Chapter 14 Statistics

Given:

The marks obtained by 30 students and their corresponding frequencies are given in the table.

Marks obtained are denoted by $x_i$ and the number of students by $f_i$.

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To Find:

The mean of the marks obtained by the students.

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Solution:

We will use the direct method to find the mean of the given data.

The formula for the mean ($\bar{x}$) using the direct method is:

$\bar{x} = \frac{\sum (f_i \times x_i)}{\sum f_i}$

First, let's prepare a table to calculate the sum of the products of marks and frequencies ($f_i \times x_i$).

Marks obtained ($x_i$)	Number of students ($f_i$)	$f_i \times x_i$
10	1	$10 \times 1 = 10$
20	1	$20 \times 1 = 20$
36	3	$36 \times 3 = 108$
40	4	$40 \times 4 = 160$
50	3	$50 \times 3 = 150$
56	2	$56 \times 2 = 112$
60	4	$60 \times 4 = 240$
70	4	$70 \times 4 = 280$
72	1	$72 \times 1 = 72$
80	1	$80 \times 1 = 80$
88	2	$88 \times 2 = 176$
92	3	$92 \times 3 = 276$
95	1	$95 \times 1 = 95$
Total	$\sum f_i$	$\sum (f_i \times x_i)$
	30	$10+20+108+160+150+112+240+280 \ $$ +72+80 \ $$ +176+276+95 = 1779$

Sum of frequencies, $\sum f_i = 30$

Sum of the products of frequencies and marks, $\sum (f_i \times x_i) = 1779$

Now, substitute these values into the mean formula:

$\bar{x} = \frac{1779}{30}$

$\bar{x} = 59.3$

The mean marks obtained by the students is $59.3$.

Given

The frequency distribution of the percentage of female teachers in primary schools of rural areas.

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To Find

The mean percentage of female teachers using the Direct Method, Assumed Mean Method, and Step-deviation Method.

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Solution

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

$x_i = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$

We also find the sum of frequencies, $\sum f_i = 6 + 11 + 7 + 4 + 4 + 2 + 1 = 35$.

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1. Direct Method

The formula for the mean is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$. We create the following table:

Percentage of female teachers	Number of States/U.T. ($f_i$)	Class Mark ($x_i$)	$f_i x_i$
15 - 25	6	20	120
25 - 35	11	30	330
35 - 45	7	40	280
45 - 55	4	50	200
55 - 65	4	60	240
65 - 75	2	70	140
75 - 85	1	80	80
Total	$\sum f_i = 35$		$\sum f_i x_i = 1390$

Mean, $\bar{x} = \frac{1390}{35} = \frac{278}{7} \approx 39.71$.

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2. Assumed Mean Method

The formula is $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$, where $a$ is the assumed mean and $d_i = x_i - a$. Let's choose the assumed mean $a=50$.

$f_i$	$x_i$	$d_i = x_i - 50$	$f_i d_i$
6	20	-30	-180
11	30	-20	-220
7	40	-10	-70
4	50	0	0
4	60	10	40
2	70	20	40
1	80	30	30
$\sum f_i = 35$			$\sum f_i d_i = -360$

Mean, $\bar{x} = 50 + \frac{-360}{35} = 50 - \frac{72}{7} = \frac{350-72}{7} = \frac{278}{7} \approx 39.71$.

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3. Step-deviation Method

The formula is $\bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right)$, where $u_i = \frac{x_i - a}{h}$ and $h$ is the class size. Here, $a=50$ and the class size $h=10$.

$f_i$	$x_i$	$d_i = x_i - 50$	$u_i = d_i/10$	$f_i u_i$
6	20	-30	-3	-18
11	30	-20	-2	-22
7	40	-10	-1	-7
4	50	0	0	0
4	60	10	1	4
2	70	20	2	4
1	80	30	3	3
$\sum f_i = 35$				$\sum f_i u_i = -36$

Mean, $\bar{x} = 50 + 10 \times \left(\frac{-36}{35}\right) = 50 - \frac{360}{35} = 50 - \frac{72}{7} = \frac{278}{7} \approx 39.71$.

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By all three methods, the mean percentage of female teachers is 39.71%.

Given

The frequency distribution of the number of wickets taken by bowlers in one-day cricket matches.

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To Find

The mean number of wickets and to explain what the mean signifies.

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Solution

The class intervals are of varying sizes (e.g., 60-20=40, 100-60=40, 150-100=50). Because the class sizes are not uniform, the step-deviation method is not suitable. The numerical values of $x_i$ and $f_i$ are large, so the Assumed Mean Method is a good choice to simplify calculations.

The formula for the mean by the Assumed Mean Method is $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$, where $a$ is the assumed mean and $d_i = x_i - a$.

Let's prepare the calculation table. Let's choose the assumed mean $a = 200$.

Number of wickets	Number of bowlers ($f_i$)	Class Mark ($x_i$)	$d_i = x_i - 200$	$f_i d_i$
20 - 60	7	40	-160	-1120
60 - 100	5	80	-120	-600
100 - 150	16	125	-75	-1200
150 - 250	12	200	0	0
250 - 350	2	300	100	200
350 - 450	3	400	200	600
Total	$\sum f_i = 45$			$\sum f_i d_i = -2120$

First, let's correct the sum of frequencies: $\sum f_i = 7+5+16+12+2+3 = 45$.

Now, let's find the sum of $f_i d_i$: $\sum f_i d_i = -1120 - 600 - 1200 + 0 + 200 + 600 = -2920 + 800 = -2120$.

Now, we can calculate the mean:

$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 200 + \frac{-2120}{45}$

$\bar{x} = 200 - \frac{424}{9} \approx 200 - 47.11$

$\bar{x} \approx 152.89$.

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The mean number of wickets taken is approximately 152.89.

Significance of the Mean: The mean of 152.89 signifies that, on average, a bowler from this group has taken approximately 153 wickets in their one-day cricket career. It represents the central tendency of the data.

Given

The frequency distribution of the number of plants in 20 houses.

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To Find

The mean number of plants per house, and to state the method used with justification.

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Solution

We will use the Direct Method because the numerical values of the frequencies ($f_i$) and the class marks ($x_i$) are small, which makes the direct calculation of the product $f_i x_i$ simple and straightforward.

The formula for the mean using the direct method is:

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$

First, we create a table to find the class marks ($x_i$) and the product $f_i x_i$ for each class.

Number of plants	Number of houses ($f_i$)	Class Mark ($x_i$)	$f_i x_i$
0 - 2	1	1	1
2 - 4	2	3	6
4 - 6	1	5	5
6 - 8	5	7	35
8 - 10	6	9	54
10 - 12	2	11	22
12 - 14	3	13	39
Total	$\sum f_i = 20$		$\sum f_i x_i = 162$

From the table, we have:

$\sum f_i = 20$

$\sum f_i x_i = 162$

Now, we can calculate the mean:

$\bar{x} = \frac{162}{20} = 8.1$

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The mean number of plants per house is 8.1.

The Direct Method was used because the data values were small, making the calculations easy.

Given

The frequency distribution of the daily wages of 50 workers.

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To Find

The mean daily wages of the workers.

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Solution

Since the numerical values of the class marks ($x_i$) are large, the Step-deviation Method is an appropriate choice to simplify the calculations.

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

$\bar{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$

where $a$ is the assumed mean, $h$ is the class size, and $u_i = \frac{x_i - a}{h}$.

From the data, the class size $h = 520 - 500 = 20$. Let's choose the assumed mean $a = 550$.

We create the following table for our calculations:

Daily wages (in $\textsf{₹}$)	Number of workers ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 550}{20}$	$f_i u_i$
500 - 520	12	510	-2	-24
520 - 540	14	530	-1	-14
540 - 560	8	550	0	0
560 - 580	6	570	1	6
580 - 600	10	590	2	20
Total	$\sum f_i = 50$			$\sum f_i u_i = -12$

From the table, we have:

$\sum f_i = 50$

$\sum f_i u_i = -24 - 14 + 0 + 6 + 20 = -12$

Now, we can calculate the mean:

$\bar{x} = 550 + 20 \left( \frac{-12}{50} \right)$

$\bar{x} = 550 - \frac{240}{50} = 550 - 4.8 = 545.2$

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The mean daily wages of the workers of the factory is ₹ 545.20.

Given

The frequency distribution of daily pocket allowance.

The mean pocket allowance, $\bar{x} = \textsf{₹} 18$.

One frequency is missing, denoted by $f$.

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To Find

The missing frequency, $f$.

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Solution

We can use any method to solve this, but the Assumed Mean Method or Step-deviation Method can simplify the algebra. Let's use the Assumed Mean Method.

The formula for the mean is $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$, where $a$ is the assumed mean and $d_i = x_i - a$.

Let's choose the assumed mean $a = 18$.

We create the following table for our calculations:

Daily pocket allowance (in $\textsf{₹}$)	Number of children ($f_i$)	Class Mark ($x_i$)	$d_i = x_i - 18$	$f_i d_i$
11 - 13	7	12	-6	-42
13 - 15	6	14	-4	-24
15 - 17	9	16	-2	-18
17 - 19	13	18	0	0
19 - 21	$f$	20	2	$2f$
21 - 23	5	22	4	20
23 - 25	4	24	6	24
Total	$\sum f_i = 44+f$			$\sum f_i d_i = 2f - 40$

From the table, we find the sums:

$\sum f_i = 7+6+9+13+f+5+4 = 44+f$.

$\sum f_i d_i = -42 - 24 - 18 + 0 + 2f + 20 + 24 = 2f - 40$.

Now, we use the mean formula with the given mean $\bar{x} = 18$ and assumed mean $a=18$:

$18 = 18 + \frac{2f - 40}{44 + f}$

$18 - 18 = \frac{2f - 40}{44 + f}$

$0 = \frac{2f - 40}{44 + f}$

For the fraction to be zero, the numerator must be zero.

$2f - 40 = 0$

$2f = 40$

$f = 20$.

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The missing frequency is 20.

Given

The frequency distribution of heartbeats per minute for 30 women.

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To Find

The mean heartbeats per minute.

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Solution

The class size is uniform ($h = 3$), and the class marks are decimals. The Step-deviation Method is a suitable choice to simplify the calculations.

The formula for the mean is $\bar{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$, where $u_i = \frac{x_i - a}{h}$.

Let's choose the assumed mean $a = 75.5$ (the middle class mark).

We create the following table for our calculations:

Heartbeats per minute	Number of women ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 75.5}{3}$	$f_i u_i$
65 - 68	2	66.5	-3	-6
68 - 71	4	69.5	-2	-8
71 - 74	3	72.5	-1	-3
74 - 77	8	75.5	0	0
77 - 80	7	78.5	1	7
80 - 83	4	81.5	2	8
83 - 86	2	84.5	3	6
Total	$\sum f_i = 30$			$\sum f_i u_i = 4$

From the table, we have:

$\sum f_i = 30$

$\sum f_i u_i = -6 - 8 - 3 + 0 + 7 + 8 + 6 = 4$.

Now, we can calculate the mean:

$\bar{x} = 75.5 + 3 \left( \frac{4}{30} \right)$

$\bar{x} = 75.5 + \frac{12}{30} = 75.5 + \frac{2}{5} = 75.5 + 0.4$

$\bar{x} = 75.9$.

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The mean heartbeats per minute for these women is 75.9.

Given

The frequency distribution of the number of mangoes in packing boxes.

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To Find

The mean number of mangoes per box and to state the method used with justification.

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Solution

The given class intervals are discontinuous. However, we can calculate the class marks as usual. The class marks are 51, 54, 57, 60, 63. The difference between consecutive class marks is uniform ($h=3$). Since the frequencies are large, the Step-deviation Method is the most suitable method to simplify calculations.

The formula for the mean is $\bar{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$, where $u_i = \frac{x_i - a}{h}$.

Let's choose the assumed mean $a = 57$ and the class size $h=3$.

We create the following table for our calculations:

Number of mangoes	Number of boxes ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 57}{3}$	$f_i u_i$
50 - 52	15	51	-2	-30
53 - 55	110	54	-1	-110
56 - 58	135	57	0	0
59 - 61	115	60	1	115
62 - 64	25	63	2	50
Total	$\sum f_i = 400$			$\sum f_i u_i = 25$

From the table, we have:

$\sum f_i = 400$

$\sum f_i u_i = -30 - 110 + 0 + 115 + 50 = 25$.

Now, we can calculate the mean:

$\bar{x} = 57 + 3 \left( \frac{25}{400} \right) = 57 + 3 \left( \frac{1}{16} \right)$

$\bar{x} = 57 + \frac{3}{16} = 57 + 0.1875 = 57.1875$.

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The mean number of mangoes kept in a packing box is 57.19 (rounded to two decimal places).

The Step-deviation Method was chosen because the frequencies were large and the class marks were uniformly spaced, which simplifies the arithmetic calculations significantly.

Given

The frequency distribution of daily expenditure on food for 25 households.

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To Find

The mean daily expenditure on food.

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Solution

The class size is uniform ($h = 50$) and the class marks are large. The Step-deviation Method is a suitable choice to simplify the calculations.

The formula for the mean is $\bar{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$, where $u_i = \frac{x_i - a}{h}$.

Let's choose the assumed mean $a = 225$.

We create the following table for our calculations:

Daily expenditure (in $\textsf{₹}$)	Number of households ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 225}{50}$	$f_i u_i$
100 - 150	4	125	-2	-8
150 - 200	5	175	-1	-5
200 - 250	12	225	0	0
250 - 300	2	275	1	2
300 - 350	2	325	2	4
Total	$\sum f_i = 25$			$\sum f_i u_i = -7$

From the table, we have:

$\sum f_i = 25$

$\sum f_i u_i = -8 - 5 + 0 + 2 + 4 = -7$.

Now, we can calculate the mean:

$\bar{x} = 225 + 50 \left( \frac{-7}{25} \right)$

$\bar{x} = 225 - (50 \times \frac{7}{25}) = 225 - (2 \times 7) = 225 - 14$

$\bar{x} = 211$.

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The mean daily expenditure on food is ₹ 211.

Given

The frequency distribution of the concentration of SO₂ in the air for 30 localities.

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To Find

The mean concentration of SO₂.

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Solution

Since the numerical values of the class marks are small decimals, the Direct Method is suitable and easy to apply.

The formula for the mean is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

We create the following table for our calculations:

Concentration of SO2 (in ppm)	Frequency ($f_i$)	Class Mark ($x_i$)	$f_i x_i$
0.00 - 0.04	4	0.02	0.08
0.04 - 0.08	9	0.06	0.54
0.08 - 0.12	9	0.10	0.90
0.12 - 0.16	2	0.14	0.28
0.16 - 0.20	4	0.18	0.72
0.20 - 0.24	2	0.22	0.44
Total	$\sum f_i = 30$		$\sum f_i x_i = 2.96$

From the table, we have:

$\sum f_i = 30$

$\sum f_i x_i = 2.96$

Now, we can calculate the mean:

$\bar{x} = \frac{2.96}{30} \approx 0.09866...$

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The mean concentration of SO₂ in the air is approximately 0.099 ppm.

Given

The frequency distribution of the number of days students were absent.

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To Find

The mean number of days a student was absent.

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Solution

The class intervals are of varying sizes, so the step-deviation method cannot be used. We will use the Assumed Mean Method to simplify calculations, as the direct method would involve larger products.

The formula for the mean is $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$, where $a$ is the assumed mean and $d_i = x_i - a$.

Let's choose the assumed mean $a = 17$.

We create the following table for our calculations:

Number of days	Number of students ($f_i$)	Class Mark ($x_i$)	$d_i = x_i - 17$	$f_i d_i$
0 - 6	11	3	-14	-154
6 - 10	10	8	-9	-90
10 - 14	7	12	-5	-35
14 - 20	4	17	0	0
20 - 28	4	24	7	28
28 - 38	3	33	16	48
38 - 40	1	39	22	22
Total	$\sum f_i = 40$			$\sum f_i d_i = -181$

From the table, we have:

$\sum f_i = 40$

$\sum f_i d_i = -154 - 90 - 35 + 0 + 28 + 48 + 22 = -279 + 98 = -181$.

Now, we can calculate the mean:

$\bar{x} = 17 + \frac{-181}{40} = 17 - 4.525$

$\bar{x} = 12.475$.

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The mean number of days a student was absent is 12.475.

Given

The frequency distribution of the literacy rate of 35 cities.

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To Find

The mean literacy rate.

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Solution

The class size is uniform ($h = 10$). We can use the Step-deviation Method to find the mean, as it will simplify the calculations.

The formula for the mean is $\bar{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$, where $u_i = \frac{x_i - a}{h}$.

Let's choose the assumed mean $a = 70$.

We create the following table for our calculations:

Literacy rate (in %)	Number of cities ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 70}{10}$	$f_i u_i$
45 - 55	3	50	-2	-6
55 - 65	10	60	-1	-10
65 - 75	11	70	0	0
75 - 85	8	80	1	8
85 - 95	3	90	2	6
Total	$\sum f_i = 35$			$\sum f_i u_i = -2$

From the table, we have:

$\sum f_i = 35$

$\sum f_i u_i = -6 - 10 + 0 + 8 + 6 = -2$.

Now, we can calculate the mean:

$\bar{x} = 70 + 10 \left( \frac{-2}{35} \right) = 70 - \frac{20}{35} = 70 - \frac{4}{7}$

$\bar{x} = \frac{490 - 4}{7} = \frac{486}{7} \approx 69.43$.

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The mean literacy rate is approximately 69.43%.

Given

The number of wickets taken by a bowler in 10 cricket matches are:

2, 6, 4, 5, 0, 2, 1, 3, 2, 3

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To Find

The mode of the data.

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Solution

The mode is the value that appears most frequently in a data set.

To find the mode, let's create a frequency table for the given data.

Number of wickets	Tally Marks	Frequency (Number of matches)
0	|	1
1	|	1
2	|||	3
3	||	2
4	|	1
5	|	1
6	|	1

From the frequency table, we can see that the number of wickets '2' has the highest frequency of 3.

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Therefore, the mode of the data is 2.

Given

A grouped frequency distribution of family size in 20 households.

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To Find

The mode of the data.

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Solution

To find the mode for grouped data, we use the formula:

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

Step 1: Identify the modal class and related values.

The modal class is the class with the highest frequency. Looking at the table, the highest frequency is 8, which corresponds to the class interval 3 - 5.

From this, we get:

• Lower limit of the modal class ($l$) = 3
• Frequency of the modal class ($f_1$) = 8
• Frequency of the class preceding the modal class ($f_0$) = 7
• Frequency of the class succeeding the modal class ($f_2$) = 2
• Class size ($h$) = 5 - 3 = 2

Step 2: Substitute the values into the formula.

Mode $= 3 + \left(\frac{8 - 7}{2(8) - 7 - 2}\right) \times 2$

Mode $= 3 + \left(\frac{1}{16 - 9}\right) \times 2$

Mode $= 3 + \left(\frac{1}{7}\right) \times 2$

Mode $= 3 + \frac{2}{7}$

Mode $= \frac{21+2}{7} = \frac{23}{7}$

Mode $\approx 3.286$.

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The mode of this data is approximately 3.286.

Given

The frequency distribution of marks of 30 students.

From Example 1, the mean of this data was calculated to be 62.

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To Find

1. The mode of the data.

2. Compare and interpret the mode and the mean.

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Solution

1. Finding the Mode:

We use the formula: Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

First, we identify the modal class. The highest frequency is 7, which corresponds to the class interval 40 - 55.

From this, we get:

• $l = 40$
• $f_1 = 7$
• $f_0 = 3$ (frequency of the class 25 - 40)
• $f_2 = 6$ (frequency of the class 55 - 70)
• $h = 55 - 40 = 15$

Substitute these values into the formula:

Mode $= 40 + \left(\frac{7 - 3}{2(7) - 3 - 6}\right) \times 15$

Mode $= 40 + \left(\frac{4}{14 - 9}\right) \times 15$

Mode $= 40 + \left(\frac{4}{5}\right) \times 15$

Mode $= 40 + (4 \times 3) = 40 + 12 = 52$.

The mode of the marks is 52.

2. Comparison and Interpretation:

We have:

Mean = 62 marks.

Mode = 52 marks.

Interpretation: The mode (52) indicates that the highest number of students scored marks in the range which has 52 as the central value. It represents the most frequent score range.

The mean (62) represents the average score of all 30 students. If the total marks were distributed equally among all students, each would get 62 marks.

The mean is greater than the mode. This suggests that while the most common group of scores is around 52, there are enough students with higher scores to pull the average up. The data is slightly skewed to the right (positively skewed).

Given:

A frequency distribution table of the ages of patients admitted to a hospital.

Age (in years)	5 - 15	15 - 25	25 - 35	35 - 45	45 - 55	55 - 65
Number of patients ($f_i$)	6	11	21	23	14	5

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To Find:

The mode and mean of the data, followed by a comparison and interpretation of the results.

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Solution:

Calculation for Mode

The formula for the mode of grouped data is:

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

First, we identify the modal class, which is the class with the highest frequency. Here, the highest frequency is 23, corresponding to the class interval 35 - 45.

From the data, we have:

• Lower limit of the modal class ($l$) = 35
• Class size ($h$) = 10
• Frequency of the modal class ($f_1$) = 23
• Frequency of the preceding class ($f_0$) = 21
• Frequency of the succeeding class ($f_2$) = 14

Substituting these values into the formula:

Mode $= 35 + \left(\frac{23 - 21}{2(23) - 21 - 14}\right) \times 10$

Mode $= 35 + \left(\frac{2}{46 - 35}\right) \times 10$

Mode $= 35 + \left(\frac{2}{11}\right) \times 10$

Mode $= 35 + \frac{20}{11} \approx 35 + 1.82$

Mode $\approx 36.82$

The mode of the data is approximately 36.82 years.

Calculation for Mean

We will use the Step-deviation method to find the mean. The formula is:

Mean $(\bar{x}) = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$

Let's create the calculation table. Let the assumed mean ($a$) be 30, and class size ($h$) is 10.

Age (in years)	Number of patients ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 30}{10}$	$f_i u_i$
5 - 15	6	10	-2	-12
15 - 25	11	20	-1	-11
25 - 35	21	30	0	0
35 - 45	23	40	1	23
45 - 55	14	50	2	28
55 - 65	5	60	3	15
Total	$\sum f_i = 80$			$\sum f_i u_i = 43$

From the table, $\sum f_i = 80$ and $\sum f_i u_i = (-12 - 11) + (23 + 28 + 15) = -23 + 66 = 43$.

Substituting these values into the formula:

$\bar{x} = 30 + \left(\frac{43}{80}\right) \times 10$

$\bar{x} = 30 + \frac{430}{80} = 30 + \frac{43}{8}$

$\bar{x} = 30 + 5.375$

$\bar{x} = 35.375$

The mean of the data is 35.375 years.

---

Comparison and Interpretation:

The mode is 36.82 years, which represents the age with the highest frequency of patients. This implies that the maximum number of patients admitted are around 37 years old.

The mean is 35.375 years, which represents the average age of all patients admitted to the hospital.

The two values are quite close. This indicates that the distribution of patient ages is nearly symmetrical. The mode being slightly higher than the mean suggests that while the average age is around 35.4, the most common age group is slightly older.

Given:

A frequency distribution table showing the lifetimes of 225 electrical components.

Lifetimes (in hours)	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120
Frequency ($f_i$)	10	35	52	61	38	29

---

To Find:

The modal lifetimes of the components.

---

Solution:

The formula for the mode of grouped data is:

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

First, we identify the modal class by finding the class with the highest frequency. The maximum frequency is 61, which corresponds to the class interval 60 - 80.

From the data, we have:

• Lower limit of the modal class ($l$) = 60
• Class size ($h$) = 20
• Frequency of the modal class ($f_1$) = 61
• Frequency of the preceding class ($f_0$) = 52
• Frequency of the succeeding class ($f_2$) = 38

Substituting these values into the formula:

Mode $= 60 + \left(\frac{61 - 52}{2(61) - 52 - 38}\right) \times 20$

Mode $= 60 + \left(\frac{9}{122 - 90}\right) \times 20$

Mode $= 60 + \left(\frac{9}{32}\right) \times 20$

Mode $= 60 + \frac{180}{32} = 60 + \frac{45}{8}$

Mode $= 60 + 5.625$

Mode $= 65.625$

Therefore, the modal lifetime of the electrical components is 65.625 hours. This means the most common lifetime for these components is approximately 65.625 hours.

Given:

A frequency distribution table of the monthly household expenditure of 200 families.

---

To Find:

The modal and mean monthly expenditure.

---

Solution:

Calculation for Mode

The formula for the mode is: Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

The highest frequency is 40, so the modal class is 1500 - 2000.

We have:

• $l = 1500$
• $h = 500$
• $f_1 = 40$
• $f_0 = 24$
• $f_2 = 33$

Substituting the values:

Mode $= 1500 + \left(\frac{40 - 24}{2(40) - 24 - 33}\right) \times 500$

Mode $= 1500 + \left(\frac{16}{80 - 57}\right) \times 500$

Mode $= 1500 + \left(\frac{16}{23}\right) \times 500$

Mode $= 1500 + \frac{8000}{23} \approx 1500 + 347.83$

Mode $\approx 1847.83$

The modal monthly expenditure is $\textsf{₹}1847.83$.

Calculation for Mean

Using the Step-deviation method: Mean $(\bar{x}) = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$

Let the assumed mean ($a$) be 2750, and class size ($h$) is 500.

Expenditure (in $\textsf{₹}$)	No. of families ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 2750}{500}$	$f_i u_i$
1000 - 1500	24	1250	-3	-72
1500 - 2000	40	1750	-2	-80
2000 - 2500	33	2250	-1	-33
2500 - 3000	28	2750	0	0
3000 - 3500	30	3250	1	30
3500 - 4000	22	3750	2	44
4000 - 4500	16	4250	3	48
4500 - 5000	7	4750	4	28
Total	$\sum f_i = 200$			$\sum f_i u_i = -35$

From the table, $\sum f_i = 200$ and $\sum f_i u_i = -185 + 150 = -35$.

Substituting the values:

$\bar{x} = 2750 + \left(\frac{-35}{200}\right) \times 500$

$\bar{x} = 2750 - \frac{35 \times 5}{2} = 2750 - \frac{175}{2}$

$\bar{x} = 2750 - 87.5$

$\bar{x} = 2662.5$

The mean monthly expenditure is $\textsf{₹}2662.50$.

Given:

A frequency distribution of the state-wise teacher-student ratio in higher secondary schools of India.

---

To Find:

The mode and mean of the data, and an interpretation of the results.

---

Solution:

Calculation for Mode

The mode formula is: Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

The highest frequency is 10, making the modal class 30 - 35.

We have:

• $l = 30$
• $h = 5$
• $f_1 = 10$
• $f_0 = 9$
• $f_2 = 3$

Substituting the values:

Mode $= 30 + \left(\frac{10 - 9}{2(10) - 9 - 3}\right) \times 5$

Mode $= 30 + \left(\frac{1}{20 - 12}\right) \times 5 = 30 + \frac{5}{8}$

Mode $= 30 + 0.625 = 30.625$

The modal ratio is 30.625.

Calculation for Mean

Using the Step-deviation method: Mean $(\bar{x}) = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$

Let the assumed mean ($a$) be 32.5, and class size ($h$) is 5.

Students per Teacher	No. of States/U.T. ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 32.5}{5}$	$f_i u_i$
15 - 20	3	17.5	-3	-9
20 - 25	8	22.5	-2	-16
25 - 30	9	27.5	-1	-9
30 - 35	10	32.5	0	0
35 - 40	3	37.5	1	3
40 - 45	0	42.5	2	0
45 - 50	0	47.5	3	0
50 - 55	2	52.5	4	8
Total	$\sum f_i = 35$			$\sum f_i u_i = -23$

From the table, $\sum f_i = 35$ and $\sum f_i u_i = -34 + 11 = -23$.

Substituting the values:

$\bar{x} = 32.5 + \left(\frac{-23}{35}\right) \times 5$

$\bar{x} = 32.5 - \frac{23 \times \cancel{5}^1}{\cancel{35}_7} = 32.5 - \frac{23}{7}$

$\bar{x} \approx 32.5 - 3.29 = 29.21$

The mean ratio is approximately 29.21.

---

Interpretation:

The mode of 30.625 implies that the most common student-teacher ratio in the states/U.T. is approximately 30.6 students per teacher.

The mean of 29.21 indicates that, on average, there are about 29.2 students per teacher across all the states/U.T.

Given:

A frequency distribution table of runs scored by top batsmen.

---

To Find:

The mode of the given data.

---

Solution:

The formula to find the mode of grouped data is:

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

From the table, the maximum frequency is 18, which corresponds to the class interval 4000 - 5000. This is our modal class.

We have:

• Lower limit of the modal class ($l$) = 4000
• Class size ($h$) = 1000
• Frequency of the modal class ($f_1$) = 18
• Frequency of the preceding class ($f_0$) = 4
• Frequency of the succeeding class ($f_2$) = 9

Substituting these values into the formula:

Mode $= 4000 + \left(\frac{18 - 4}{2(18) - 4 - 9}\right) \times 1000$

Mode $= 4000 + \left(\frac{14}{36 - 13}\right) \times 1000$

Mode $= 4000 + \left(\frac{14}{23}\right) \times 1000$

Mode $= 4000 + \frac{14000}{23} \approx 4000 + 608.70$

Mode $\approx 4608.70$

The mode of the data is approximately 4608.7 runs, indicating the most common range of scores for top batsmen.

Given:

A frequency distribution of the number of cars passing a spot in 100 periods.

---

To Find:

The mode of the data.

---

Solution:

The formula to find the mode of grouped data is:

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

The highest frequency in the data is 20, which corresponds to the class interval 40 - 50. This is the modal class.

From the data, we have:

• Lower limit of the modal class ($l$) = 40
• Class size ($h$) = 10
• Frequency of the modal class ($f_1$) = 20
• Frequency of the preceding class ($f_0$) = 12
• Frequency of the succeeding class ($f_2$) = 11

Substituting these values into the formula:

Mode $= 40 + \left(\frac{20 - 12}{2(20) - 12 - 11}\right) \times 10$

Mode $= 40 + \left(\frac{8}{40 - 23}\right) \times 10$

Mode $= 40 + \left(\frac{8}{17}\right) \times 10$

Mode $= 40 + \frac{80}{17} \approx 40 + 4.71$

Mode $\approx 44.71$

Therefore, the mode of the data is approximately 44.71 cars. This represents the most frequent number of cars passing the spot during a 3-minute period.

Given:

A 'less than' type cumulative frequency distribution of the heights of 51 girls.

---

To Find:

The median height of the girls.

---

Solution:

Step 1: Convert the cumulative frequency distribution to a standard frequency distribution.

The given data is of 'less than' type. We need to create class intervals and find the frequency for each class. The class size is constant ($145-140=5$).

Height (in cm)	Number of girls (Frequency $f_i$)	Cumulative Frequency (CF)
135 - 140	4	4
140 - 145	$11 - 4 = 7$	11
145 - 150	$29 - 11 = 18$	29
150 - 155	$40 - 29 = 11$	40
155 - 160	$46 - 40 = 6$	46
160 - 165	$51 - 46 = 5$	51
Total	$\sum f_i = N = 51$

Step 2: Identify the median class.

The total number of observations is $N = 51$.

We need to find the value of $\frac{N}{2} = \frac{51}{2} = 25.5$.

Now, we locate the class whose cumulative frequency is just greater than or equal to 25.5. From the table, this cumulative frequency is 29, which corresponds to the class interval 145 - 150.

Therefore, the median class is $145 - 150$.

Step 3: Calculate the median using the formula.

The formula for the median is:

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

Here,

• $l$ = Lower limit of the median class = $145$
• $N$ = Total frequency = $51$
• $CF$ = Cumulative frequency of the class preceding the median class = $11$
• $f$ = Frequency of the median class = $18$
• $h$ = Class size = $5$

Substituting these values into the formula:

Median $= 145 + \left(\frac{25.5 - 11}{18}\right) \times 5$

Median $= 145 + \left(\frac{14.5}{18}\right) \times 5$

Median $= 145 + \frac{72.5}{18}$

Median $\approx 145 + 4.03$

Median $\approx 149.03$

Hence, the median height of the girls is approximately 149.03 cm.

Given:

• A grouped frequency distribution with two unknown frequencies, x and y.
• The median of the distribution is 525.
• The total frequency ($N$) is 100.

---

To Find:

The values of x and y.

---

Solution:

Step 1: Construct the cumulative frequency table and form the first equation.

We are given that the sum of all frequencies is 100.

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

The sum of frequencies is $2+5+x+12+17+20+y+9+7+4 = 76 + x + y$.

Since the total frequency is 100, we have:

$76 + x + y = 100$

$x + y = 100 - 76$

Step 2: Identify the median class and apply the median formula to find x.

The given median is 525. This value lies in the class interval 500 - 600. So, this is our median class.

The formula for the median is: Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

Here,

• $l$ = Lower limit of the median class = $500$
• $N$ = Total frequency = $100$, so $\frac{N}{2} = 50$
• $CF$ = Cumulative frequency of the class preceding the median class = $36 + x$
• $f$ = Frequency of the median class = $20$
• $h$ = Class size = $100$

Substituting these values into the formula:

$525 = 500 + \left(\frac{50 - (36 + x)}{20}\right) \times 100$

$525 - 500 = \left(\frac{50 - 36 - x}{20}\right) \times 100$

$25 = \left(\frac{14 - x}{\cancel{20}_1}\right) \times \cancel{100}^5$

$25 = (14 - x) \times 5$

$\frac{25}{5} = 14 - x$

$5 = 14 - x$

$x = 14 - 5$

$x = 9$

Step 3: Find the value of y.

Substitute the value of $x = 9$ into equation (i):

$9 + y = 24$

$y = 24 - 9$

$y = 15$

Hence, the values of the missing frequencies are x = 9 and y = 15.

Given:

A frequency distribution of the monthly electricity consumption of 68 consumers.

---

To Find:

The median, mean, and mode of the data and to compare them.

---

Solution:

First, we construct a comprehensive table with all necessary calculations for mean, median, and mode.

Let the assumed mean, $a = 135$. Class size, $h = 20$.

Monthly Consumption (units)	No. of Consumers ($f_i$)	Class Mark ($x_i$)	Cumulative Frequency (CF)	$u_i = \frac{x_i - 135}{20}$	$f_i u_i$
65 - 85	4	75	4	-3	-12
85 - 105	5	95	9	-2	-10
105 - 125	13	115	22	-1	-13
125 - 145	20	135	42	0	0
145 - 165	14	155	56	1	14
165 - 185	8	175	64	2	16
185 - 205	4	195	68	3	12
Total	$N=\sum f_i = 68$				$\sum f_i u_i = 7$

---

Calculation for Median

Here, $N = 68$, so $\frac{N}{2} = \frac{68}{2} = 34$.

The cumulative frequency just greater than 34 is 42, which corresponds to the class interval 125 - 145. This is the median class.

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

Here, $l=125$, $CF=22$, $f=20$, $h=20$.

Median $= 125 + \left(\frac{34 - 22}{20}\right) \times 20 = 125 + \frac{12}{20} \times 20 = 125 + 12 = 137$.

Median = 137 units.

---

Calculation for Mean

Using the step-deviation method: Mean $(\bar{x}) = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$.

From the table, $\sum f_i u_i = 7$ and $\sum f_i = 68$.

$\bar{x} = 135 + \left(\frac{7}{68}\right) \times 20 = 135 + \frac{140}{68} = 135 + \frac{35}{17} \approx 135 + 2.059$.

Mean $\approx 137.06$ units.

---

Calculation for Mode

The highest frequency is 20, so the modal class is 125 - 145.

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

Here, $l=125$, $f_1=20$, $f_0=13$, $f_2=14$, $h=20$.

Mode $= 125 + \left(\frac{20 - 13}{2(20) - 13 - 14}\right) \times 20 = 125 + \left(\frac{7}{40 - 27}\right) \times 20 = 125 + \frac{140}{13} \ $$ \approx 125 + 10.77$.

Mode $\approx 135.77$ units.

---

Comparison of Results:

• Mean $\approx 137.06$
• Median $= 137$
• Mode $\approx 135.77$

The three measures of central tendency are very close to each other. This indicates that the data distribution is nearly symmetrical and bell-shaped, with a central value around 137 units.

Given:

• A frequency distribution with two unknown frequencies, x and y.
• The median of the distribution is 28.5.
• The total frequency, $N = 60$.

---

To Find:

The values of x and y.

---

Solution:

Since there are two unknown variables, we need to form two independent equations to solve for them.

Step 1: Form the first equation using the total frequency.

We first construct the cumulative frequency (CF) table.

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

The sum of all frequencies is given as 60. From the table, this sum is $45 + x + y$.

$45 + x + y = 60$

$x + y = 60 - 45$

Step 2: Form the second equation using the given median.

The median is given as 28.5, which lies in the class interval 20 - 30. So, this is the median class.

The median formula is: Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

Here, $l=20$, $N=60$ (so $\frac{N}{2}=30$), $CF = 5+x$ (CF of the preceding class), $f=20$, and $h=10$.

Substituting these values:

$28.5 = 20 + \left(\frac{30 - (5 + x)}{20}\right) \times 10$

$28.5 - 20 = \left(\frac{25 - x}{20}\right) \times 10$

$8.5 = \frac{25 - x}{2}$

$8.5 \times 2 = 25 - x$

$17 = 25 - x$

$x = 25 - 17 \implies x = 8$.

Step 3: Solve for y.

Substitute $x=8$ into equation (i):

$8 + y = 15$

$y = 15 - 8 \implies y = 7$.

---

The values of the missing frequencies are x = 8 and y = 7.

Given:

A 'less than' type cumulative frequency distribution of the ages of 100 policy holders, aged 18 years and above.

---

To Find:

The median age of the policy holders.

---

Solution:

Step 1: Convert the given data into a frequency distribution table.

The given data is cumulative. We need to find the frequency of each class interval. Since policies start at age 18, the first class interval will be 18-20.

Age (in years)	Number of policy holders (Frequency $f_i$)	Cumulative Frequency (CF)
18 - 20	2	2
20 - 25	$6 - 2 = 4$	6
25 - 30	$24 - 6 = 18$	24
30 - 35	$45 - 24 = 21$	45
35 - 40	$78 - 45 = 33$	78
40 - 45	$89 - 78 = 11$	89
45 - 50	$92 - 89 = 3$	92
50 - 55	$98 - 92 = 6$	98
55 - 60	$100 - 98 = 2$	100
Total	$N = 100$

Step 2: Identify the median class.

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

The cumulative frequency just greater than 50 is 78, which corresponds to the class interval 35 - 40. This is the median class.

Step 3: Calculate the median.

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

Here, $l=35$, $CF=45$ (CF of the preceding class), $f=33$, and $h=5$.

Median $= 35 + \left(\frac{50 - 45}{33}\right) \times 5$

Median $= 35 + \frac{5}{33} \times 5 = 35 + \frac{25}{33}$

Median $\approx 35 + 0.7576$

Median $\approx 35.76$

The median age of the policy holders is approximately 35.76 years.

Given:

A frequency distribution of the lengths of 40 leaves with discontinuous class intervals.

---

To Find:

The median length of the leaves.

---

Solution:

Step 1: Convert the discontinuous classes to continuous classes.

The median formula requires continuous class intervals. There is a gap of 1 mm between the upper limit of one class and the lower limit of the next (e.g., 126 and 127). The adjustment factor is half the gap, i.e., $\frac{1}{2} = 0.5$. We subtract 0.5 from each lower limit and add 0.5 to each upper limit.

Length (in mm)	No. of Leaves ($f_i$)	Cumulative Frequency (CF)
117.5 - 126.5	3	3
126.5 - 135.5	5	8
135.5 - 144.5	9	17
144.5 - 153.5	12	29
153.5 - 162.5	5	34
162.5 - 171.5	4	38
171.5 - 180.5	2	40
Total	$N = 40$

Step 2: Identify the median class.

Here, $N = 40$, so $\frac{N}{2} = \frac{40}{2} = 20$.

The cumulative frequency just greater than 20 is 29, corresponding to the class 144.5 - 153.5. This is the median class.

Step 3: Calculate the median.

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

Here, $l=144.5$, $CF=17$, $f=12$, and the class size $h = 153.5 - 144.5 = 9$.

Median $= 144.5 + \left(\frac{20 - 17}{12}\right) \times 9$

Median $= 144.5 + \left(\frac{3}{12}\right) \times 9 = 144.5 + \frac{1}{4} \times 9$

Median $= 144.5 + \frac{9}{4} = 144.5 + 2.25$

Median $= 146.75$

The median length of the leaves is 146.75 mm.

Given:

The frequency distribution of the lifetime of 400 neon lamps.

---

To Find:

The median lifetime of a lamp.

---

Solution:

Step 1: Construct the cumulative frequency (CF) table.

Life time (in hours)	No. of lamps ($f_i$)	Cumulative Frequency (CF)
1500 - 2000	14	14
2000 - 2500	56	70
2500 - 3000	60	130
3000 - 3500	86	216
3500 - 4000	74	290
4000 - 4500	62	352
4500 - 5000	48	400
Total	$N = 400$

Step 2: Identify the median class.

Total frequency $N = 400$, so $\frac{N}{2} = \frac{400}{2} = 200$.

The cumulative frequency just greater than 200 is 216, which corresponds to the class interval 3000 - 3500. This is the median class.

Step 3: Calculate the median.

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

Here, $l=3000$, $CF=130$, $f=86$, and $h=500$.

Median $= 3000 + \left(\frac{200 - 130}{86}\right) \times 500$

Median $= 3000 + \left(\frac{70}{86}\right) \times 500 = 3000 + \frac{35000}{86}$

Median $\approx 3000 + 406.98$

Median $\approx 3406.98$

The median lifetime of a lamp is approximately 3406.98 hours.

Given:

A frequency distribution of the number of letters in 100 surnames.

---

To Find:

The median, mean, and mode for the number of letters in the surnames.

---

Solution:

First, we create a master table for all necessary calculations. Let the assumed mean, $a = 8.5$. Class size, $h = 3$.

No. of Letters	No. of Surnames ($f_i$)	Class Mark ($x_i$)	CF	$u_i = \frac{x_i - 8.5}{3}$	$f_i u_i$
1 - 4	6	2.5	6	-2	-12
4 - 7	30	5.5	36	-1	-30
7 - 10	40	8.5	76	0	0
10 - 13	16	11.5	92	1	16
13 - 16	4	14.5	96	2	8
16 - 19	4	17.5	100	3	12
Total	$N = 100$				$\sum f_i u_i = -6$

---

Median Calculation

Here, $N = 100$, so $\frac{N}{2} = 50$. The CF just greater than 50 is 76, so the median class is 7 - 10.

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h = 7 + \left(\frac{50 - 36}{40}\right) \times 3 = 7 + \frac{14 \times 3}{40} \ $$ = 7 + \frac{42}{40} \ $$ = 7 + 1.05$.

Median = 8.05 letters.

---

Mean Calculation

Mean $(\bar{x}) = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 8.5 + \left(\frac{-6}{100}\right) \times 3 = 8.5 - \frac{18}{100} \ $$ = 8.5 - 0.18$.

Mean = 8.32 letters.

---

Mode Calculation

The highest frequency is 40, so the modal class is 7 - 10.

Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h = 7 + \left(\frac{40 - 30}{2(40) - 30 - 16}\right) \times 3 \ $$ = 7 + \left(\frac{10}{80 - 46}\right) \times 3 \ $$ = 7 + \frac{30}{34} \approx 7 + 0.88$.

Mode $\approx 7.88$ letters.

Given:

The frequency distribution of the weights of 30 students.

---

To Find:

The median weight of the students.

---

Solution:

Step 1: Create the cumulative frequency (CF) table.

Weight (in kg)	No. of students ($f_i$)	Cumulative Frequency (CF)
40 - 45	2	2
45 - 50	3	5
50 - 55	8	13
55 - 60	6	19
60 - 65	6	25
65 - 70	3	28
70 - 75	2	30
Total	$N = 30$

Step 2: Identify the median class.

Total frequency $N = 30$, so $\frac{N}{2} = \frac{30}{2} = 15$.

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

Step 3: Calculate the median.

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

Here, $l=55$, $CF=13$, $f=6$, and $h=5$.

Median $= 55 + \left(\frac{15 - 13}{6}\right) \times 5$

Median $= 55 + \frac{2}{6} \times 5 = 55 + \frac{1}{3} \times 5$

Median $= 55 + \frac{5}{3} \approx 55 + 1.67$

Median $\approx 56.67$

The median weight of the students is approximately 56.67 kg.

Given:

A 'more than or equal to' type cumulative frequency distribution of the annual profits of 30 shops.

---

To Do:

Draw both 'less than' and 'more than' ogives for the data and find the median profit from the graph.

---

Solution:

Step 1: Convert the given data into a frequency distribution and prepare data for both ogives.

We first construct a frequency distribution table from the 'more than' data. Then, we find the 'less than' cumulative frequencies.

Profit (in lakhs of ₹)	Frequency ($f_i$)	'Less than' Cumulative Frequency	'More than' Cumulative Frequency
5 - 10	$30 - 28 = 2$	2	30
10 - 15	$28 - 16 = 12$	14	28
15 - 20	$16 - 14 = 2$	16	16
20 - 25	$14 - 10 = 4$	20	14
25 - 30	$10 - 7 = 3$	23	10
30 - 35	$7 - 3 = 4$	27	7
35 - 40	$3 - 0 = 3$	30	3
Total	$N=30$

Step 2: Determine the points for plotting the ogives.

For the 'Less than' Ogive, we plot the upper class limits against their corresponding cumulative frequencies:

Points: (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27), (40, 30). We also add the point (5, 0).

For the 'More than' Ogive, we plot the lower class limits against their corresponding cumulative frequencies:

Points: (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7), (35, 3).

Step 3: Draw the ogives and find the median.

We plot these points on a graph with Profit (in lakhs of ₹) on the x-axis and the Number of Shops (cumulative frequency) on the y-axis. We then join the points for each ogive with a smooth curve.

The two ogives intersect at a point. We draw a perpendicular from this intersection point to the x-axis. The x-coordinate of this point gives the median of the data.

From the graph, the intersection point is approximately (17.5, 15).

The x-coordinate is 17.5.

Hence, the median profit is $\textsf{₹}17.5$ lakhs.

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Verification by Formula

Total frequency $N = 30$, so $\frac{N}{2} = 15$.

From the 'less than' CF column, the CF just greater than 15 is 16. The corresponding class is 15 - 20, which is the median class.

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

Here, $l=15$, $CF=14$, $f=2$, and $h=5$.

Median $= 15 + \left(\frac{15 - 14}{2}\right) \times 5 = 15 + \frac{1}{2} \times 5 = 15 + 2.5 = 17.5$.

The calculated median matches the value obtained from the graph.

Given:

A frequency distribution of the daily income of 50 factory workers.

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To Do:

1. Convert the distribution to a 'less than' type cumulative frequency distribution.

2. Draw the 'less than' ogive for this data.

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Solution:

Step 1: Create the 'less than' type cumulative frequency distribution.

We use the upper limits of the class intervals to define the 'less than' categories and calculate the cumulative frequencies by successively adding the frequencies.

Daily income (in $\textsf{₹}$)	Number of workers (Cumulative Frequency)	Coordinates for Ogive (Upper Limit, CF)
Less than 120	12	(120, 12)
Less than 140	$12 + 14 = 26$	(140, 26)
Less than 160	$26 + 8 = 34$	(160, 34)
Less than 180	$34 + 6 = 40$	(180, 40)
Less than 200	$40 + 10 = 50$	(200, 50)

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Step 2: Draw the 'less than' ogive.

To draw the ogive, we plot the coordinates derived in the table on a graph. The x-axis represents the Daily Income (in $\textsf{₹}$), and the y-axis represents the Number of Workers (Cumulative Frequency).

The points to be plotted are: (120, 12), (140, 26), (160, 34), (180, 40), (200, 50).

We also add the point (100, 0) corresponding to the lower limit of the first class interval, as there are no workers with income less than $\textsf{₹}100$.

By plotting these points and joining them with a smooth freehand curve, we obtain the 'less than' ogive.

Given:

A 'less than' type cumulative frequency distribution of the weights of 35 students.

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To Do:

1. Draw a 'less than' type ogive for the data.

2. Obtain the median weight from the graph.

3. Verify the result using the median formula.

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Solution:

Step 1: Draw the 'less than' ogive.

The data is already in the required format. The points for the ogive are the upper limits of weight and the corresponding cumulative frequencies.

Points to plot: (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35).

Plot these points on a graph with Weight (in kg) on the x-axis and Number of Students (CF) on the y-axis. Join them with a smooth curve to get the ogive.

Step 2: Obtain the median from the graph.

Total number of students, $N = 35$. The median position is $\frac{N}{2} = \frac{35}{2} = 17.5$.

On the graph, locate 17.5 on the y-axis. Draw a horizontal line from this point to intersect the ogive. From the intersection point, draw a perpendicular to the x-axis. The point where this perpendicular meets the x-axis is the median.

From the graph, this value is found to be 46.5 kg.

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Step 3: Verify the median using the formula.

First, we create a standard frequency distribution from the given cumulative data.

Weight (in kg)	Frequency ($f_i$)	Cumulative Frequency (CF)
Below 38	0	0
38 - 40	$3 - 0 = 3$	3
40 - 42	$5 - 3 = 2$	5
42 - 44	$9 - 5 = 4$	9
44 - 46	$14 - 9 = 5$	14
46 - 48	$28 - 14 = 14$	28
48 - 50	$32 - 28 = 4$	32
50 - 52	$35 - 32 = 3$	35

Since $\frac{N}{2} = 17.5$, the CF just greater than 17.5 is 28. The corresponding class, 46 - 48, is the median class.

Using the formula: Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

Here, $l=46$, $CF=14$, $f=14$, and $h=2$.

Median $= 46 + \left(\frac{17.5 - 14}{14}\right) \times 2 = 46 + \frac{3.5}{14} \times 2 = 46 + \frac{7}{14} \ $$ = 46 + 0.5 \ $$ = 46.5$.

The median weight calculated using the formula is 46.5 kg, which successfully verifies the result obtained from the graph.

Given:

A frequency distribution of the production yield for 100 farms.

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To Do:

1. Change the distribution to a 'more than' type distribution.

2. Draw its 'more than' ogive.

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Solution:

Step 1: Create the 'more than' type cumulative frequency distribution.

We use the lower limits of the class intervals and calculate the cumulative frequency by starting with the total frequency (100) and successively subtracting the frequencies of the preceding classes.

Production yield (in kg/ha)	Number of farms (Cumulative Frequency)	Coordinates for Ogive (Lower Limit, CF)
More than or equal to 50	100	(50, 100)
More than or equal to 55	$100 - 2 = 98$	(55, 98)
More than or equal to 60	$98 - 8 = 90$	(60, 90)
More than or equal to 65	$90 - 12 = 78$	(65, 78)
More than or equal to 70	$78 - 24 = 54$	(70, 54)
More than or equal to 75	$54 - 38 = 16$	(75, 16)

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Step 2: Draw the 'more than' ogive.

To draw the 'more than' ogive, we plot the coordinates derived above on a graph. The x-axis represents the Production Yield (in kg/ha), and the y-axis represents the Number of Farms (Cumulative Frequency).

The points to be plotted are: (50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16).

By plotting these points and joining them with a smooth freehand curve, we obtain the 'more than' ogive.