive: This ogive is constructed by plotting the upper limits of the class intervals on the x-axis and the 'less than' cumulative frequencies on the y-axis. The points are connected by a smooth curve, and it starts from the lower limit of the first class on the x-axis and ends at the upper limit of the last class. It is an increasing curve.

2. More Than Ogive: This ogive is constructed by plotting the lower limits of the class intervals on the x-axis and the 'more than' cumulative frequencies on the y-axis. The points are connected by a smooth curve, and it starts from the lower limit of the first class with the total frequency and ends at the upper limit of the last class. It is a decreasing curve.

Definition of Cumulative Frequency:

Cumulative frequency is the running total of frequencies. It is obtained by adding the frequency of a class interval to the sum of frequencies of all preceding class intervals. Cumulative frequency distributions show the number of observations that fall below (or above) a certain value or class boundary.

There are two types of cumulative frequency:

• 'Less than' cumulative frequency: The total number of observations less than the upper boundary of a class interval.
• 'More than' cumulative frequency: The total number of observations greater than or equal to the lower boundary of a class interval.

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How Cumulative Frequency is Used to Find the Median:

For grouped data, cumulative frequency is essential for finding the median. The steps involved are:

1. Calculate Cumulative Frequencies: Prepare a cumulative frequency distribution (usually 'less than' type) by adding the frequencies of each class to the cumulative frequency of the previous class.

2. Find Total Number of Observations: Calculate the total number of observations, $N$, which is the sum of all frequencies. This is also the last cumulative frequency in the 'less than' distribution.

3. Determine Median Position: Calculate $\frac{N}{2}$. This value indicates the position of the median observation in the ordered data set.

4. Identify the Median Class: Locate the class interval in the cumulative frequency distribution whose cumulative frequency is just greater than or equal to $\frac{N}{2}$. This class interval is called the median class. The median value lies within this class.

5. Apply the Median Formula: Once the median class is identified, the median is calculated using the formula:

Here, $l$ is the lower limit of the median class, $cf$ is the cumulative frequency of the class preceding the median class, $f$ is the frequency of the median class, and $h$ is the class size.

In summary, cumulative frequency helps us identify the class interval where the median lies by indicating which interval contains the $\frac{N}{2}$-th observation. Once the median class is found, its lower limit, the cumulative frequency before it, and its own frequency are used in the formula to pinpoint the exact median value within that class.

Given:

The frequency distribution is given as:

Class Interval	Frequency ($f$)
0-10	2
10-20	3
20-30	5
30-40	p
40-50	2

The mean of the distribution is $\overline{x} = 25$.

---

To Find:

The value of p.

---

Solution:

We will use the Direct Method formula for calculating the mean ($\overline{x}$) of a grouped frequency distribution:

Where $f_i$ is the frequency and $x_i$ is the class mark (midpoint) of the $i$-th class interval.

We need to calculate the class mark ($x_i$) for each interval and the product $f_i x_i$.

Class Interval	Frequency ($f_i$)	Class Mark ($x_i$)	$f_i x_i$
0-10	2	$\frac{0+10}{2} = 5$	$2 \times 5 = 10$
10-20	3	$\frac{10+20}{2} = 15$	$3 \times 15 = 45$
20-30	5	$\frac{20+30}{2} = 25$	$5 \times 25 = 125$
30-40	p	$\frac{30+40}{2} = 35$	$p \times 35 = 35p$
40-50	2	$\frac{40+50}{2} = 45$	$2 \times 45 = 90$
Total	$\sum f_i = 2+3+5+p+2 = 12+p$		$\sum f_i x_i = 10+45+125+35p+90 = 270 + 35p$

From the table, we have:

We are given that the mean $\overline{x} = 25$. Substitute these values into the formula (i):

Multiply both sides of equation (ii) by $(12+p)$:

Expand the left side of equation (iii):

Rearrange equation (iv) to solve for p. Subtract $25p$ from both sides and subtract 270 from both sides:

Divide both sides of equation (v) by 10:

Therefore, the value of p is 3.

The final answer is 3.

Yes, the mean, median, and mode of a data set can be equal.

---

This happens particularly when the data distribution is perfectly or nearly perfectly symmetric. In a perfectly symmetric distribution, the centre of the distribution, the middle value, and the most frequent value coincide. A common example of a perfectly symmetric distribution where this occurs is the Normal Distribution.

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Example:

Consider the following data set: 2, 3, 3, 3, 4.

Let's calculate the Mean, Median, and Mode for this data:

Mean: The sum of the observations divided by the number of observations.

Median: The middle value when the data is arranged in ascending or descending order.

The data is already arranged in ascending order: 2, 3, 3, 3, 4.

There are 5 observations ($n=5$), which is an odd number. The median is the $\left(\frac{n+1}{2}\right)$-th observation.

The 3rd observation in the ordered data is 3.

Mode: The observation that occurs most frequently.

In the data set (2, 3, 3, 3, 4), the value 3 appears 3 times, which is more frequent than any other value.

In this example, Mean = 3, Median = 3, and Mode = 3. Thus, they are all equal.

Definition of Class Mark:

The class mark (or midpoint) of a class interval is the average of its lower limit and its upper limit. It represents the central value of that class interval.

---

Formula for Class Mark:

The formula for calculating the class mark is:

---

Calculation for the interval 35-45:

For the given class interval $35 - 45$:

Lower Limit = 35

Upper Limit = 45

Substitute these values into the formula (i):

Simplify equation (ii):

The class mark of the class interval $35 - 45$ is 40.

The final answer is 40.

Given:

The frequency distribution of heights of students is given as:

Height (cm)	Number of Students (Frequency, $f_i$)
150-155	10
155-160	18
160-165	25
165-170	12
170-175	5

---

To Find:

The median class for the given frequency distribution.

---

Solution:

To find the median class, we need to calculate the cumulative frequencies and the total number of observations ($N$).

Let's construct a table with frequencies and cumulative frequencies:

Height (cm)	Frequency ($f_i$)	Cumulative Frequency (cf)
150-155	10	10
155-160	18	$10 + 18 = 28$
160-165	25	$28 + 25 = 53$
165-170	12	$53 + 12 = 65$
170-175	5	$65 + 5 = 70$
Total	$N = \sum f_i = 70$

The total number of observations (students) is $N = 70$.

We need to find the class where the $\frac{N}{2}$-th observation lies.

Now, we look for the cumulative frequency that is just greater than or equal to 35 in the cumulative frequency column.

The cumulative frequency 10 is less than 35.

The cumulative frequency 28 is less than 35.

The cumulative frequency 53 is greater than 35.

The class interval corresponding to the cumulative frequency 53 is 160-165.

Therefore, the median class is 160-165.

The final answer is 160-165.

Answer: No.

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Justification:

The data provided represents the number of students in different ranges of school classes (e.g., Class I to V, Class VI to VIII, etc.). These ranges are not numerical class intervals with a defined lower limit and upper limit in the sense required for calculating mean, median, or mode using the standard formulas for grouped numerical data as taught in Class 10.

The methods for grouped data (Direct Method, Assumed Mean Method, Step Deviation Method for Mean; formula involving cumulative frequency for Median; formula involving preceding and succeeding frequencies for Mode) are designed for distributions where the variable (like marks, height, wage, etc.) is numerical and grouped into class intervals with defined numerical boundaries. These methods require calculations involving class marks (midpoints), lower limits, cumulative frequencies based on numerical values, etc., which cannot be determined from the given categorical/ordinal class ranges like "I-V" or "VI-VIII".

Therefore, this data is not suitable for applying the techniques of finding the mean, median, or mode of grouped numerical data as taught in Class 10.

Purpose of Drawing Ogives:

Ogives are graphical representations of cumulative frequency distributions. Their main purposes are:

1. To visualize the cumulative distribution of data. A 'less than' ogive shows the number or percentage of observations below certain values, while a 'more than' ogive shows the number or percentage of observations above certain values.

2. To easily determine certain positional measures such as the median, quartiles (Q1, Q3), deciles, and percentiles directly from the graph without needing to use formulas.

3. To compare two or more distributions graphically.

4. To estimate the number or proportion of observations that lie between two given values.

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Estimation of Median from an Ogive:

The median is the value that divides the data into two equal halves. For a grouped frequency distribution with total observations $N$, the median is the value below which $\frac{N}{2}$ observations lie.

The median can be estimated from an ogive (or from both types of ogives plotted on the same graph) using the following steps:

1. Calculate $\frac{N}{2}$: Find the total number of observations, $N$, and calculate $\frac{N}{2}$.

2. Using a Single Ogive (usually 'Less Than' Ogive):

a. Locate the value $\frac{N}{2}$ on the cumulative frequency (y-axis).

b. From this point on the y-axis, draw a horizontal line parallel to the x-axis to intersect the 'less than' ogive.

c. From the point of intersection on the ogive, draw a vertical line parallel to the y-axis to intersect the x-axis.

d. The point where the vertical line intersects the x-axis gives the estimated median value.

3. Using Both Ogives ('Less Than' and 'More Than' Ogives):

a. Plot both the 'less than' ogive and the 'more than' ogive on the same graph with the same scales for the x-axis (class boundaries) and y-axis (cumulative frequency).

b. The two ogives will intersect at a point.

c. From the point of intersection, draw a vertical line parallel to the y-axis to intersect the x-axis.

d. The point where this vertical line intersects the x-axis gives the estimated median value. The y-coordinate of the intersection point will be $\frac{N}{2}$.

This graphical method provides a visual estimate of the median, which should be close to the value calculated using the median formula for grouped data.

Given:

The frequency distribution of marks obtained by 100 students is given as:

Marks	Number of Students (Frequency, $f$)
0-10	5
10-20	8
20-30	12
30-40	15
40-50	18
50-60	16
60-70	10
70-80	6
80-90	4
90-100	6

Total number of students, $N = 100$.

---

To Find:

The mean, median, and mode of the given data.

---

Solution (Mean):

We will calculate the mean using the Direct Method. The formula is:

First, calculate the class mark ($x_i$) for each interval and the product $f_i x_i$.

Marks	Frequency ($f_i$)	Class Mark ($x_i$)	$f_i x_i$
0-10	5	$\frac{0+10}{2} = 5$	$5 \times 5 = 25$
10-20	8	$\frac{10+20}{2} = 15$	$8 \times 15 = 120$
20-30	12	$\frac{20+30}{2} = 25$	$12 \times 25 = 300$
30-40	15	$\frac{30+40}{2} = 35$	$15 \times 35 = 525$
40-50	18	$\frac{40+50}{2} = 45$	$18 \times 45 = 810$
50-60	16	$\frac{50+60}{2} = 55$	$16 \times 55 = 880$
60-70	10	$\frac{60+70}{2} = 65$	$10 \times 65 = 650$
70-80	6	$\frac{70+80}{2} = 75$	$6 \times 75 = 450$
80-90	4	$\frac{80+90}{2} = 85$	$4 \times 85 = 340$
90-100	6	$\frac{90+100}{2} = 95$	$6 \times 95 = 570$
Total	$\sum f_i = 100$		$\sum f_i x_i = 25+120+300+525+810+880+650+450+340+570 = 4670$

Using formula (1):

The mean of the data is $\mathbf{46.7}$.

---

Solution (Median):

To find the median, we first calculate the cumulative frequencies.

Marks	Frequency ($f_i$)	Cumulative Frequency (cf)
0-10	5	5
10-20	8	$5 + 8 = 13$
20-30	12	$13 + 12 = 25$
30-40	15	$25 + 15 = 40$
40-50	18	$40 + 18 = 58$
50-60	16	$58 + 16 = 74$
60-70	10	$74 + 10 = 84$
70-80	6	$84 + 6 = 90$
80-90	4	$90 + 4 = 94$
90-100	6	$94 + 6 = 100$
Total	$N = 100$

The total number of observations is $N = 100$.

We find the position of the median: $\frac{N}{2} = \frac{100}{2} = 50$.

The cumulative frequency just greater than or equal to 50 is 58, which corresponds to the class interval 40-50.

So, the median class is 40-50.

The formula for the median of grouped data is:

From the median class (40-50), we have:

Substitute these values into formula (3):

Simplify equation (4):

The median of the data is approximately $\mathbf{45.56}$.

---

Solution (Mode):

To find the mode, we identify the modal class, which is the class with the highest frequency.

From the frequency table, the highest frequency is 18, which corresponds to the class interval 40-50.

So, the modal class is 40-50.

The formula for the mode of grouped data is:

From the modal class (40-50), we have:

Substitute these values into formula (5):

Simplify equation (6):

The mode of the data is $\mathbf{46}$.

---

Summary of Results:

Mean $\approx 46.7$

Median $\approx 45.56$

Mode $= 46$

Note: The empirical relationship (Mode = 3 Median - 2 Mean) can be used to check the consistency of the values.

The empirical formula gives a value of approximately 43.28, while the calculated mode is 46. There is a slight difference, which is expected as the empirical formula is an approximate relationship for moderately skewed distributions, and this distribution may not be perfectly moderately skewed. The calculated values directly from the data are correct.

The final answers are: Mean $\approx$ 46.7, Median $\approx$ 45.56, Mode = 46.

Given:

The frequency distribution of daily income of 50 workers is given as:

Daily Income ($\textsf{₹}$)	Number of Workers (Frequency, $f$)
200-250	12
250-300	14
300-350	8
350-400	6
400-450	10

Total number of workers, $N = 50$.

---

To Find:

The mean daily income of the workers using the Step-deviation method.

---

Solution:

The Step-deviation method formula for calculating the mean ($\overline{x}$) of a grouped frequency distribution is:

Where:

$a$ is the assumed mean.

$f_i$ is the frequency of the $i$-th class interval.

$u_i = \frac{x_i - a}{h}$, where $x_i$ is the class mark of the $i$-th class interval and $h$ is the class size.

$\sum f_i$ is the sum of all frequencies ($N$).

$\sum f_i u_i$ is the sum of the products of frequencies and their respective $u_i$ values.

First, we calculate the class mark ($x_i$) for each interval. The class mark is the midpoint of the class interval.

We choose an assumed mean ($a$). Let's choose the class mark of the class 300-350, which is $\frac{300+350}{2} = 325$. So, $a = 325$.

The class size ($h$) is the difference between consecutive lower limits (or upper limits). $h = 250 - 200 = 50$.

Now, we construct a table including frequency, class mark, $u_i$, and the product of frequency and $u_i$:

Daily Income ($\textsf{₹}$)	Number of Workers ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 325}{50}$	$f_i u_i$
200-250	12	$\frac{200+250}{2} = 225$	$\frac{225 - 325}{50} = \frac{-100}{50} = -2$	$12 \times (-2) = -24$
250-300	14	$\frac{250+300}{2} = 275$	$\frac{275 - 325}{50} = \frac{-50}{50} = -1$	$14 \times (-1) = -14$
300-350	8	$\frac{300+350}{2} = 325$	$\frac{325 - 325}{50} = \frac{0}{50} = 0$	$8 \times 0 = 0$
350-400	6	$\frac{350+400}{2} = 375$	$\frac{375 - 325}{50} = \frac{50}{50} = 1$	$6 \times 1 = 6$
400-450	10	$\frac{400+450}{2} = 425$	$\frac{425 - 325}{50} = \frac{100}{50} = 2$	$10 \times 2 = 20$
Total	$\sum f_i = 50$			$\sum f_i u_i = -24 - 14 + 0 + 6 + 20 = -38 + 26 = -12$

From the table, we have:

Now, substitute these values into the Step-deviation formula (1):

Simplify equation (2):

The mean daily income of the workers is $\textsf{₹}$ 313.

The final answer is $\textsf{₹}$ 313.

Given:

The frequency distribution of marks obtained by students is given as:

Marks Obtained	Number of Students (Frequency, $f$)
0-10	5
10-20	8
20-30	12
30-40	15
40-50	18
50-60	16
60-70	10

---

To Draw:

A 'less than' type cumulative frequency distribution ogive.

To Estimate:

The median marks from the ogive.

---

Solution (Drawing the Ogive):

To draw a 'less than' ogive, we first prepare the 'less than' cumulative frequency distribution. We list the upper limits of the class intervals and their corresponding cumulative frequencies. We also include a point at the lower limit of the first class with a cumulative frequency of 0.

Marks (Upper Limit)	Number of Students ('Less than' Cumulative Frequency, cf)	Point to Plot (x, y)
Less than 0	0	(0, 0)
Less than 10	5	(10, 5)
Less than 20	$5 + 8 = 13$	(20, 13)
Less than 30	$13 + 12 = 25$	(30, 25)
Less than 40	$25 + 15 = 40$	(40, 40)
Less than 50	$40 + 18 = 58$	(50, 58)
Less than 60	$58 + 16 = 74$	(60, 74)
Less than 70	$74 + 10 = 84$	(70, 84)
Total	$N = 84$

The points to plot for the 'less than' ogive are (0, 0), (10, 5), (20, 13), (30, 25), (40, 40), (50, 58), (60, 74), and (70, 84).

To draw the ogive:

1. Draw the x-axis representing the 'Marks Obtained' (specifically the upper limits, starting from 0).

2. Draw the y-axis representing the 'Cumulative Frequency'.

3. Plot the points from the table: (0, 0), (10, 5), (20, 13), (30, 25), (40, 40), (50, 58), (60, 74), (70, 84).

4. Connect the plotted points with a smooth curve. This curve is the 'less than' ogive.

(Note: The actual drawing of the ogive is not possible in this text-based format.)

---

Solution (Estimating Median from Ogive):

The median is the value corresponding to the cumulative frequency $\frac{N}{2}$.

From the table, the total number of students is $N = 84$.

To estimate the median from the 'less than' ogive:

1. Locate the value 42 on the cumulative frequency axis (y-axis).

2. From the point representing 42 on the y-axis, draw a horizontal line parallel to the x-axis until it intersects the 'less than' ogive.

3. From the point of intersection on the ogive, draw a vertical line parallel to the y-axis downwards to the x-axis.

4. The point where this vertical line touches the x-axis gives the estimated median marks.

Looking at the cumulative frequency table, 42 lies between cumulative frequencies 40 (corresponding to mark 40) and 58 (corresponding to mark 50). The median must be between 40 and 50.

Using the median formula for grouped data as a reference, for the median class 40-50 ($l=40, cf=40, f=18, h=10$), the median is:

Therefore, the vertical line drawn from the intersection point on the ogive at cf=42 is expected to intersect the x-axis at approximately 41.11.

The estimated median marks from the ogive is approximately $\mathbf{41.11}$.

The final answer is approximately 41.11.

Given:

The frequency distribution of daily pocket allowance is given as:

Daily Pocket Allowance ($\textsf{₹}$)	Number of Children (Frequency, $f_i$)
11-13	7
13-15	6
15-17	9
17-19	13
19-21	f
21-23	5
23-25	4

The mean pocket allowance is $\overline{x} = 18$.

---

To Find:

The missing frequency, $f$.

---

Solution:

We will use the Direct Method to calculate the mean. The formula is:

Where $f_i$ is the frequency and $x_i$ is the class mark (midpoint) of the $i$-th class interval.

First, calculate the class mark ($x_i$) for each interval and the product $f_i x_i$.

Daily Pocket Allowance ($\textsf{₹}$)	Number of Children ($f_i$)	Class Mark ($x_i$)	$f_i x_i$
11-13	7	$\frac{11+13}{2} = 12$	$7 \times 12 = 84$
13-15	6	$\frac{13+15}{2} = 14$	$6 \times 14 = 84$
15-17	9	$\frac{15+17}{2} = 16$	$9 \times 16 = 144$
17-19	13	$\frac{17+19}{2} = 18$	$13 \times 18 = 234$
19-21	f	$\frac{19+21}{2} = 20$	$f \times 20 = 20f$
21-23	5	$\frac{21+23}{2} = 22$	$5 \times 22 = 110$
23-25	4	$\frac{23+25}{2} = 24$	$4 \times 24 = 96$
Total	$\sum f_i = 7+6+9+13+f+5+4 = 44+f$		$\sum f_i x_i = 84+84+144+234+20f+110+96 = 752 + 20f$

From the table, we have:

We are given that the mean $\overline{x} = 18$. Substitute these values into the formula (1):

Multiply both sides of equation (2) by $(44+f)$:

Expand the left side:

Rearrange equation (3) to isolate the terms with $f$ on one side and constant terms on the other. Subtract $18f$ from both sides and subtract 752 from both sides:

Simplify both sides:

Divide both sides of equation (4) by 2:

Therefore, the missing frequency $f$ is 20.

The final answer is 20.

Given:

The frequency distribution is given with missing frequencies x and y:

Class Interval	Frequency ($f_i$)
0-100	2
100-200	5
200-300	x
300-400	12
400-500	17
500-600	20
600-700	y
700-800	9
800-900	7
900-1000	4
Total	100

The median of the data is 525.

Total frequency, $N = 100$.

---

To Find:

The values of x and y.

---

Solution:

First, we calculate the cumulative frequencies.

Class Interval	Frequency ($f_i$)	Cumulative Frequency (cf)
0-100	2	2
100-200	5	$2 + 5 = 7$
200-300	x	$7 + x$
300-400	12	$7 + x + 12 = 19 + x$
400-500	17	$19 + x + 17 = 36 + x$
500-600	20	$36 + x + 20 = 56 + x$
600-700	y	$56 + x + y$
700-800	9	$56 + x + y + 9 = 65 + x + y$
800-900	7	$65 + x + y + 7 = 72 + x + y$
900-1000	4	$72 + x + y + 4 = 76 + x + y$
Total	$N = 100$	$76 + x + y$

We are given that the total frequency is 100. From the table, the total frequency is also $76 + x + y$.

We are given that the median is 525. Since 525 lies in the class interval 500-600, the median class is 500-600.

For the median class 500-600:

Lower limit, $l = 500$

Frequency of the median class, $f = 20$

Cumulative frequency of the class preceding the median class, $cf = 36 + x$ (from the table, cf for 400-500)

Class size, $h = 600 - 500 = 100$

Total frequency, $N = 100$, so $\frac{N}{2} = \frac{100}{2} = 50$.

The formula for the median of grouped data is:

Substitute the given median value and the values from the median class into formula (2):

Simplify equation (3):

Divide both sides of equation (4) by 5:

Rearrange equation (5) to solve for x:

Now that we have the value of x, substitute it into equation (1) to find the value of y:

Therefore, the values of x and y are 9 and 15 respectively.

The final answer is x = 9 and y = 15.

Given:

The frequency distribution is given as:

Class Interval	Frequency ($f$)
10-20	3
20-30	5
30-40	8
40-50	10
50-60	12
60-70	9
70-80	4

The total frequency is $N = 3 + 5 + 8 + 10 + 12 + 9 + 4 = 51$.

---

To Draw:

'Less than' and 'more than' cumulative frequency ogives.

To Find:

The median from the ogives.

---

Solution:

To draw the ogives, we first prepare the cumulative frequency distributions of 'less than' and 'more than' types.

'Less Than' Cumulative Frequency Distribution:

We consider the upper limits of the class intervals. We also include the lower limit of the first class with a cumulative frequency of 0.

Class Upper Limit	'Less than' Cumulative Frequency (cf)	Point to Plot (x, y)
10	0	(10, 0)
20	3	(20, 3)
30	$3 + 5 = 8$	(30, 8)
40	$8 + 8 = 16$	(40, 16)
50	$16 + 10 = 26$	(50, 26)
60	$26 + 12 = 38$	(60, 38)
70	$38 + 9 = 47$	(70, 47)
80	$47 + 4 = 51$	(80, 51)

The points to plot for the 'less than' ogive are (10, 0), (20, 3), (30, 8), (40, 16), (50, 26), (60, 38), (70, 47), and (80, 51).

---

'More Than' Cumulative Frequency Distribution:

We consider the lower limits of the class intervals. The cumulative frequency for the first lower limit is the total frequency. The cumulative frequency for subsequent lower limits is obtained by subtracting the frequency of the preceding class from the previous cumulative frequency.

Class Lower Limit	'More than' Cumulative Frequency (cf)	Point to Plot (x, y)
10	51	(10, 51)
20	$51 - 3 = 48$	(20, 48)
30	$48 - 5 = 43$	(30, 43)
40	$43 - 8 = 35$	(40, 35)
50	$35 - 10 = 25$	(50, 25)
60	$25 - 12 = 13$	(60, 13)
70	$13 - 9 = 4$	(70, 4)
80	$4 - 4 = 0$	(80, 0)

The points to plot for the 'more than' ogive are (10, 51), (20, 48), (30, 43), (40, 35), (50, 25), (60, 13), (70, 4), and (80, 0).

---

Drawing the Ogives:

1. Draw the x-axis and label it 'Class Interval' or the variable name (e.g., Marks, Height). Use a scale that accommodates the class boundaries (from 10 to 80).

2. Draw the y-axis and label it 'Cumulative Frequency'. Use a scale that accommodates the cumulative frequencies (from 0 to 51).

3. Plot the points for the 'less than' ogive and connect them with a smooth curve. This curve starts from (10, 0) and goes upwards to (80, 51).

4. Plot the points for the 'more than' ogive on the same graph and connect them with a smooth curve. This curve starts from (10, 51) and goes downwards to (80, 0).

(Note: The actual drawing of the ogives is not possible in this text-based format.)

---

Finding the Median from the Ogives:

The median is the value on the x-axis corresponding to the cumulative frequency $\frac{N}{2}$.

To estimate the median graphically:

1. Locate the value 25.5 on the cumulative frequency axis (y-axis).

2. From this point (0, 25.5) on the y-axis, draw a horizontal line parallel to the x-axis.

3. This horizontal line will intersect both the 'less than' ogive and the 'more than' ogive at the same point (or very close if the curves are smooth). Alternatively, find the intersection point of the two drawn ogives directly.

4. From the point of intersection, draw a vertical line parallel to the y-axis downwards to the x-axis.

5. The point where this vertical line intersects the x-axis gives the estimated median value.

Observing the tables, the cumulative frequency 25.5 lies between 16 (at x=40) and 26 (at x=50) in the 'less than' table. It lies between 35 (at x=40) and 25 (at x=50) in the 'more than' table. This indicates the median lies in the class interval 40-50.

The x-coordinate of the intersection point of the two ogives is the median. Based on the cumulative frequencies around 25.5, the intersection is expected to be slightly less than 50.

Using interpolation or precise plotting, the intersection point will be at approximately $x = 49.5$.

The estimated median from the ogives is approximately $\mathbf{49.5}$.

The final answer is approximately 49.5.

Given:

The frequency distribution of daily wages of 100 workers is given as:

Daily Wages ($\textsf{₹}$)	Number of Workers (Frequency, $f$)
50-60	15
60-70	18
70-80	25
80-90	20
90-100	12
100-110	10

Total number of workers, $N = 100$.

---

To Find:

The mean and mode of the given data.

---

Solution (Mean):

We will calculate the mean using the Direct Method. The formula is:

First, calculate the class mark ($x_i$) for each interval and the product $f_i x_i$.

Daily Wages ($\textsf{₹}$)	Number of Workers ($f_i$)	Class Mark ($x_i$)	$f_i x_i$
50-60	15	$\frac{50+60}{2} = 55$	$15 \times 55 = 825$
60-70	18	$\frac{60+70}{2} = 65$	$18 \times 65 = 1170$
70-80	25	$\frac{70+80}{2} = 75$	$25 \times 75 = 1875$
80-90	20	$\frac{80+90}{2} = 85$	$20 \times 85 = 1700$
90-100	12	$\frac{90+100}{2} = 95$	$12 \times 95 = 1140$
100-110	10	$\frac{100+110}{2} = 105$	$10 \times 105 = 1050$
Total	$\sum f_i = 100$		$\sum f_i x_i = 825+1170+1875+1700+1140+1050 = 7760$

Using formula (1):

The mean daily wage is $\textsf{₹}$ $\mathbf{77.6}$.

---

Solution (Mode):

To find the mode, we identify the modal class, which is the class with the highest frequency.

From the frequency table, the highest frequency is 25, which corresponds to the class interval 70-80.

So, the modal class is 70-80.

The formula for the mode of grouped data is:

From the modal class (70-80), we have:

Substitute these values into formula (3):

Simplify equation (4):

The mode of the data is approximately $\textsf{₹}$ $\mathbf{75.83}$.

---

Summary of Results:

Mean = $\textsf{₹}$ 77.6

Mode $\approx$ $\textsf{₹}$ 75.83

The final answers are: Mean = $\textsf{₹}$ 77.6, Mode $\approx$ $\textsf{₹}$ 75.83.

Given:

The frequency distribution is given as:

Class	Frequency ($f$)
10-25	6
25-40	20
40-55	44
55-70	26
70-85	3
85-100	1

The total frequency is $N = 6 + 20 + 44 + 26 + 3 + 1 = 100$.

---

To Find:

The median and mode of the given frequency distribution.

---

Solution (Median):

To find the median, we first calculate the cumulative frequencies.

Class	Frequency ($f_i$)	Cumulative Frequency (cf)
10-25	6	6
25-40	20	$6 + 20 = 26$
40-55	44	$26 + 44 = 70$
55-70	26	$70 + 26 = 96$
70-85	3	$96 + 3 = 99$
85-100	1	$99 + 1 = 100$
Total	$N = 100$

The total number of observations is $N = 100$.

We find the position of the median:

The cumulative frequency just greater than or equal to 50 is 70, which corresponds to the class interval 40-55.

So, the median class is 40-55.

The formula for the median of grouped data is:

From the median class (40-55), we have:

Substitute these values into formula (1):

Simplify equation (2):

The median of the data is approximately $\mathbf{48.18}$.

---

Solution (Mode):

To find the mode, we identify the modal class, which is the class with the highest frequency.

From the frequency table, the highest frequency is 44, which corresponds to the class interval 40-55.

So, the modal class is 40-55.

The formula for the mode of grouped data is:

From the modal class (40-55), we have:

Substitute these values into formula (3):

Simplify equation (4):

The mode of the data is approximately $\mathbf{48.57}$.

---

Summary of Results:

Median $\approx 48.18$

Mode $\approx 48.57$

The final answers are: Median $\approx$ 48.18, Mode $\approx$ 48.57.

Given:

The frequency distribution of weights of 30 students is given as:

Weight (in kg)	Number of Students (Frequency, $f$)
40-45	2
45-50	3
50-55	8
55-60	6
60-65	6
65-70	3
70-75	2

Total number of students, $N = 30$.

---

To Find:

The median weight of the students.

---

Solution:

To find the median, we first calculate the cumulative frequencies.

Weight (in kg)	Number of Students ($f_i$)	Cumulative Frequency (cf)
40-45	2	2
45-50	3	$2 + 3 = 5$
50-55	8	$5 + 8 = 13$
55-60	6	$13 + 6 = 19$
60-65	6	$19 + 6 = 25$
65-70	3	$25 + 3 = 28$
70-75	2	$28 + 2 = 30$
Total	$N = 30$

The total number of observations is $N = 30$.

We find the position of the median:

The cumulative frequency just greater than or equal to 15 is 19, which corresponds to the class interval 55-60.

So, the median class is 55-60.

The formula for the median of grouped data is:

From the median class (55-60), we have:

Substitute these values into formula (1):

Simplify equation (2):

The median weight of the students is approximately $\mathbf{56.67}$ kg.

The final answer is approximately 56.67 kg.

Given:

The frequency distribution is given with missing frequencies x and y:

Class Interval	Frequency ($f_i$)
0-10	5
10-20	x
20-30	20
30-40	15
40-50	y
50-60	5
Total	60

The median of the distribution is 28.5.

Total frequency, $N = 60$.

---

To Find:

The values of x and y.

---

Solution:

First, we calculate the cumulative frequencies.

Class Interval	Frequency ($f_i$)	Cumulative Frequency (cf)
0-10	5	5
10-20	x	$5 + x$
20-30	20	$5 + x + 20 = 25 + x$
30-40	15	$25 + x + 15 = 40 + x$
40-50	y	$40 + x + y$
50-60	5	$40 + x + y + 5 = 45 + x + y$
Total	$N = 60$	$45 + x + y$

We are given that the total frequency is 60. From the table, the total frequency is also $45 + x + y$.

We are given that the median is 28.5. Since 28.5 lies in the class interval 20-30, the median class is 20-30.

For the median class 20-30:

Lower limit, $l = 20$

Frequency of the median class, $f = 20$

Cumulative frequency of the class preceding the median class, $cf = 5 + x$ (from the table, cf for 10-20)

Class size, $h = 30 - 20 = 10$

Total frequency, $N = 60$, so $\frac{N}{2} = \frac{60}{2} = 30$.

The formula for the median of grouped data is:

Substitute the given median value and the values from the median class into formula (2):

Simplify equation (3):

Multiply both sides of equation (4) by 2:

Rearrange equation (5) to solve for x:

Now that we have the value of x, substitute it into equation (1) to find the value of y:

Therefore, the values of x and y are 8 and 7 respectively.

The final answer is x = 8 and y = 7.

Given:

The marks obtained by students are given in a 'less than' cumulative frequency distribution:

Marks	Number of Students (Cumulative Frequency, cf)
Below 10	5
Below 20	13
Below 30	20
Below 40	30
Below 50	38
Below 60	50

The total number of students is $N = 50$.

---

To Find:

The mean, median, and mode of the given data.

---

Solution:

First, we convert the 'less than' cumulative frequency distribution into a simple frequency distribution with class intervals. The upper limits of the classes are 10, 20, 30, 40, 50, and 60. Assuming the lower limit of the first class is 0 (since marks cannot be negative in this context unless specified), the class intervals are 0-10, 10-20, etc. The frequency of each class is found by subtracting the cumulative frequency of the previous upper limit from the cumulative frequency of the current upper limit.

Class Interval	Cumulative Frequency (cf)	Frequency ($f_i$)
0-10	5	$5 - 0 = 5$
10-20	13	$13 - 5 = 8$
20-30	20	$20 - 13 = 7$
30-40	30	$30 - 20 = 10$
40-50	38	$38 - 30 = 8$
50-60	50	$50 - 38 = 12$
Total		$\sum f_i = 50$

The frequency distribution is:

Class Interval	Frequency ($f$)
0-10	5
10-20	8
20-30	7
30-40	10
40-50	8
50-60	12
Total	$N = 50$

---

Mean:

We calculate the mean using the Direct Method.

Calculate class marks ($x_i$) and $f_i x_i$:

Class Interval	Frequency ($f_i$)	Class Mark ($x_i$)	$f_i x_i$
0-10	5	5	25
10-20	8	15	120
20-30	7	25	175
30-40	10	35	350
40-50	8	45	360
50-60	12	55	660
Total	$\sum f_i = 50$		$\sum f_i x_i = 1690$

Using formula (1):

The mean is $\mathbf{33.8}$.

---

Median:

We find the median class using the cumulative frequency distribution. $N = 50$, so $\frac{N}{2} = \frac{50}{2} = 25$.

The cumulative frequency just greater than or equal to 25 is 30, which corresponds to the class interval 30-40.

So, the median class is 30-40.

The formula for the median of grouped data is:

From the median class (30-40): $l = 30$, $\frac{N}{2} = 25$, cf = 20 (cf of 20-30), $f = 10$ (frequency of 30-40), $h = 10$ (class size).

Substitute these values into formula (2):

The median is $\mathbf{35}$.

---

Mode:

We find the modal class, which is the class with the highest frequency. From the frequency table (5, 8, 7, 10, 8, 12), the highest frequency is 12, corresponding to the class interval 50-60.

So, the modal class is 50-60.

The formula for the mode of grouped data is:

From the modal class (50-60): $l = 50$ (lower limit), $f_1 = 12$ (frequency), $f_0 = 8$ (frequency of 40-50), $f_2 = 0$ (frequency of 60-70, as this is the last class), $h = 10$ (class size).

Substitute these values into formula (3):

The mode is $\mathbf{52.5}$.

---

Summary of Results:

Mean = 33.8

Median = 35

Mode = 52.5

The final answers are: Mean = 33.8, Median = 35, Mode = 52.5.

Given:

The frequency distribution of electric gadgets sold is given as:

Number of Gadgets	Number of Days (Frequency, $f$)
10-20	6
20-30	8
30-40	15
40-50	11
50-60	10

The total number of days (total frequency) is $N = 6 + 8 + 15 + 11 + 10 = 50$.

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To Draw:

A 'more than' type cumulative frequency ogive.

To Estimate:

The median number of gadgets sold from the ogive.

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Solution (Drawing the Ogive):

To draw a 'more than' ogive, we first prepare the 'more than' cumulative frequency distribution. We list the lower limits of the class intervals and their corresponding cumulative frequencies. The cumulative frequency for the lowest lower limit is the total frequency.

Number of Gadgets ($\ge$ Lower Limit)	'More than' Cumulative Frequency (cf)	Point to Plot (x, y)
$\ge 10$	$50$	(10, 50)
$\ge 20$	$50 - 6 = 44$	(20, 44)
$\ge 30$	$44 - 8 = 36$	(30, 36)
$\ge 40$	$36 - 15 = 21$	(40, 21)
$\ge 50$	$21 - 11 = 10$	(50, 10)
$\ge 60$	$10 - 10 = 0$	(60, 0)
Total	$N = 50$

The points to plot for the 'more than' ogive are (10, 50), (20, 44), (30, 36), (40, 21), (50, 10), and (60, 0).

To draw the ogive:

1. Draw the x-axis representing the 'Number of Gadgets' (specifically the lower limits).

2. Draw the y-axis representing the 'Cumulative Frequency'.

3. Plot the points from the table: (10, 50), (20, 44), (30, 36), (40, 21), (50, 10), and (60, 0).

4. Connect the plotted points with a smooth curve. This curve starts from (10, 50) and goes downwards to (60, 0).

(Note: The actual drawing of the ogive is not possible in this text-based format.)

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Solution (Estimating Median from Ogive):

The median is the value on the x-axis corresponding to the cumulative frequency $\frac{N}{2}$ in a cumulative frequency distribution.

The total number of days is $N = 50$.

To estimate the median from the 'more than' ogive:

1. Locate the value 25 on the cumulative frequency axis (y-axis).

2. From the point representing 25 on the y-axis, draw a horizontal line parallel to the x-axis until it intersects the 'more than' ogive.

3. From the point of intersection on the ogive, draw a vertical line parallel to the y-axis downwards to the x-axis.

4. The point where this vertical line touches the x-axis gives the estimated median value.

Looking at the 'more than' cumulative frequency table, a cumulative frequency of 25 lies between 36 (at x=30) and 21 (at x=40). Therefore, the median must lie between 30 and 40. Graphically locating the point on the ogive where the y-coordinate is 25 and dropping a perpendicular to the x-axis, we can estimate the median.

Using interpolation (or a precise graph), the x-value corresponding to a cumulative frequency of 25 on the 'more than' ogive can be estimated. This value should be approximately 37.33 (as calculated using the median formula for grouped data, though this calculation is not required by the prompt).

The estimated median number of gadgets sold from the ogive is approximately $\mathbf{37.33}$.

The final answer is approximately 37.33.