Chapter 13 Statistics

Given data set: $6, 7, 10, 12, 13, 4, 8, 12$.

We need to find the mean deviation about the mean for this data.

---

Step 1: Calculate the mean ($\overline{x}$) of the data.

The number of observations is $n = 8$.

The sum of the observations is $\sum x_i = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72$.

The mean is $\overline{x} = \frac{\sum x_i}{n}$.

From (i), $\overline{x} = 9$.

---

Step 2: Find the absolute deviation of each observation from the mean, $|x_i - \overline{x}|$.

We calculate $|x_i - 9|$ for each observation $x_i$.

Observation ($x_i$)	Mean ($\overline{x}$)	Deviation ($x_i - \overline{x}$)	Absolute Deviation ($|x_i - \overline{x}|$)
6	9	$6 - 9 = -3$	$|-3| = 3$
7	9	$7 - 9 = -2$	$|-2| = 2$
10	9	$10 - 9 = 1$	$|1| = 1$
12	9	$12 - 9 = 3$	$|3| = 3$
13	9	$13 - 9 = 4$	$|4| = 4$
4	9	$4 - 9 = -5$	$|-5| = 5$
8	9	$8 - 9 = -1$	$|-1| = 1$
12	9	$12 - 9 = 3$	$|3| = 3$

---

Step 3: Sum the absolute deviations, $\sum |x_i - \overline{x}|$.

$\sum |x_i - \overline{x}| = 3 + 2 + 1 + 3 + 4 + 5 + 1 + 3 = 22$.

---

Step 4: Calculate the Mean Deviation about the Mean.

The formula for Mean Deviation about the Mean is M.D.($\overline{x}$) = $\frac{\sum |x_i - \overline{x}|}{n}$.

From (ii), M.D.($\overline{x}$) = $\frac{11}{4} = 2.75$.

---

Thus, the mean deviation about the mean for the given data is $2.75$.

Given:

The data points are: 12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5.

---

To Find:

The mean deviation about the mean for the given data.

---

Solution:

First, we need to calculate the mean ($\bar{x}$) of the data.

The formula for the mean of raw data is $\bar{x} = \frac{\sum x_i}{n}$, where $\sum x_i$ is the sum of all observations and $n$ is the number of observations.

The number of observations is $n = 20$.

The sum of the observations is:

$\sum x_i = 12 + 3 + 18 + 17 + 4 + 9 + 17 + 19 + 20 + 15 + 8 + 17 + 2 + 3 + 16 + 11 + 3 + 1 + 0 + 5$

$\sum x_i = 200$

Now, calculate the mean:

$\bar{x} = \frac{200}{20} = 10$

---

Next, we find the absolute deviation of each observation from the mean, i.e., $|x_i - \bar{x}| = |x_i - 10|$ for each data point $x_i$.

The absolute deviations $|x_i - \bar{x}| = |x_i - 10|$ for each data point $x_i$ are:

$|12 - 10| = 2$

$|3 - 10| = 7$

$|18 - 10| = 8$

$|17 - 10| = 7$

$|4 - 10| = 6$

$|9 - 10| = 1$

$|17 - 10| = 7$

$|19 - 10| = 9$

$|20 - 10| = 10$

$|15 - 10| = 5$

$|8 - 10| = 2$

$|17 - 10| = 7$

$|2 - 10| = 8$

$|3 - 10| = 7$

$|16 - 10| = 6$

$|11 - 10| = 1$

$|3 - 10| = 7$

$|1 - 10| = 9$

$|0 - 10| = 10$

$|5 - 10| = 5$

We can list the observations and their absolute deviations in a table:

$x_i$	$|x_i - 10|$	$x_i$	$|x_i - 10|$
12	2	8	2
3	7	17	7
18	8	2	8
17	7	3	7
4	6	16	6
9	1	11	1
17	7	3	7
19	9	1	9
20	10	0	10
15	5	5	5

---

Now, we calculate the sum of the absolute deviations, $\sum |x_i - \bar{x}| = \sum |x_i - 10|$.

$\sum |x_i - 10| = 2 + 7 + 8 + 7 + 6 + 1 + 7 + 9 + 10 + 5 + 2 + 7 + 8 + 7 + 6 + 1 + 7 + 9 + 10 + 5$

$\sum |x_i - 10| = 124$

---

Finally, we calculate the mean deviation about the mean using the formula:

Mean Deviation ($\text{M.D.}(\bar{x})$) $= \frac{\sum |x_i - \bar{x}|}{n}$

Substitute the values of $\sum |x_i - \bar{x}|$ and $n$:

$\text{M.D.}(\bar{x}) = \frac{124}{20}$

Simplify the fraction:

$\text{M.D.}(\bar{x}) = \frac{\cancel{124}^{62}}{\cancel{20}_{10}} = \frac{62}{10} = 6.2$

---

Answer:

The mean deviation about the mean for the given data is $\mathbf{6.2}$.

Given data set: $3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21$.

We need to find the mean deviation about the median for this data.

---

Step 1: Arrange the data in ascending order.

Arranged data: $3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21$.

The number of observations is $n = 11$.

---

Step 2: Find the median (M) of the data.

Since $n$ is odd, the median is the $\left(\frac{n+1}{2}\right)^{\text{th}}$ observation.

Median (M) = $\left(\frac{11+1}{2}\right)^{\text{th}}$ observation = $\left(\frac{12}{2}\right)^{\text{th}}$ observation = $6^{\text{th}}$ observation.

The $6^{\text{th}}$ observation in the arranged data is $9$.

So, the median M = $9$.

---

Step 3: Find the absolute deviation of each observation from the median, $|x_i - M|$.

We calculate $|x_i - 9|$ for each observation $x_i$.

Observation ($x_i$)	Median (M)	Deviation ($x_i - M$)	Absolute Deviation ($|x_i - M|$)
3	9	$3 - 9 = -6$	$|-6| = 6$
3	9	$3 - 9 = -6$	$|-6| = 6$
4	9	$4 - 9 = -5$	$|-5| = 5$
5	9	$5 - 9 = -4$	$|-4| = 4$
7	9	$7 - 9 = -2$	$|-2| = 2$
9	9	$9 - 9 = 0$	$|0| = 0$
10	9	$10 - 9 = 1$	$|1| = 1$
12	9	$12 - 9 = 3$	$|3| = 3$
18	9	$18 - 9 = 9$	$|9| = 9$
19	9	$19 - 9 = 10$	$|10| = 10$
21	9	$21 - 9 = 12$	$|12| = 12$

---

Step 4: Sum the absolute deviations, $\sum |x_i - M|$.

$\sum |x_i - M| = 6 + 6 + 5 + 4 + 2 + 0 + 1 + 3 + 9 + 10 + 12 = 58$.

---

Step 5: Calculate the Mean Deviation about the Median.

The formula for Mean Deviation about the Median is M.D.(M) = $\frac{\sum |x_i - M|}{n}$.

From (i), M.D.(M) $\approx 5.27$ (approximately).

---

Thus, the mean deviation about the median for the given data is $\frac{58}{11}$ or approximately $5.27$.

Given data is a discrete frequency distribution with observations $x_i$ and corresponding frequencies $f_i$.

---

Step 1: Calculate the mean ($\overline{x}$) of the data.

The mean for a discrete frequency distribution is given by $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$.

We first calculate the product $f_i x_i$ for each observation and the total frequency $\sum f_i$.

$x_i$	$f_i$	$f_i x_i$
2	2	$2 \times 2 = 4$
5	8	$8 \times 5 = 40$
6	10	$10 \times 6 = 60$
8	7	$7 \times 8 = 56$
10	8	$8 \times 10 = 80$
12	5	$5 \times 12 = 60$
Total	$\sum f_i = 40$	$\sum f_i x_i = 300$

The mean is:

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{300}{40}$

$\overline{x} = 7.5$

---

Step 2: Find the absolute deviation of each observation from the mean, $|x_i - \overline{x}|$.

We calculate $|x_i - 7.5|$ for each $x_i$.

---

Step 3: Calculate $f_i |x_i - \overline{x}|$ for each observation.

We calculate the product of the frequency and the absolute deviation.

Step 4: Sum the products $\sum f_i |x_i - \overline{x}|$.

We organize the calculations in a table:

$x_i$	$f_i$	$|x_i - \overline{x}| = |x_i - 7.5|$	$f_i |x_i - \overline{x}|$
2	2	$|2 - 7.5| = 5.5$	$2 \times 5.5 = 11.0$
5	8	$|5 - 7.5| = 2.5$	$8 \times 2.5 = 20.0$
6	10	$|6 - 7.5| = 1.5$	$10 \times 1.5 = 15.0$
8	7	$|8 - 7.5| = 0.5$	$7 \times 0.5 = 3.5$
10	8	$|10 - 7.5| = 2.5$	$8 \times 2.5 = 20.0$
12	5	$|12 - 7.5| = 4.5$	$5 \times 4.5 = 22.5$
Total	$\sum f_i = 40$		$\sum f_i |x_i - \overline{x}| = 92.0$

---

Step 5: Calculate the Mean Deviation about the Mean.

The formula for Mean Deviation about the Mean for a discrete frequency distribution is M.D.($\overline{x}$) = $\frac{\sum f_i |x_i - \overline{x}|}{\sum f_i}$.

M.D.($\overline{x}$) = $\frac{92.0}{40}$

M.D.($\overline{x}$) = $\frac{92}{40} = \frac{23}{10} = 2.3$

---

Thus, the mean deviation about the mean for the given data is $2.3$.

Given data is a discrete frequency distribution with observations $x_i$ and corresponding frequencies $f_i$. The observations are already in ascending order.

---

Step 1: Calculate the total frequency ($\text{N} = \sum f_i$).

$\text{N} = 3 + 4 + 5 + 2 + 4 + 5 + 4 + 3 = 30$.

---

Step 2: Find the median (M) of the data.

Since $\text{N} = 30$ is even, the median is the average of the $\left(\frac{\text{N}}{2}\right)^{\text{th}}$ and $\left(\frac{\text{N}}{2} + 1\right)^{\text{th}}$ observations.

The positions are $\left(\frac{30}{2}\right)^{\text{th}} = 15^{\text{th}}$ and $\left(\frac{30}{2} + 1\right)^{\text{th}} = 16^{\text{th}}$.

We need to find the observations at the $15^{\text{th}}$ and $16^{\text{th}}$ positions using the cumulative frequencies.

$x_i$	$f_i$	Cumulative Frequency (c.f.)
3	3	3
6	4	$3 + 4 = 7$
9	5	$7 + 5 = 12$
12	2	$12 + 2 = 14$
13	4	$14 + 4 = 18$
15	5	$18 + 5 = 23$
21	4	$23 + 4 = 27$
22	3	$27 + 3 = 30$

The $15^{\text{th}}$ observation falls in the class with cumulative frequency $18$, which corresponds to $x_i = 13$.

The $16^{\text{th}}$ observation falls in the class with cumulative frequency $18$, which also corresponds to $x_i = 13$.

Median (M) = $\frac{15^{\text{th}} \text{ observation} + 16^{\text{th}} \text{ observation}}{2} = \frac{13 + 13}{2} = \frac{26}{2} = 13$.

So, the median M = $13$.

---

Step 3: Find the absolute deviation of each observation from the median, $|x_i - M|$.

We calculate $|x_i - 13|$ for each $x_i$.

---

Step 4: Calculate $f_i |x_i - M|$ for each observation.

We calculate the product of the frequency and the absolute deviation.

Step 5: Sum the products $\sum f_i |x_i - M|$.

We organize the calculations in a table:

$x_i$	$f_i$	Median (M)	$|x_i - M| = |x_i - 13|$	$f_i |x_i - M|$
3	3	13	$|3 - 13| = 10$	$3 \times 10 = 30$
6	4	13	$|6 - 13| = 7$	$4 \times 7 = 28$
9	5	13	$|9 - 13| = 4$	$5 \times 4 = 20$
12	2	13	$|12 - 13| = 1$	$2 \times 1 = 2$
13	4	13	$|13 - 13| = 0$	$4 \times 0 = 0$
15	5	13	$|15 - 13| = 2$	$5 \times 2 = 10$
21	4	13	$|21 - 13| = 8$	$4 \times 8 = 32$
22	3	13	$|22 - 13| = 9$	$3 \times 9 = 27$
Total	$\sum f_i = 30$			$\sum f_i |x_i - M| = 149$

---

Step 6: Calculate the Mean Deviation about the Median.

The formula for Mean Deviation about the Median for a discrete frequency distribution is M.D.(M) = $\frac{\sum f_i |x_i - M|}{\sum f_i}$.

M.D.(M) = $\frac{149}{30}$

M.D.(M) $\approx 4.97$ (approximately).

---

Thus, the mean deviation about the median for the given data is $\frac{149}{30}$ or approximately $4.97$.

Given data is a continuous frequency distribution with class intervals and corresponding frequencies $f_i$.

---

Step 1: Find the mid-point ($x_i$) of each class interval.

The mid-point is calculated as $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

---

Step 2: Calculate the mean ($\overline{x}$) of the data.

The mean for a continuous frequency distribution is given by $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$.

We first calculate the product $f_i x_i$ for each class and the total frequency $\sum f_i$.

Class Interval	Frequency ($f_i$)	Mid-point ($x_i$)	$f_i x_i$
10 - 20	2	$\frac{10+20}{2} = 15$	$2 \times 15 = 30$
20 - 30	3	$\frac{20+30}{2} = 25$	$3 \times 25 = 75$
30 - 40	8	$\frac{30+40}{2} = 35$	$8 \times 35 = 280$
40 - 50	14	$\frac{40+50}{2} = 45$	$14 \times 45 = 630$
50 - 60	8	$\frac{50+60}{2} = 55$	$8 \times 55 = 440$
60 - 70	3	$\frac{60+70}{2} = 65$	$3 \times 65 = 195$
70 - 80	2	$\frac{70+80}{2} = 75$	$2 \times 75 = 150$
Total	$\sum f_i = 40$		$\sum f_i x_i = 1800$

The mean is:

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1800}{40}$

$\overline{x} = 45$

---

Step 3: Find the absolute deviation of each mid-point from the mean, $|x_i - \overline{x}|$.

We calculate $|x_i - 45|$ for each $x_i$.

---

Step 4: Calculate $f_i |x_i - \overline{x}|$ for each class.

We calculate the product of the frequency and the absolute deviation.

Step 5: Sum the products $\sum f_i |x_i - \overline{x}|$.

We organize the calculations in a table:

Class Interval	$f_i$	$x_i$	$|x_i - \overline{x}| = |x_i - 45|$	$f_i |x_i - \overline{x}|$
10 - 20	2	15	$|15 - 45| = 30$	$2 \times 30 = 60$
20 - 30	3	25	$|25 - 45| = 20$	$3 \times 20 = 60$
30 - 40	8	35	$|35 - 45| = 10$	$8 \times 10 = 80$
40 - 50	14	45	$|45 - 45| = 0$	$14 \times 0 = 0$
50 - 60	8	55	$|55 - 45| = 10$	$8 \times 10 = 80$
60 - 70	3	65	$|65 - 45| = 20$	$3 \times 20 = 60$
70 - 80	2	75	$|75 - 45| = 30$	$2 \times 30 = 60$
Total	$\sum f_i = 40$			$\sum f_i |x_i - \overline{x}| = 400$

---

Step 6: Calculate the Mean Deviation about the Mean.

The formula for Mean Deviation about the Mean for a continuous frequency distribution is M.D.($\overline{x}$) = $\frac{\sum f_i |x_i - \overline{x}|}{\sum f_i}$.

M.D.($\overline{x}$) = $\frac{400}{40}$

M.D.($\overline{x}$) = $10$

---

Thus, the mean deviation about the mean for the given data is $10$.

Given data is a continuous frequency distribution with class intervals and corresponding frequencies $f_i$.

---

To Find: Mean deviation about the median.

---

Solution:

Step 1: Calculate the total frequency ($\text{N} = \sum f_i$).

We also need the cumulative frequency (c.f.) to find the median.

Class	Frequency ($f_i$)	Cumulative Frequency (c.f.)
0 - 10	6	6
10 - 20	7	$6 + 7 = 13$
20 - 30	15	$13 + 15 = 28$
30 - 40	16	$28 + 16 = 44$
40 - 50	4	$44 + 4 = 48$
50 - 60	2	$48 + 2 = 50$
Total	$\text{N} = \sum f_i = 50$

The total frequency is $\text{N} = 50$.

---

Step 2: Find the median (M) of the data.

We need to find the class that contains the $\left(\frac{\text{N}}{2}\right)^{\text{th}}$ observation.

$\frac{\text{N}}{2} = \frac{50}{2} = 25$.

Looking at the cumulative frequency column, the $25^{\text{th}}$ observation lies in the class 20 - 30 (since its cumulative frequency is 28, which is the first c.f. greater than or equal to 25).

This is the median class.

For the median class (20 - 30):

Lower boundary ($L$) = 20

Frequency of the median class ($f$) = 15

Cumulative frequency of the class preceding the median class ($c.f.$) = 13 (c.f. of 10 - 20)

Class size ($h$) = $10 - 0 = 10$

The formula for the median of a continuous frequency distribution is:

$M = L + \frac{\frac{N}{2} - c.f.}{f} \times h$

$M = 20 + \frac{25 - 13}{15} \times 10$

$M = 20 + \frac{12}{15} \times 10$

$M = 20 + \frac{4}{5} \times 10$

$M = 20 + 4 \times 2$

$M = 20 + 8$

$M = 28$

The median is M = $28$.

---

Step 3: Find the mid-point ($x_i$) of each class interval and the absolute deviation from the median, $|x_i - M|$.

Step 4: Calculate $f_i |x_i - M|$ for each class and sum them ($\sum f_i |x_i - M|$).

We organize the calculations in a table:

Class	$f_i$	Mid-point ($x_i$)	$|x_i - M| = |x_i - 28|$	$f_i |x_i - M|$
0 - 10	6	$\frac{0+10}{2} = 5$	$|5 - 28| = |-23| = 23$	$6 \times 23 = 138$
10 - 20	7	$\frac{10+20}{2} = 15$	$|15 - 28| = |-13| = 13$	$7 \times 13 = 91$
20 - 30	15	$\frac{20+30}{2} = 25$	$|25 - 28| = |-3| = 3$	$15 \times 3 = 45$
30 - 40	16	$\frac{30+40}{2} = 35$	$|35 - 28| = |7| = 7$	$16 \times 7 = 112$
40 - 50	4	$\frac{40+50}{2} = 45$	$|45 - 28| = |17| = 17$	$4 \times 17 = 68$
50 - 60	2	$\frac{50+60}{2} = 55$	$|55 - 28| = |27| = 27$	$2 \times 27 = 54$
Total	$\sum f_i = 50$			$\sum f_i |x_i - M| = 508$

---

Step 5: Calculate the Mean Deviation about the Median.

The formula for Mean Deviation about the Median for a continuous frequency distribution is M.D.(M) = $\frac{\sum f_i |x_i - M|}{\sum f_i}$.

M.D.(M) = $\frac{508}{50}$

M.D.(M) = $10.16$

---

Thus, the mean deviation about the median for the given data is $10.16$.

Given data set: $4, 7, 8, 9, 10, 12, 13, 17$.

We need to find the mean deviation about the mean for this data.

---

Step 1: Calculate the mean ($\overline{x}$) of the data.

The number of observations is $n = 8$.

The sum of the observations is $\sum x_i = 4 + 7 + 8 + 9 + 10 + 12 + 13 + 17 = 80$.

The mean is $\overline{x} = \frac{\sum x_i}{n}$.

From (i), $\overline{x} = 10$.

---

Step 2: Find the absolute deviation of each observation from the mean, $|x_i - \overline{x}|$.

We calculate $|x_i - 10|$ for each observation $x_i$.

Observation ($x_i$)	Mean ($\overline{x}$)	Deviation ($x_i - \overline{x}$)	Absolute Deviation ($|x_i - \overline{x}|$)
4	10	$4 - 10 = -6$	$|-6| = 6$
7	10	$7 - 10 = -3$	$|-3| = 3$
8	10	$8 - 10 = -2$	$|-2| = 2$
9	10	$9 - 10 = -1$	$|-1| = 1$
10	10	$10 - 10 = 0$	$|0| = 0$
12	10	$12 - 10 = 2$	$|2| = 2$
13	10	$13 - 10 = 3$	$|3| = 3$
17	10	$17 - 10 = 7$	$|7| = 7$

---

Step 3: Sum the absolute deviations, $\sum |x_i - \overline{x}|$.

$\sum |x_i - \overline{x}| = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24$.

---

Step 4: Calculate the Mean Deviation about the Mean.

The formula for Mean Deviation about the Mean is M.D.($\overline{x}$) = $\frac{\sum |x_i - \overline{x}|}{n}$.

From (ii), M.D.($\overline{x}$) = $3$.

---

Thus, the mean deviation about the mean for the given data is $3$.

Given data set: $38, 70, 48, 40, 42, 55, 63, 46, 54, 44$.

We need to find the mean deviation about the mean for this data.

---

Step 1: Calculate the mean ($\overline{x}$) of the data.

The number of observations is $n = 10$.

The sum of the observations is $\sum x_i = 38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44 = 500$.

The mean is $\overline{x} = \frac{\sum x_i}{n}$.

From (i), $\overline{x} = 50$.

---

Step 2: Find the absolute deviation of each observation from the mean, $|x_i - \overline{x}|$.

We calculate $|x_i - 50|$ for each observation $x_i$.

Observation ($x_i$)	Mean ($\overline{x}$)	Deviation ($x_i - \overline{x}$)	Absolute Deviation ($|x_i - \overline{x}|$)
38	50	$38 - 50 = -12$	$|-12| = 12$
70	50	$70 - 50 = 20$	$|20| = 20$
48	50	$48 - 50 = -2$	$|-2| = 2$
40	50	$40 - 50 = -10$	$|-10| = 10$
42	50	$42 - 50 = -8$	$|-8| = 8$
55	50	$55 - 50 = 5$	$|5| = 5$
63	50	$63 - 50 = 13$	$|13| = 13$
46	50	$46 - 50 = -4$	$|-4| = 4$
54	50	$54 - 50 = 4$	$|4| = 4$
44	50	$44 - 50 = -6$	$|-6| = 6$

---

Step 3: Sum the absolute deviations, $\sum |x_i - \overline{x}|$.

$\sum |x_i - \overline{x}| = 12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6 = 84$.

---

Step 4: Calculate the Mean Deviation about the Mean.

The formula for Mean Deviation about the Mean is M.D.($\overline{x}$) = $\frac{\sum |x_i - \overline{x}|}{n}$.

From (ii), M.D.($\overline{x}$) = $8.4$.

---

Thus, the mean deviation about the mean for the given data is $8.4$.

Given data set: $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$.

We need to find the mean deviation about the median for this data.

---

Step 1: Arrange the data in ascending order.

Arranged data: $10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18$.

The number of observations is $n = 12$.

---

Step 2: Find the median (M) of the data.

Since $n = 12$ is even, the median is the average of the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ observations.

The positions are $\left(\frac{12}{2}\right)^{\text{th}} = 6^{\text{th}}$ and $\left(\frac{12}{2} + 1\right)^{\text{th}} = 7^{\text{th}}$.

The $6^{\text{th}}$ observation in the arranged data is $13$.

The $7^{\text{th}}$ observation in the arranged data is $14$.

Median (M) = $\frac{6^{\text{th}} \text{ observation} + 7^{\text{th}} \text{ observation}}{2} = \frac{13 + 14}{2} = \frac{27}{2}$.

So, the median M = $13.5$.

---

Step 3: Find the absolute deviation of each observation from the median, $|x_i - M|$.

We calculate $|x_i - 13.5|$ for each observation $x_i$.

Observation ($x_i$)	Median (M)	Deviation ($x_i - M$)	Absolute Deviation ($|x_i - M|$)
10	13.5	$10 - 13.5 = -3.5$	$|-3.5| = 3.5$
11	13.5	$11 - 13.5 = -2.5$	$|-2.5| = 2.5$
11	13.5	$11 - 13.5 = -2.5$	$|-2.5| = 2.5$
12	13.5	$12 - 13.5 = -1.5$	$|-1.5| = 1.5$
13	13.5	$13 - 13.5 = -0.5$	$|-0.5| = 0.5$
13	13.5	$13 - 13.5 = -0.5$	$|-0.5| = 0.5$
14	13.5	$14 - 13.5 = 0.5$	$|0.5| = 0.5$
16	13.5	$16 - 13.5 = 2.5$	$|2.5| = 2.5$
16	13.5	$16 - 13.5 = 2.5$	$|2.5| = 2.5$
17	13.5	$17 - 13.5 = 3.5$	$|3.5| = 3.5$
17	13.5	$17 - 13.5 = 3.5$	$|3.5| = 3.5$
18	13.5	$18 - 13.5 = 4.5$	$|4.5| = 4.5$

---

Step 4: Sum the absolute deviations, $\sum |x_i - M|$.

$\sum |x_i - M| = 3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5 = 28$.

---

Step 5: Calculate the Mean Deviation about the Median.

The formula for Mean Deviation about the Median is M.D.(M) = $\frac{\sum |x_i - M|}{n}$.

From (i), M.D.(M) = $\frac{7}{3} \approx 2.33$ (approximately).

---

Thus, the mean deviation about the median for the given data is $\frac{28}{12}$ or $\frac{7}{3}$ or approximately $2.33$.

Given data set: $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$.

We need to find the mean deviation about the median for this data.

---

Step 1: Arrange the data in ascending order.

Arranged data: $36, 42, 45, 46, 46, 49, 51, 53, 60, 72$.

The number of observations is $n = 10$.

---

Step 2: Find the median (M) of the data.

Since $n = 10$ is even, the median is the average of the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ observations.

The positions are $\left(\frac{10}{2}\right)^{\text{th}} = 5^{\text{th}}$ and $\left(\frac{10}{2} + 1\right)^{\text{th}} = 6^{\text{th}}$.

The $5^{\text{th}}$ observation in the arranged data is $46$.

The $6^{\text{th}}$ observation in the arranged data is $49$.

Median (M) = $\frac{5^{\text{th}} \text{ observation} + 6^{\text{th}} \text{ observation}}{2} = \frac{46 + 49}{2} = \frac{95}{2}$.

So, the median M = $47.5$.

---

Step 3: Find the absolute deviation of each observation from the median, $|x_i - M|$.

We calculate $|x_i - 47.5|$ for each observation $x_i$.

Observation ($x_i$)	Median (M)	Deviation ($x_i - M$)	Absolute Deviation ($|x_i - M|$)
36	47.5	$36 - 47.5 = -11.5$	$|-11.5| = 11.5$
42	47.5	$42 - 47.5 = -5.5$	$|-5.5| = 5.5$
45	47.5	$45 - 47.5 = -2.5$	$|-2.5| = 2.5$
46	47.5	$46 - 47.5 = -1.5$	$|-1.5| = 1.5$
46	47.5	$46 - 47.5 = -1.5$	$|-1.5| = 1.5$
49	47.5	$49 - 47.5 = 1.5$	$|1.5| = 1.5$
51	47.5	$51 - 47.5 = 3.5$	$|3.5| = 3.5$
53	47.5	$53 - 47.5 = 5.5$	$|5.5| = 5.5$
60	47.5	$60 - 47.5 = 12.5$	$|12.5| = 12.5$
72	47.5	$72 - 47.5 = 24.5$	$|24.5| = 24.5$

---

Step 4: Sum the absolute deviations, $\sum |x_i - M|$.

$\sum |x_i - M| = 11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5 = 70$.

---

Step 5: Calculate the Mean Deviation about the Median.

The formula for Mean Deviation about the Median is M.D.(M) = $\frac{\sum |x_i - M|}{n}$.

From (i), M.D.(M) = $7$.

---

Thus, the mean deviation about the median for the given data is $7$.

Given data is a discrete frequency distribution with observations $x_i$ and corresponding frequencies $f_i$.

---

Step 1: Calculate the mean ($\overline{x}$) of the data.

The mean for a discrete frequency distribution is given by $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$.

We first calculate the product $f_i x_i$ for each observation and the total frequency $\sum f_i$.

$x_i$	$f_i$	$f_i x_i$
5	7	$7 \times 5 = 35$
10	4	$4 \times 10 = 40$
15	6	$6 \times 15 = 90$
20	3	$3 \times 20 = 60$
25	5	$5 \times 25 = 125$
Total	$\sum f_i = 25$	$\sum f_i x_i = 350$

The mean is:

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{350}{25}$

$\overline{x} = 14$

---

Step 2: Find the absolute deviation of each observation from the mean, $|x_i - \overline{x}|$.

We calculate $|x_i - 14|$ for each $x_i$.

---

Step 3: Calculate $f_i |x_i - \overline{x}|$ for each observation.

We calculate the product of the frequency and the absolute deviation.

Step 4: Sum the products $\sum f_i |x_i - \overline{x}|$.

We organize the calculations in a table:

$x_i$	$f_i$	$|x_i - \overline{x}| = |x_i - 14|$	$f_i |x_i - \overline{x}|$
5	7	$|5 - 14| = 9$	$7 \times 9 = 63$
10	4	$|10 - 14| = 4$	$4 \times 4 = 16$
15	6	$|15 - 14| = 1$	$6 \times 1 = 6$
20	3	$|20 - 14| = 6$	$3 \times 6 = 18$
25	5	$|25 - 14| = 11$	$5 \times 11 = 55$
Total	$\sum f_i = 25$		$\sum f_i |x_i - \overline{x}| = 158$

---

Step 5: Calculate the Mean Deviation about the Mean.

The formula for Mean Deviation about the Mean for a discrete frequency distribution is M.D.($\overline{x}$) = $\frac{\sum f_i |x_i - \overline{x}|}{\sum f_i}$.

M.D.($\overline{x}$) = $\frac{158}{25}$

M.D.($\overline{x}$) = $6.32$

---

Thus, the mean deviation about the mean for the given data is $6.32$.

Given data is a discrete frequency distribution with observations $x_i$ and corresponding frequencies $f_i$.

---

Step 1: Calculate the mean ($\overline{x}$) of the data.

The mean for a discrete frequency distribution is given by $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$.

We first calculate the product $f_i x_i$ for each observation and the total frequency $\sum f_i$.

$x_i$	$f_i$	$f_i x_i$
10	4	$4 \times 10 = 40$
30	24	$24 \times 30 = 720$
50	28	$28 \times 50 = 1400$
70	16	$16 \times 70 = 1120$
90	8	$8 \times 90 = 720$
Total	$\sum f_i = 80$	$\sum f_i x_i = 4000$

The mean is:

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4000}{80}$

$\overline{x} = 50$

---

Step 2: Find the absolute deviation of each observation from the mean, $|x_i - \overline{x}|$.

We calculate $|x_i - 50|$ for each $x_i$.

---

Step 3: Calculate $f_i |x_i - \overline{x}|$ for each observation.

We calculate the product of the frequency and the absolute deviation.

Step 4: Sum the products $\sum f_i |x_i - \overline{x}|$.

We organize the calculations in a table:

$x_i$	$f_i$	$|x_i - \overline{x}| = |x_i - 50|$	$f_i |x_i - \overline{x}|$
10	4	$|10 - 50| = 40$	$4 \times 40 = 160$
30	24	$|30 - 50| = 20$	$24 \times 20 = 480$
50	28	$|50 - 50| = 0$	$28 \times 0 = 0$
70	16	$|70 - 50| = 20$	$16 \times 20 = 320$
90	8	$|90 - 50| = 40$	$8 \times 40 = 320$
Total	$\sum f_i = 80$		$\sum f_i |x_i - \overline{x}| = 1280$

---

Step 5: Calculate the Mean Deviation about the Mean.

The formula for Mean Deviation about the Mean for a discrete frequency distribution is M.D.($\overline{x}$) = $\frac{\sum f_i |x_i - \overline{x}|}{\sum f_i}$.

M.D.($\overline{x}$) = $\frac{1280}{80}$

M.D.($\overline{x}$) = $16$

---

Thus, the mean deviation about the mean for the given data is $16$.

Given data is a discrete frequency distribution with observations $x_i$ and corresponding frequencies $f_i$. The observations are already in ascending order.

---

Step 1: Calculate the total frequency ($\text{N} = \sum f_i$) and Cumulative Frequency (c.f.).

$x_i$	$f_i$	Cumulative Frequency (c.f.)
5	8	8
7	6	$8 + 6 = 14$
9	2	$14 + 2 = 16$
10	2	$16 + 2 = 18$
12	2	$18 + 2 = 20$
15	6	$20 + 6 = 26$
Total	$\text{N} = \sum f_i = 26$

The total frequency is $\text{N} = 26$.

---

Step 2: Find the median (M) of the data.

Since $\text{N} = 26$ is even, the median is the average of the $\left(\frac{\text{N}}{2}\right)^{\text{th}}$ and $\left(\frac{\text{N}}{2} + 1\right)^{\text{th}}$ observations.

The positions are $\left(\frac{26}{2}\right)^{\text{th}} = 13^{\text{th}}$ and $\left(\frac{26}{2} + 1\right)^{\text{th}} = 14^{\text{th}}$.

From the cumulative frequency column:

The $13^{\text{th}}$ observation falls in the row where c.f. is $14$, which corresponds to $x_i = 7$.

The $14^{\text{th}}$ observation falls in the row where c.f. is $14$, which also corresponds to $x_i = 7$.

Median (M) = $\frac{13^{\text{th}} \text{ observation} + 14^{\text{th}} \text{ observation}}{2} = \frac{7 + 7}{2} = \frac{14}{2} = 7$.

So, the median M = $7$.

---

Step 3: Find the absolute deviation of each observation from the median, $|x_i - M|$, and calculate $f_i |x_i - M|$.

Step 4: Sum the products $\sum f_i |x_i - M|$.

We organize the calculations in a table:

$x_i$	$f_i$	Median (M)	$|x_i - M| = |x_i - 7|$	$f_i |x_i - M|$
5	8	7	$|5 - 7| = |-2| = 2$	$8 \times 2 = 16$
7	6	7	$|7 - 7| = |0| = 0$	$6 \times 0 = 0$
9	2	7	$|9 - 7| = |2| = 2$	$2 \times 2 = 4$
10	2	7	$|10 - 7| = |3| = 3$	$2 \times 3 = 6$
12	2	7	$|12 - 7| = |5| = 5$	$2 \times 5 = 10$
15	6	7	$|15 - 7| = |8| = 8$	$6 \times 8 = 48$
Total	$\sum f_i = 26$			$\sum f_i |x_i - M| = 84$

---

Step 5: Calculate the Mean Deviation about the Median.

The formula for Mean Deviation about the Median for a discrete frequency distribution is M.D.(M) = $\frac{\sum f_i |x_i - M|}{\sum f_i}$.

M.D.(M) = $\frac{84}{26}$

M.D.(M) = $\frac{42}{13} \approx 3.23$ (approximately).

---

Thus, the mean deviation about the median for the given data is $\frac{84}{26}$ or $\frac{42}{13}$ or approximately $3.23$.

Given data is a discrete frequency distribution with observations $x_i$ and corresponding frequencies $f_i$. The observations are already in ascending order.

---

Step 1: Calculate the total frequency ($\text{N} = \sum f_i$) and Cumulative Frequency (c.f.).

$x_i$	$f_i$	Cumulative Frequency (c.f.)
15	3	3
21	5	$3 + 5 = 8$
27	6	$8 + 6 = 14$
30	7	$14 + 7 = 21$
35	8	$21 + 8 = 29$
Total	$\text{N} = \sum f_i = 29$

The total frequency is $\text{N} = 29$.

---

Step 2: Find the median (M) of the data.

Since $\text{N} = 29$ is odd, the median is the $\left(\frac{\text{N}+1}{2}\right)^{\text{th}}$ observation.

The median position is $\left(\frac{29+1}{2}\right)^{\text{th}} = \left(\frac{30}{2}\right)^{\text{th}} = 15^{\text{th}}$ observation.

From the cumulative frequency column, the $15^{\text{th}}$ observation falls in the row where c.f. is $21$ (since 21 is the first c.f. greater than or equal to 15).

This corresponds to $x_i = 30$.

So, the median M = $30$.

---

Step 3: Find the absolute deviation of each observation from the median, $|x_i - M|$, and calculate $f_i |x_i - M|$.

Step 4: Sum the products $\sum f_i |x_i - M|$.

We organize the calculations in a table:

$x_i$	$f_i$	Median (M)	$|x_i - M| = |x_i - 30|$	$f_i |x_i - M|$
15	3	30	$|15 - 30| = |-15| = 15$	$3 \times 15 = 45$
21	5	30	$|21 - 30| = |-9| = 9$	$5 \times 9 = 45$
27	6	30	$|27 - 30| = |-3| = 3$	$6 \times 3 = 18$
30	7	30	$|30 - 30| = |0| = 0$	$7 \times 0 = 0$
35	8	30	$|35 - 30| = |5| = 5$	$8 \times 5 = 40$
Total	$\sum f_i = 29$			$\sum f_i |x_i - M| = 148$

---

Step 5: Calculate the Mean Deviation about the Median.

The formula for Mean Deviation about the Median for a discrete frequency distribution is M.D.(M) = $\frac{\sum f_i |x_i - M|}{\sum f_i}$.

M.D.(M) = $\frac{148}{29}$

M.D.(M) $\approx 5.10$ (approximately).

---

Thus, the mean deviation about the median for the given data is $\frac{148}{29}$ or approximately $5.10$.

Given data is a continuous frequency distribution with class intervals (Income per day in $\textsf{₹}$) and corresponding frequencies $f_i$ (Number of persons).

---

To Find: Mean deviation about the mean.

---

Solution:

Step 1: Find the mid-point ($x_i$) of each class interval.

The mid-point is calculated as $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

---

Step 2: Calculate the mean ($\overline{x}$) of the data.

The mean for a continuous frequency distribution is given by $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$.

We first calculate the product $f_i x_i$ for each class and the total frequency $\sum f_i$.

Class Interval (Income)	Frequency ($f_i$)	Mid-point ($x_i$)	$f_i x_i$
0 - 100	4	$\frac{0+100}{2} = 50$	$4 \times 50 = 200$
100 - 200	8	$\frac{100+200}{2} = 150$	$8 \times 150 = 1200$
200 - 300	9	$\frac{200+300}{2} = 250$	$9 \times 250 = 2250$
300 - 400	10	$\frac{300+400}{2} = 350$	$10 \times 350 = 3500$
400 - 500	7	$\frac{400+500}{2} = 450$	$7 \times 450 = 3150$
500 - 600	5	$\frac{500+600}{2} = 550$	$5 \times 550 = 2750$
600 - 700	4	$\frac{600+700}{2} = 650$	$4 \times 650 = 2600$
700 - 800	3	$\frac{700+800}{2} = 750$	$3 \times 750 = 2250$
Total	$\sum f_i = 50$		$\sum f_i x_i = 17900$

The mean is:

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{17900}{50}$

$\overline{x} = 358$

---

Step 3: Find the absolute deviation of each mid-point from the mean, $|x_i - \overline{x}|$.

We calculate $|x_i - 358|$ for each $x_i$.

---

Step 4: Calculate $f_i |x_i - \overline{x}|$ for each class and sum them ($\sum f_i |x_i - \overline{x}|$).

We organize the calculations in a table:

Class Interval	$f_i$	$x_i$	$|x_i - \overline{x}| = |x_i - 358|$	$f_i |x_i - \overline{x}|$
0 - 100	4	50	$|50 - 358| = |-308| = 308$	$4 \times 308 = 1232$
100 - 200	8	150	$|150 - 358| = |-208| = 208$	$8 \times 208 = 1664$
200 - 300	9	250	$|250 - 358| = |-108| = 108$	$9 \times 108 = 972$
300 - 400	10	350	$|350 - 358| = |-8| = 8$	$10 \times 8 = 80$
400 - 500	7	450	$|450 - 358| = |92| = 92$	$7 \times 92 = 644$
500 - 600	5	550	$|550 - 358| = |192| = 192$	$5 \times 192 = 960$
600 - 700	4	650	$|650 - 358| = |292| = 292$	$4 \times 292 = 1168$
700 - 800	3	750	$|750 - 358| = |392| = 392$	$3 \times 392 = 1176$
Total	$\sum f_i = 50$			$\sum f_i |x_i - \overline{x}| = 7896$

---

Step 5: Calculate the Mean Deviation about the Mean.

The formula for Mean Deviation about the Mean for a continuous frequency distribution is M.D.($\overline{x}$) = $\frac{\sum f_i |x_i - \overline{x}|}{\sum f_i}$.

M.D.($\overline{x}$) = $\frac{7896}{50}$

M.D.($\overline{x}$) = $157.92$

Thus, the mean deviation about the mean for the given data is $\textsf{₹ } 157.92$.

Given data is a continuous frequency distribution with class intervals (Height in cms) and corresponding frequencies $f_i$ (Number of boys).

---

To Find: Mean deviation about the mean.

---

Solution:

Step 1: Find the mid-point ($x_i$) of each class interval.

The mid-point is calculated as $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

---

Step 2: Calculate the mean ($\overline{x}$) of the data.

The mean for a continuous frequency distribution is given by $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$.

We first calculate the product $f_i x_i$ for each class and the total frequency $\sum f_i$.

Class Interval (Height)	Frequency ($f_i$)	Mid-point ($x_i$)	$f_i x_i$
95 - 105	9	$\frac{95+105}{2} = 100$	$9 \times 100 = 900$
105 - 115	13	$\frac{105+115}{2} = 110$	$13 \times 110 = 1430$
115 - 125	26	$\frac{115+125}{2} = 120$	$26 \times 120 = 3120$
125 - 135	30	$\frac{125+135}{2} = 130$	$30 \times 130 = 3900$
135 - 145	12	$\frac{135+145}{2} = 140$	$12 \times 140 = 1680$
145 - 155	10	$\frac{145+155}{2} = 150$	$10 \times 150 = 1500$
Total	$\sum f_i = 100$		$\sum f_i x_i = 12530$

The mean is:

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{12530}{100}$

$\overline{x} = 125.3$

---

Step 3: Find the absolute deviation of each mid-point from the mean, $|x_i - \overline{x}|$.

We calculate $|x_i - 125.3|$ for each $x_i$.

---

Step 4: Calculate $f_i |x_i - \overline{x}|$ for each class and sum them ($\sum f_i |x_i - \overline{x}|$).

We organize the calculations in a table:

Class Interval	$f_i$	$x_i$	$|x_i - \overline{x}| = |x_i - 125.3|$	$f_i |x_i - \overline{x}|$
95 - 105	9	100	$|100 - 125.3| = |-25.3| = 25.3$	$9 \times 25.3 = 227.7$
105 - 115	13	110	$|110 - 125.3| = |-15.3| = 15.3$	$13 \times 15.3 = 198.9$
115 - 125	26	120	$|120 - 125.3| = |-5.3| = 5.3$	$26 \times 5.3 = 137.8$
125 - 135	30	130	$|130 - 125.3| = |4.7| = 4.7$	$30 \times 4.7 = 141.0$
135 - 145	12	140	$|140 - 125.3| = |14.7| = 14.7$	$12 \times 14.7 = 176.4$
145 - 155	10	150	$|150 - 125.3| = |24.7| = 24.7$	$10 \times 24.7 = 247.0$
Total	$\sum f_i = 100$			$\sum f_i |x_i - \overline{x}| = 1128.8$

---

Step 5: Calculate the Mean Deviation about the Mean.

The formula for Mean Deviation about the Mean for a continuous frequency distribution is M.D.($\overline{x}$) = $\frac{\sum f_i |x_i - \overline{x}|}{\sum f_i}$.

M.D.($\overline{x}$) = $\frac{1128.8}{100}$

M.D.($\overline{x}$) = $11.288$

Thus, the mean deviation about the mean for the given data is $11.288$ cms.

Given data is a continuous frequency distribution with class intervals (Marks) and corresponding frequencies $f_i$ (Number of Girls).

---

To Find: Mean deviation about the median.

---

Solution:

Step 1: Calculate the total frequency ($\text{N} = \sum f_i$) and Cumulative Frequency (c.f.).

Class Interval (Marks)	Frequency ($f_i$)	Cumulative Frequency (c.f.)
0 - 10	6	6
10 - 20	8	$6 + 8 = 14$
20 - 30	14	$14 + 14 = 28$
30 - 40	16	$28 + 16 = 44$
40 - 50	4	$44 + 4 = 48$
50 - 60	2	$48 + 2 = 50$
Total	$\text{N} = \sum f_i = 50$

The total frequency is $\text{N} = 50$.

---

Step 2: Find the median (M) of the data.

We need to find the class that contains the $\left(\frac{\text{N}}{2}\right)^{\text{th}}$ observation.

$\frac{\text{N}}{2} = \frac{50}{2} = 25$.

Looking at the cumulative frequency column, the $25^{\text{th}}$ observation lies in the class 20 - 30 (since its cumulative frequency is 28, which is the first c.f. greater than or equal to 25).

This is the median class.

For the median class (20 - 30):

Lower boundary ($L$) = 20

Frequency of the median class ($f$) = 14

Cumulative frequency of the class preceding the median class ($c.f.$) = 14 (c.f. of 10 - 20)

Class size ($h$) = $10 - 0 = 10$

The formula for the median of a continuous frequency distribution is:

$M = L + \frac{\frac{N}{2} - c.f.}{f} \times h$

$M = 20 + \frac{25 - 14}{14} \times 10$

$M = 20 + \frac{11}{14} \times 10$

$M = 20 + \frac{110}{14}$

$M = 20 + \frac{55}{7}$

$M = \frac{140}{7} + \frac{55}{7} = \frac{195}{7}$

So, the median M = $\frac{195}{7} \approx 27.86$.

---

Step 3: Find the mid-point ($x_i$) of each class interval and the absolute deviation from the median, $|x_i - M|$.

Step 4: Calculate $f_i |x_i - M|$ for each class and sum them ($\sum f_i |x_i - M|$).

We organize the calculations in a table:

Class Interval	$f_i$	Mid-point ($x_i$)	$|x_i - M| = |x_i - \frac{195}{7}|$	$f_i |x_i - M|$
0 - 10	6	5	$|5 - \frac{195}{7}| = |\frac{35 - 195}{7}| = |-\frac{160}{7}| = \frac{160}{7}$	$6 \times \frac{160}{7} = \frac{960}{7}$
10 - 20	8	15	$|15 - \frac{195}{7}| = |\frac{105 - 195}{7}| = |-\frac{90}{7}| = \frac{90}{7}$	$8 \times \frac{90}{7} = \frac{720}{7}$
20 - 30	14	25	$|25 - \frac{195}{7}| = |\frac{175 - 195}{7}| = |-\frac{20}{7}| = \frac{20}{7}$	$14 \times \frac{20}{7} = 2 \times 20 = 40$
30 - 40	16	35	$|35 - \frac{195}{7}| = |\frac{245 - 195}{7}| = |\frac{50}{7}| = \frac{50}{7}$	$16 \times \frac{50}{7} = \frac{800}{7}$
40 - 50	4	45	$|45 - \frac{195}{7}| = |\frac{315 - 195}{7}| = |\frac{120}{7}| = \frac{120}{7}$	$4 \times \frac{120}{7} = \frac{480}{7}$
50 - 60	2	55	$|55 - \frac{195}{7}| = |\frac{385 - 195}{7}| = |\frac{190}{7}| = \frac{190}{7}$	$2 \times \frac{190}{7} = \frac{380}{7}$
Total	$\sum f_i = 50$			$\sum f_i |x_i - M| = \frac{960+720+280+800+480+380}{7} = \frac{3620}{7}$

---

Step 5: Calculate the Mean Deviation about the Median.

The formula for Mean Deviation about the Median for a continuous frequency distribution is M.D.(M) = $\frac{\sum f_i |x_i - M|}{\sum f_i}$.

M.D.(M) = $\frac{\frac{3620}{7}}{50} = \frac{3620}{7 \times 50} = \frac{3620}{350}$

Cancel out the common factor 10:

M.D.(M) = $\frac{362}{35}$

M.D.(M) $\approx 10.34$ (approximately).

---

Thus, the mean deviation about the median for the given data is $\frac{362}{35}$ or approximately $10.34$.

Given data is a discrete frequency distribution. We need to convert it into a continuous frequency distribution as per the hint.

Subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class interval.

---

To Find: Mean deviation about the median age.

---

Solution:

Step 1: Convert to continuous frequency distribution and calculate Cumulative Frequency (c.f.).

Continuous Class Interval (Age)	Frequency ($f_i$)	Cumulative Frequency (c.f.)
15.5 - 20.5	5	5
20.5 - 25.5	6	$5 + 6 = 11$
25.5 - 30.5	12	$11 + 12 = 23$
30.5 - 35.5	14	$23 + 14 = 37$
35.5 - 40.5	26	$37 + 26 = 63$
40.5 - 45.5	12	$63 + 12 = 75$
45.5 - 50.5	16	$75 + 16 = 91$
50.5 - 55.5	9	$91 + 9 = 100$
Total	$\text{N} = \sum f_i = 100$

The total frequency is $\text{N} = 100$.

---

Step 2: Find the median (M) of the data.

We need to find the class that contains the $\left(\frac{\text{N}}{2}\right)^{\text{th}}$ observation.

$\frac{\text{N}}{2} = \frac{100}{2} = 50$.

Looking at the cumulative frequency column, the $50^{\text{th}}$ observation lies in the class 35.5 - 40.5 (since its cumulative frequency is 63, which is the first c.f. greater than or equal to 50).

This is the median class.

For the median class (35.5 - 40.5):

Lower boundary ($L$) = 35.5

Frequency of the median class ($f$) = 26

Cumulative frequency of the class preceding the median class ($c.f.$) = 37 (c.f. of 30.5 - 35.5)

Class size ($h$) = $20.5 - 15.5 = 5$

The formula for the median of a continuous frequency distribution is:

$M = L + \frac{\frac{N}{2} - c.f.}{f} \times h$

$M = 35.5 + \frac{50 - 37}{26} \times 5$

$M = 35.5 + \frac{13}{26} \times 5$

$M = 35.5 + \frac{1}{2} \times 5$

$M = 35.5 + 2.5$

$M = 38$

The median is M = $38$ years.

---

Step 3: Find the mid-point ($x_i$) of each continuous class interval and the absolute deviation from the median, $|x_i - M|$.

Step 4: Calculate $f_i |x_i - M|$ for each class and sum them ($\sum f_i |x_i - M|$).

We organize the calculations in a table:

Continuous Class Interval	$f_i$	Mid-point ($x_i$)	Median (M)	$|x_i - M| = |x_i - 38|$	$f_i |x_i - M|$
15.5 - 20.5	5	$\frac{15.5+20.5}{2} = 18$	38	$|18 - 38| = |-20| = 20$	$5 \times 20 = 100$
20.5 - 25.5	6	$\frac{20.5+25.5}{2} = 23$	38	$|23 - 38| = |-15| = 15$	$6 \times 15 = 90$
25.5 - 30.5	12	$\frac{25.5+30.5}{2} = 28$	38	$|28 - 38| = |-10| = 10$	$12 \times 10 = 120$
30.5 - 35.5	14	$\frac{30.5+35.5}{2} = 33$	38	$|33 - 38| = |-5| = 5$	$14 \times 5 = 70$
35.5 - 40.5	26	$\frac{35.5+40.5}{2} = 38$	38	$|38 - 38| = |0| = 0$	$26 \times 0 = 0$
40.5 - 45.5	12	$\frac{40.5+45.5}{2} = 43$	38	$|43 - 38| = |5| = 5$	$12 \times 5 = 60$
45.5 - 50.5	16	$\frac{45.5+50.5}{2} = 48$	38	$|48 - 38| = |10| = 10$	$16 \times 10 = 160$
50.5 - 55.5	9	$\frac{50.5+55.5}{2} = 53$	38	$|53 - 38| = |15| = 15$	$9 \times 15 = 135$
Total	$\sum f_i = 100$				$\sum f_i |x_i - M| = 735$

---

Step 5: Calculate the Mean Deviation about the Median.

The formula for Mean Deviation about the Median for a continuous frequency distribution is M.D.(M) = $\frac{\sum f_i |x_i - M|}{\sum f_i}$.

M.D.(M) = $\frac{735}{100}$

M.D.(M) = $7.35$

Thus, the mean deviation about the median age for the given data is $7.35$ years.

Given Data:

6, 8, 10, 12, 14, 16, 18, 20, 22, 24

---

To Find:

The variance of the given data.

---

Solution:

Let the given data points be $x_i$. The number of data points is $n = 10$.

First, calculate the mean ($\bar{x}$) of the data.

The mean is given by the formula $\bar{x} = \frac{\sum x_i}{n}$.

Sum of the data points:

$\sum x_i = 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 = 150$

Calculate the mean:

$\bar{x} = \frac{150}{10} = 15$

The mean is $15$.

---

Next, calculate the deviation ($x_i - \bar{x}$) and the squared deviation ($(x_i - \bar{x})^2$) for each data point. We can organize these calculations in a table:

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
6	$6 - 15 = -9$	$(-9)^2 = 81$
8	$8 - 15 = -7$	$(-7)^2 = 49$
10	$10 - 15 = -5$	$(-5)^2 = 25$
12	$12 - 15 = -3$	$(-3)^2 = 9$
14	$14 - 15 = -1$	$(-1)^2 = 1$
16	$16 - 15 = 1$	$1^2 = 1$
18	$18 - 15 = 3$	$3^2 = 9$
20	$20 - 15 = 5$	$5^2 = 25$
22	$22 - 15 = 7$	$7^2 = 49$
24	$24 - 15 = 9$	$9^2 = 81$
Total	$\sum (x_i - \bar{x}) = 0$	$\sum (x_i - \bar{x})^2 = 330$

The sum of squared deviations is $\sum (x_i - \bar{x})^2 = 330$.

---

Finally, calculate the variance ($\sigma^2$). For a finite population, the variance is given by the formula:

$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$

Substitute the values:

$\sigma^2 = \frac{330}{10}$

$\sigma^2 = 33$

The variance of the given data is $\mathbf{33}$.

Given:

The data is given in the form of values ($x_i$) and their corresponding frequencies ($f_i$).

$x_i$	4	8	11	17	20	24	32
$f_i$	3	5	9	5	4	3	1

---

To Find:

The variance ($\sigma^2$) and the standard deviation ($\sigma$) of the data.

---

Solution:

First, we need to calculate the mean ($\bar{x}$) of the grouped data. The formula for the mean is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

Let's prepare a table to calculate the necessary sums.

$x_i$	$f_i$	$f_i x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$	$f_i (x_i - \bar{x})^2$
4	3	$3 \times 4 = 12$	$4 - \bar{x}$	$(4 - \bar{x})^2$	$3 \times (4 - \bar{x})^2$
8	5	$5 \times 8 = 40$	$8 - \bar{x}$	$(8 - \bar{x})^2$	$5 \times (8 - \bar{x})^2$
11	9	$9 \times 11 = 99$	$11 - \bar{x}$	$(11 - \bar{x})^2$	$9 \times (11 - \bar{x})^2$
17	5	$5 \times 17 = 85$	$17 - \bar{x}$	$(17 - \bar{x})^2$	$5 \times (17 - \bar{x})^2$
20	4	$4 \times 20 = 80$	$20 - \bar{x}$	$(20 - \bar{x})^2$	$4 \times (20 - \bar{x})^2$
24	3	$3 \times 24 = 72$	$24 - \bar{x}$	$(24 - \bar{x})^2$	$3 \times (24 - \bar{x})^2$
32	1	$1 \times 32 = 32$	$32 - \bar{x}$	$(32 - \bar{x})^2$	$1 \times (32 - \bar{x})^2$
Total	$\sum f_i = 30$	$\sum f_i x_i = 420$			$\sum f_i (x_i - \bar{x})^2$

Calculate the mean:

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{420}{30} = 14$

Now complete the table with deviations from the mean ($\bar{x} = 14$).

$x_i$	$f_i$	$f_i x_i$	$x_i - 14$	$(x_i - 14)^2$	$f_i (x_i - 14)^2$
4	3	12	$4 - 14 = -10$	$(-10)^2 = 100$	$3 \times 100 = 300$
8	5	40	$8 - 14 = -6$	$(-6)^2 = 36$	$5 \times 36 = 180$
11	9	99	$11 - 14 = -3$	$(-3)^2 = 9$	$9 \times 9 = 81$
17	5	85	$17 - 14 = 3$	$3^2 = 9$	$5 \times 9 = 45$
20	4	80	$20 - 14 = 6$	$6^2 = 36$	$4 \times 36 = 144$
24	3	72	$24 - 14 = 10$	$10^2 = 100$	$3 \times 100 = 300$
32	1	32	$32 - 14 = 18$	$18^2 = 324$	$1 \times 324 = 324$
Total	30	420			$\sum f_i (x_i - 14)^2 = 300 + 180 + 81 + 45 + 144 + 300 + 324 = 1374$

---

Calculate the variance ($\sigma^2$). The formula for the variance of grouped data is $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$.

$\sigma^2 = \frac{1374}{30}$

$\sigma^2 = 45.8$

---

Calculate the standard deviation ($\sigma$). The standard deviation is the square root of the variance.

$\sigma = \sqrt{\sigma^2} = \sqrt{45.8}$

$\sigma \approx 6.76756...$

Rounding to two decimal places:

$\sigma \approx 6.77$

---

Answer:

The variance is $\mathbf{45.8}$ and the standard deviation is approximately $\mathbf{6.77}$.

Given Data:

The given data is a grouped frequency distribution:

Class	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100
Frequency ($f_i$)	3	7	12	15	8	3	2

---

To Find:

The mean, variance, and standard deviation for the given distribution.

---

Solution:

For grouped data, we first find the mid-point ($x_i$) of each class interval. The mid-point is the average of the lower and upper limits of the class.

We then calculate the mean, variance, and standard deviation using the formulas for grouped data.

Total frequency, $n = \sum f_i$.

Mean, $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

Variance, $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$.

Standard Deviation, $\sigma = \sqrt{\sigma^2}$.

Let's create a table to facilitate the calculations:

Class	Frequency ($f_i$)	Mid-point ($x_i$)	$f_i x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$	$f_i (x_i - \bar{x})^2$
30 - 40	3	$\frac{30+40}{2} = 35$	$3 \times 35 = 105$	$35 - 62 = -27$	$(-27)^2 = 729$	$3 \times 729 = 2187$
40 - 50	7	$\frac{40+50}{2} = 45$	$7 \times 45 = 315$	$45 - 62 = -17$	$(-17)^2 = 289$	$7 \times 289 = 2023$
50 - 60	12	$\frac{50+60}{2} = 55$	$12 \times 55 = 660$	$55 - 62 = -7$	$(-7)^2 = 49$	$12 \times 49 = 588$
60 - 70	15	$\frac{60+70}{2} = 65$	$15 \times 65 = 975$	$65 - 62 = 3$	$3^2 = 9$	$15 \times 9 = 135$
70 - 80	8	$\frac{70+80}{2} = 75$	$8 \times 75 = 600$	$75 - 62 = 13$	$13^2 = 169$	$8 \times 169 = 1352$
80 - 90	3	$\frac{80+90}{2} = 85$	$3 \times 85 = 255$	$85 - 62 = 23$	$23^2 = 529$	$3 \times 529 = 1587$
90 - 100	2	$\frac{90+100}{2} = 95$	$2 \times 95 = 190$	$95 - 62 = 33$	$33^2 = 1089$	$2 \times 1089 = 2178$
Total	$\sum f_i = 50$		$\sum f_i x_i = 3100$			$\sum f_i (x_i - \bar{x})^2 = 10050$

From the table:

Total frequency, $\sum f_i = 50$.

Sum of $f_i x_i$, $\sum f_i x_i = 3100$.

Sum of $f_i (x_i - \bar{x})^2$, $\sum f_i (x_i - \bar{x})^2 = 10050$.

---

Calculate the mean:

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3100}{50} = 62$

The mean is $\mathbf{62}$.

---

Calculate the variance:

$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} = \frac{10050}{50}$

$\sigma^2 = 201$

The variance is $\mathbf{201}$.

---

Calculate the standard deviation:

$\sigma = \sqrt{\sigma^2} = \sqrt{201}$

Using a calculator, $\sqrt{201} \approx 14.17744$

The standard deviation is approximately $\mathbf{14.18}$.

Given Data:

The given data is in the form of a frequency distribution:

$x_i$	3	8	13	18	23
$f_i$	7	10	15	10	6

---

To Find:

The standard deviation for the given data.

---

Solution:

Let $n = \sum f_i$ be the total number of observations.

First, we calculate the mean ($\bar{x}$) of the data using the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

Let's create a table to calculate $f_i x_i$ and the required terms for the variance and standard deviation:

$x_i$	$f_i$	$f_i x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$	$f_i (x_i - \bar{x})^2$
3	7	$7 \times 3 = 21$	$3 - \frac{307}{24} = -\frac{235}{24}$	$(-\frac{235}{24})^2 = \frac{55225}{576}$	$7 \times \frac{55225}{576} = \frac{386575}{576}$
8	10	$10 \times 8 = 80$	$8 - \frac{307}{24} = -\frac{115}{24}$	$(-\frac{115}{24})^2 = \frac{13225}{576}$	$10 \times \frac{13225}{576} = \frac{132250}{576}$
13	15	$15 \times 13 = 195$	$13 - \frac{307}{24} = \frac{5}{24}$	$(\frac{5}{24})^2 = \frac{25}{576}$	$15 \times \frac{25}{576} = \frac{375}{576}$
18	10	$10 \times 18 = 180$	$18 - \frac{307}{24} = \frac{125}{24}$	$(\frac{125}{24})^2 = \frac{15625}{576}$	$10 \times \frac{15625}{576} = \frac{156250}{576}$
23	6	$6 \times 23 = 138$	$23 - \frac{307}{24} = \frac{245}{24}$	$(\frac{245}{24})^2 = \frac{60025}{576}$	$6 \times \frac{60025}{576} = \frac{360150}{576}$
Total	$\sum f_i = 48$	$\sum f_i x_i = 614$			$\sum f_i (x_i - \bar{x})^2 = \frac{1035600}{576}$

From the table:

Total frequency, $\sum f_i = 48$.

Sum of $f_i x_i$, $\sum f_i x_i = 614$.

Sum of $f_i (x_i - \bar{x})^2$, $\sum f_i (x_i - \bar{x})^2 = \frac{1035600}{576} = \frac{21575}{576}$.

---

Calculate the mean:

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{614}{48} = \frac{307}{24}$

The mean is $\frac{307}{24} \approx 12.7917$.

---

Calculate the variance ($\sigma^2$) using the formula:

$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$

Substitute the values:

$\sigma^2 = \frac{21575/576}{48} = \frac{21575}{576 \times 48} = \frac{21575}{27648}$

Simplifying the fraction: $\sigma^2 = \frac{21575}{27648}$ (Dividing numerator and denominator by their greatest common divisor, which is 1 in this case). However, the sum of squared deviations was $\frac{1035600}{576}$. Let's check the total sum of squares of deviations:

$\sum f_i (x_i - \bar{x})^2 = \frac{386575 + 132250 + 375 + 156250 + 360150}{576} = \frac{1035600}{576}$

Variance $\sigma^2 = \frac{1035600/576}{48} = \frac{1035600}{576 \times 48} = \frac{1035600}{27648}$.

Simplifying this fraction by dividing numerator and denominator by $48$: $\frac{1035600 \div 48}{27648 \div 48} = \frac{21575}{576}$.

The variance is $\sigma^2 = \frac{21575}{576}$.

$\sigma^2 \approx 37.4566$

---

Calculate the standard deviation ($\sigma$), which is the square root of the variance.

$\sigma = \sqrt{\sigma^2} = \sqrt{\frac{21575}{576}}$

$\sigma = \frac{\sqrt{21575}}{\sqrt{576}} = \frac{\sqrt{21575}}{24}$

Using a calculator, $\sqrt{21575} \approx 146.88499...$

$\sigma \approx \frac{146.88499}{24} \approx 6.120187...$

The standard deviation is approximately $\mathbf{6.12}$.

Given Data:

The given data is a grouped frequency distribution:

Class	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100
Frequency ($f_i$)	3	7	12	15	8	3	2

---

To Find:

The mean, variance, and standard deviation for the given distribution.

---

Solution:

For grouped data, we first find the mid-point ($x_i$) of each class interval. The mid-point is the average of the lower and upper limits of the class.

We then calculate the mean, variance, and standard deviation using the formulas for grouped data.

Total frequency, $n = \sum f_i$.

Mean, $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

Variance, $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$.

Standard Deviation, $\sigma = \sqrt{\sigma^2}$.

Let's create a table to facilitate the calculations:

Class	Frequency ($f_i$)	Mid-point ($x_i$)	$f_i x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$	$f_i (x_i - \bar{x})^2$
30 - 40	3	$\frac{30+40}{2} = 35$	$3 \times 35 = 105$	$35 - 62 = -27$	$(-27)^2 = 729$	$3 \times 729 = 2187$
40 - 50	7	$\frac{40+50}{2} = 45$	$7 \times 45 = 315$	$45 - 62 = -17$	$(-17)^2 = 289$	$7 \times 289 = 2023$
50 - 60	12	$\frac{50+60}{2} = 55$	$12 \times 55 = 660$	$55 - 62 = -7$	$(-7)^2 = 49$	$12 \times 49 = 588$
60 - 70	15	$\frac{60+70}{2} = 65$	$15 \times 65 = 975$	$65 - 62 = 3$	$3^2 = 9$	$15 \times 9 = 135$
70 - 80	8	$\frac{70+80}{2} = 75$	$8 \times 75 = 600$	$75 - 62 = 13$	$13^2 = 169$	$8 \times 169 = 1352$
80 - 90	3	$\frac{80+90}{2} = 85$	$3 \times 85 = 255$	$85 - 62 = 23$	$23^2 = 529$	$3 \times 529 = 1587$
90 - 100	2	$\frac{90+100}{2} = 95$	$2 \times 95 = 190$	$95 - 62 = 33$	$33^2 = 1089$	$2 \times 1089 = 2178$
Total	$\sum f_i = 50$		$\sum f_i x_i = 3100$			$\sum f_i (x_i - \bar{x})^2 = 10050$

From the table:

Total frequency, $\sum f_i = 50$.

Sum of $f_i x_i$, $\sum f_i x_i = 3100$.

Sum of $f_i (x_i - \bar{x})^2$, $\sum f_i (x_i - \bar{x})^2 = 10050$.

---

Calculate the mean:

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3100}{50} = 62$

The mean is $\mathbf{62}$.

---

Calculate the variance:

$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} = \frac{10050}{50}$

$\sigma^2 = 201$

The variance is $\mathbf{201}$.

---

Calculate the standard deviation:

$\sigma = \sqrt{\sigma^2} = \sqrt{201}$

Using a calculator, $\sqrt{201} \approx 14.17744$

The standard deviation is approximately $\mathbf{14.18}$.

Given Data:

6, 7, 10, 12, 13, 4, 8, 12

---

To Find:

The mean and variance of the given data.

---

Solution:

Let the given data points be $x_i$. The number of data points is $n = 8$.

First, calculate the mean ($\bar{x}$) of the data.

The mean is given by the formula $\bar{x} = \frac{\sum x_i}{n}$.

Sum of the data points:

$\sum x_i = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72$

Calculate the mean:

$\bar{x} = \frac{72}{8} = 9$

The mean is $\mathbf{9}$.

---

Next, calculate the deviation ($x_i - \bar{x}$) and the squared deviation ($(x_i - \bar{x})^2$) for each data point. We can organize these calculations in a table:

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
6	$6 - 9 = -3$	$(-3)^2 = 9$
7	$7 - 9 = -2$	$(-2)^2 = 4$
10	$10 - 9 = 1$	$1^2 = 1$
12	$12 - 9 = 3$	$3^2 = 9$
13	$13 - 9 = 4$	$4^2 = 16$
4	$4 - 9 = -5$	$(-5)^2 = 25$
8	$8 - 9 = -1$	$(-1)^2 = 1$
12	$12 - 9 = 3$	$3^2 = 9$
Total	$\sum (x_i - \bar{x}) = 0$	$\sum (x_i - \bar{x})^2 = 74$

The sum of squared deviations is $\sum (x_i - \bar{x})^2 = 74$.

---

Finally, calculate the variance ($\sigma^2$). For a finite population, the variance is given by the formula:

$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$

Substitute the values:

$\sigma^2 = \frac{74}{8}$

$\sigma^2 = \frac{\cancel{74}^{37}}{\cancel{8}_{4}}$

$\sigma^2 = \frac{37}{4} = 9.25$

The variance of the given data is $\mathbf{9.25}$.

Given Data:

The first $n$ natural numbers: $1, 2, 3, \dots, n$.

---

To Find:

The mean and variance of the first $n$ natural numbers.

---

Solution:

Let the data points be $x_i = i$, for $i = 1, 2, \dots, n$. The number of data points is $n$.

First, calculate the mean ($\bar{x}$) of the data.

The sum of the first $n$ natural numbers is given by the formula $\sum\limits_{i=1}^n i = \frac{n(n+1)}{2}$.

The mean is $\bar{x} = \frac{\sum\limits_{i=1}^n x_i}{n} = \frac{\sum\limits_{i=1}^n i}{n}$.

Substitute the sum formula:

$\bar{x} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n(n+1)}{2n}$

$\bar{x} = \frac{\cancel{n}(n+1)}{2\cancel{n}}$

$\bar{x} = \frac{n+1}{2}$

The mean of the first $n$ natural numbers is $\mathbf{\frac{n+1}{2}}$.

---

Next, calculate the variance ($\sigma^2$). The formula for variance is $\sigma^2 = \frac{\sum\limits_{i=1}^n (x_i - \bar{x})^2}{n}$.

We need to calculate $\sum\limits_{i=1}^n (i - \frac{n+1}{2})^2$.

$\sum\limits_{i=1}^n (i - \frac{n+1}{2})^2 = \sum\limits_{i=1}^n \left(i^2 - 2i\left(\frac{n+1}{2}\right) + \left(\frac{n+1}{2}\right)^2\right)$

$= \sum\limits_{i=1}^n \left(i^2 - i(n+1) + \frac{(n+1)^2}{4}\right)$

$= \sum\limits_{i=1}^n i^2 - (n+1)\sum\limits_{i=1}^n i + \sum\limits_{i=1}^n \frac{(n+1)^2}{4}$

Using standard summation formulas ($\sum i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum i = \frac{n(n+1)}{2}$):

$= \frac{n(n+1)(2n+1)}{6} - (n+1)\left(\frac{n(n+1)}{2}\right) + n\left(\frac{(n+1)^2}{4}\right)$

$= n(n+1)\left[\frac{2n+1}{6} - \frac{n+1}{2} + \frac{n+1}{4}\right]$

Find a common denominator (12) for the terms inside the square brackets:

$= n(n+1)\left[\frac{2(2n+1)}{12} - \frac{6(n+1)}{12} + \frac{3(n+1)}{12}\right]$

$= n(n+1)\left[\frac{4n+2 - (6n+6) + (3n+3)}{12}\right]$

$= n(n+1)\left[\frac{4n+2 - 6n-6 + 3n+3}{12}\right]$

$= n(n+1)\left[\frac{(4n - 6n + 3n) + (2 - 6 + 3)}{12}\right]$

$= n(n+1)\left[\frac{n - 1}{12}\right]$

The sum of squared deviations is $\sum\limits_{i=1}^n (x_i - \bar{x})^2 = \frac{n(n+1)(n-1)}{12} = \frac{n(n^2-1)}{12}$.

Now, calculate the variance:

$\sigma^2 = \frac{\sum\limits_{i=1}^n (x_i - \bar{x})^2}{n} = \frac{\frac{n(n^2-1)}{12}}{n}$

$\sigma^2 = \frac{n(n^2-1)}{12n} = \frac{\cancel{n}(n^2-1)}{12\cancel{n}}$

$\sigma^2 = \frac{n^2-1}{12}$

The variance of the first $n$ natural numbers is $\mathbf{\frac{n^2-1}{12}}$.

Given Data:

The first 10 multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.

---

To Find:

The mean and variance of the given data.

---

Solution:

Let the given data points be $x_i$. The number of data points is $n = 10$.

First, calculate the mean ($\bar{x}$) of the data.

The mean is given by the formula $\bar{x} = \frac{\sum x_i}{n}$.

Sum of the data points:

$\sum x_i = 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30 = 165$

Calculate the mean:

$\bar{x} = \frac{165}{10} = 16.5$

The mean is $\mathbf{16.5}$.

---

Next, calculate the deviation ($x_i - \bar{x}$) and the squared deviation ($(x_i - \bar{x})^2$) for each data point. We can organize these calculations in a table:

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
3	$3 - 16.5 = -13.5$	$(-13.5)^2 = 182.25$
6	$6 - 16.5 = -10.5$	$(-10.5)^2 = 110.25$
9	$9 - 16.5 = -7.5$	$(-7.5)^2 = 56.25$
12	$12 - 16.5 = -4.5$	$(-4.5)^2 = 20.25$
15	$15 - 16.5 = -1.5$	$(-1.5)^2 = 2.25$
18	$18 - 16.5 = 1.5$	$(1.5)^2 = 2.25$
21	$21 - 16.5 = 4.5$	$(4.5)^2 = 20.25$
24	$24 - 16.5 = 7.5$	$(7.5)^2 = 56.25$
27	$27 - 16.5 = 10.5$	$(10.5)^2 = 110.25$
30	$30 - 16.5 = 13.5$	$(13.5)^2 = 182.25$
Total	$\sum (x_i - \bar{x}) = 0$	$\sum (x_i - \bar{x})^2 = 742.50$

The sum of squared deviations is $\sum (x_i - \bar{x})^2 = 742.50$.

---

Finally, calculate the variance ($\sigma^2$). For a finite population, the variance is given by the formula:

$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$

Substitute the values:

$\sigma^2 = \frac{742.50}{10}$

$\sigma^2 = 74.25$

The variance of the given data is $\mathbf{74.25}$.

Given:

The data is provided in the form of a frequency distribution table:

$x_i$	6	10	14	18	24	28	30
$f_i$	2	4	7	12	8	4	3

---

To Find:

The mean ($\bar{x}$) and the variance ($\sigma^2$) of the given data.

---

Solution:

First, we calculate the mean ($\bar{x}$) of the grouped data. The formula for the mean is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

Let's prepare a table to calculate the necessary sums for the mean and variance.

$x_i$	$f_i$	$f_i x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$	$f_i (x_i - \bar{x})^2$
6	2	$2 \times 6 = 12$	$6 - 19 = -13$	$(-13)^2 = 169$	$2 \times 169 = 338$
10	4	$4 \times 10 = 40$	$10 - 19 = -9$	$(-9)^2 = 81$	$4 \times 81 = 324$
14	7	$7 \times 14 = 98$	$14 - 19 = -5$	$(-5)^2 = 25$	$7 \times 25 = 175$
18	12	$12 \times 18 = 216$	$18 - 19 = -1$	$(-1)^2 = 1$	$12 \times 1 = 12$
24	8	$8 \times 24 = 192$	$24 - 19 = 5$	$5^2 = 25$	$8 \times 25 = 200$
28	4	$4 \times 28 = 112$	$28 - 19 = 9$	$9^2 = 81$	$4 \times 81 = 324$
30	3	$3 \times 30 = 90$	$30 - 19 = 11$	$11^2 = 121$	$3 \times 121 = 363$
Total	$\sum f_i = 40$	$\sum f_i x_i = 760$			$\sum f_i (x_i - \bar{x})^2 = 1736$

From the table:

Total number of observations, $\sum f_i = 40$.

Sum of the products of observations and frequencies, $\sum f_i x_i = 760$.

Sum of the weighted squared deviations from the mean, $\sum f_i (x_i - \bar{x})^2 = 1736$.

---

Calculate the mean ($\bar{x}$):

The mean of the data is $\mathbf{19}$.

---

Calculate the variance ($\sigma^2$) using the formula:

$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$

Substitute the values from the table ($\sum f_i (x_i - \bar{x})^2 = 1736$ and $\sum f_i = 40$):

The variance of the data is $\mathbf{43.4}$.

---

Answer:

The mean is $\mathbf{19}$ and the variance is $\mathbf{43.4}$.

Given Data:

The given data is in the form of a frequency distribution:

$x_i$	92	93	97	98	102	104	109
$f_i$	3	2	3	2	6	3	3

---

To Find:

The mean and variance of the given data.

---

Solution:

Let $n = \sum f_i$ be the total number of observations.

First, we calculate the mean ($\bar{x}$) of the data using the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

Let's create a table to calculate $f_i x_i$ and the required terms for variance:

$x_i$	$f_i$	$f_i x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$	$f_i (x_i - \bar{x})^2$
92	3	$3 \times 92 = 276$	$92 - 100 = -8$	$(-8)^2 = 64$	$3 \times 64 = 192$
93	2	$2 \times 93 = 186$	$93 - 100 = -7$	$(-7)^2 = 49$	$2 \times 49 = 98$
97	3	$3 \times 97 = 291$	$97 - 100 = -3$	$(-3)^2 = 9$	$3 \times 9 = 27$
98	2	$2 \times 98 = 196$	$98 - 100 = -2$	$(-2)^2 = 4$	$2 \times 4 = 8$
102	6	$6 \times 102 = 612$	$102 - 100 = 2$	$2^2 = 4$	$6 \times 4 = 24$
104	3	$3 \times 104 = 312$	$104 - 100 = 4$	$4^2 = 16$	$3 \times 16 = 48$
109	3	$3 \times 109 = 327$	$109 - 100 = 9$	$9^2 = 81$	$3 \times 81 = 243$
Total	$\sum f_i = 22$	$\sum f_i x_i = 2200$			$\sum f_i (x_i - \bar{x})^2 = 640$

From the table:

Total frequency, $\sum f_i = 22$.

Sum of $f_i x_i$, $\sum f_i x_i = 2200$.

Sum of $f_i (x_i - \bar{x})^2$, $\sum f_i (x_i - \bar{x})^2 = 640$.

---

Calculate the mean:

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2200}{22} = 100$

The mean is $\mathbf{100}$.

---

Calculate the variance ($\sigma^2$) using the formula:

$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$

Substitute the values:

$\sigma^2 = \frac{640}{22}$

$\sigma^2 = \frac{\cancel{640}^{320}}{\cancel{22}_{11}} = \frac{320}{11}$

The variance of the given data is $\mathbf{\frac{320}{11}}$.

As a decimal, $\sigma^2 \approx 29.0909...$

The variance is approximately $\mathbf{29.09}$.

Given Data:

The given data is in the form of a frequency distribution:

$x_i$	60	61	62	63	64	65	66	67	68
$f_i$	2	1	12	29	25	12	10	4	5

---

To Find:

The mean and standard deviation using the short-cut method.

---

Solution:

Let $n = \sum f_i$ be the total number of observations.

We use the short-cut method (Assumed Mean Method) to calculate the mean and variance.

Let the assumed mean be $A$. We choose $A=64$ (the value corresponding to the highest frequency).

Let $d_i = x_i - A = x_i - 64$ be the deviation of each observation from the assumed mean.

The formula for the mean using the short-cut method is $\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}$.

The formula for the variance using the short-cut method is $\sigma^2 = \frac{\sum f_i d_i^2}{\sum f_i} - \left(\frac{\sum f_i d_i}{\sum f_i}\right)^2$.

The standard deviation is $\sigma = \sqrt{\sigma^2}$.

Let's create a table to calculate the necessary values:

$x_i$	$f_i$	$d_i = x_i - 64$	$f_i d_i$	$d_i^2$	$f_i d_i^2$
60	2	-4	$2 \times (-4) = -8$	16	$2 \times 16 = 32$
61	1	-3	$1 \times (-3) = -3$	9	$1 \times 9 = 9$
62	12	-2	$12 \times (-2) = -24$	4	$12 \times 4 = 48$
63	29	-1	$29 \times (-1) = -29$	1	$29 \times 1 = 29$
64	25	0	$25 \times 0 = 0$	0	$25 \times 0 = 0$
65	12	1	$12 \times 1 = 12$	1	$12 \times 1 = 12$
66	10	2	$10 \times 2 = 20$	4	$10 \times 4 = 40$
67	4	3	$4 \times 3 = 12$	9	$4 \times 9 = 36$
68	5	4	$5 \times 4 = 20$	16	$5 \times 16 = 80$
Total	$\sum f_i = 100$		$\sum f_i d_i = 0$		$\sum f_i d_i^2 = 286$

From the table:

Total frequency, $\sum f_i = 100$.

Sum of $f_i d_i$, $\sum f_i d_i = 0$.

Sum of $f_i d_i^2$, $\sum f_i d_i^2 = 286$.

---

Calculate the mean:

$\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}$

$\bar{x} = 64 + \frac{0}{100}$

$\bar{x} = 64 + 0 = 64$

The mean is $\mathbf{64}$.

---

Calculate the variance:

$\sigma^2 = \frac{\sum f_i d_i^2}{\sum f_i} - \left(\frac{\sum f_i d_i}{\sum f_i}\right)^2$

$\sigma^2 = \frac{286}{100} - \left(\frac{0}{100}\right)^2$

$\sigma^2 = 2.86 - 0^2$

$\sigma^2 = 2.86$

The variance is $\mathbf{2.86}$.

---

Calculate the standard deviation:

$\sigma = \sqrt{\sigma^2} = \sqrt{2.86}$

Using a calculator, $\sqrt{2.86} \approx 1.69115$

The standard deviation is approximately $\mathbf{1.69}$.

Given Data:

The given data is a grouped frequency distribution:

Classes	0 - 30	30 - 60	60 - 90	90 - 120	120 - 150	150 - 180	180 - 210
Frequencies ($f_i$)	2	3	5	10	3	5	2

---

To Find:

The mean and variance for the given distribution using the short-cut method.

---

Solution:

Let $n = \sum f_i$ be the total number of observations.

First, we find the mid-point ($x_i$) of each class interval. The mid-point is the average of the lower and upper limits of the class.

We use the short-cut method (Assumed Mean Method) to calculate the mean and variance.

Let the assumed mean be $A$. We choose $A=105$ (the mid-point of the class 90 - 120, which has the highest frequency).

Let $d_i = x_i - A = x_i - 105$ be the deviation of each mid-point from the assumed mean.

The formula for the mean using the short-cut method is $\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}$.

The formula for the variance using the short-cut method is $\sigma^2 = \frac{\sum f_i d_i^2}{\sum f_i} - \left(\frac{\sum f_i d_i}{\sum f_i}\right)^2$.

Let's create a table to calculate the necessary values:

Classes	Frequency ($f_i$)	Mid-point ($x_i$)	$d_i = x_i - 105$	$f_i d_i$	$d_i^2$	$f_i d_i^2$
0 - 30	2	$\frac{0+30}{2} = 15$	$15 - 105 = -90$	$2 \times (-90) = -180$	$(-90)^2 = 8100$	$2 \times 8100 = 16200$
30 - 60	3	$\frac{30+60}{2} = 45$	$45 - 105 = -60$	$3 \times (-60) = -180$	$(-60)^2 = 3600$	$3 \times 3600 = 10800$
60 - 90	5	$\frac{60+90}{2} = 75$	$75 - 105 = -30$	$5 \times (-30) = -150$	$(-30)^2 = 900$	$5 \times 900 = 4500$
90 - 120	10	$\frac{90+120}{2} = 105$	$105 - 105 = 0$	$10 \times 0 = 0$	$0^2 = 0$	$10 \times 0 = 0$
120 - 150	3	$\frac{120+150}{2} = 135$	$135 - 105 = 30$	$3 \times 30 = 90$	$30^2 = 900$	$3 \times 900 = 2700$
150 - 180	5	$\frac{150+180}{2} = 165$	$165 - 105 = 60$	$5 \times 60 = 300$	$60^2 = 3600$	$5 \times 3600 = 18000$
180 - 210	2	$\frac{180+210}{2} = 195$	$195 - 105 = 90$	$2 \times 90 = 180$	$90^2 = 8100$	$2 \times 8100 = 16200$
Total	$\sum f_i = 30$			$\sum f_i d_i = 60$		$\sum f_i d_i^2 = 68400$

From the table:

Total frequency, $\sum f_i = 30$.

Sum of $f_i d_i$, $\sum f_i d_i = 60$.

Sum of $f_i d_i^2$, $\sum f_i d_i^2 = 68400$.

---

Calculate the mean:

$\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}$

$\bar{x} = 105 + \frac{60}{30}$

$\bar{x} = 105 + 2 = 107$

The mean is $\mathbf{107}$.

---

Calculate the variance:

$\sigma^2 = \frac{\sum f_i d_i^2}{\sum f_i} - \left(\frac{\sum f_i d_i}{\sum f_i}\right)^2$

$\sigma^2 = \frac{68400}{30} - \left(\frac{60}{30}\right)^2$

$\sigma^2 = 2280 - (2)^2$

$\sigma^2 = 2280 - 4 = 2276$

The variance is $\mathbf{2276}$.

Given Data:

The given data is a grouped frequency distribution:

Classes	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Frequencies ($f_i$)	5	8	15	16	6

---

To Find:

The mean and variance for the given distribution using the short-cut method.

---

Solution:

Let $n = \sum f_i$ be the total number of observations.

First, we find the mid-point ($x_i$) of each class interval. The mid-point is the average of the lower and upper limits of the class.

We use the short-cut method (Assumed Mean Method) to calculate the mean and variance.

Let the assumed mean be $A$. We choose $A=35$ (the mid-point of the class 30 - 40, which has the highest frequency).

Let $d_i = x_i - A = x_i - 35$ be the deviation of each mid-point from the assumed mean.

The formula for the mean using the short-cut method is $\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}$.

The formula for the variance using the short-cut method is $\sigma^2 = \frac{\sum f_i d_i^2}{\sum f_i} - \left(\frac{\sum f_i d_i}{\sum f_i}\right)^2$.

Let's create a table to calculate the necessary values:

Classes	Frequency ($f_i$)	Mid-point ($x_i$)	$d_i = x_i - 35$	$f_i d_i$	$d_i^2$	$f_i d_i^2$
0 - 10	5	$\frac{0+10}{2} = 5$	$5 - 35 = -30$	$5 \times (-30) = -150$	$(-30)^2 = 900$	$5 \times 900 = 4500$
10 - 20	8	$\frac{10+20}{2} = 15$	$15 - 35 = -20$	$8 \times (-20) = -160$	$(-20)^2 = 400$	$8 \times 400 = 3200$
20 - 30	15	$\frac{20+30}{2} = 25$	$25 - 35 = -10$	$15 \times (-10) = -150$	$(-10)^2 = 100$	$15 \times 100 = 1500$
30 - 40	16	$\frac{30+40}{2} = 35$	$35 - 35 = 0$	$16 \times 0 = 0$	$0^2 = 0$	$16 \times 0 = 0$
40 - 50	6	$\frac{40+50}{2} = 45$	$45 - 35 = 10$	$6 \times 10 = 60$	$10^2 = 100$	$6 \times 100 = 600$
Total	$\sum f_i = 50$			$\sum f_i d_i = -400$		$\sum f_i d_i^2 = 9800$

From the table:

Total frequency, $\sum f_i = 50$.

Sum of $f_i d_i$, $\sum f_i d_i = -400$.

Sum of $f_i d_i^2$, $\sum f_i d_i^2 = 9800$.

---

Calculate the mean:

$\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}$

$\bar{x} = 35 + \frac{-400}{50}$

$\bar{x} = 35 + \frac{-40}{5}$

$\bar{x} = 35 - 8 = 27$

The mean is $\mathbf{27}$.

---

Calculate the variance:

$\sigma^2 = \frac{\sum f_i d_i^2}{\sum f_i} - \left(\frac{\sum f_i d_i}{\sum f_i}\right)^2$

$\sigma^2 = \frac{9800}{50} - \left(\frac{-400}{50}\right)^2$

$\sigma^2 = \frac{980}{5} - (-8)^2$

$\sigma^2 = 196 - 64 = 132$

The variance is $\mathbf{132}$.

Given Data:

The given data is a grouped frequency distribution:

Height in cms (Classes)	70 - 75	75 - 80	80 - 85	85 - 90	90 - 95	95 - 100	100 - 105	105 - 110	110 - 115
No. of children (Frequency, $f_i$)	3	4	7	7	15	9	6	6	3

---

To Find:

The mean, variance, and standard deviation using the short-cut method.

---

Solution:

Let $n = \sum f_i$ be the total number of observations.

First, we find the mid-point ($x_i$) of each class interval. The mid-point is the average of the lower and upper limits of the class.

The class intervals have equal width, $h = 5$. We use the Step-Deviation Method, which is a short-cut method for grouped data with equal class intervals.

Let the assumed mean be $A$. We choose $A=92.5$ (the mid-point of the class 90 - 95, which has the highest frequency).

Let $u_i = \frac{x_i - A}{h} = \frac{x_i - 92.5}{5}$.

The formula for the mean using the step-deviation method is $\bar{x} = A + h \left(\frac{\sum f_i u_i}{\sum f_i}\right)$.

The formula for the variance using the step-deviation method is $\sigma^2 = h^2 \left[\frac{\sum f_i u_i^2}{\sum f_i} - \left(\frac{\sum f_i u_i}{\sum f_i}\right)^2\right]$.

The standard deviation is $\sigma = \sqrt{\sigma^2}$.

Let's create a table to calculate the necessary values:

Classes	Frequency ($f_i$)	Mid-point ($x_i$)	$u_i = \frac{x_i - 92.5}{5}$	$f_i u_i$	$u_i^2$	$f_i u_i^2$
70 - 75	3	72.5	-4	$3 \times (-4) = -12$	16	$3 \times 16 = 48$
75 - 80	4	77.5	-3	$4 \times (-3) = -12$	9	$4 \times 9 = 36$
80 - 85	7	82.5	-2	$7 \times (-2) = -14$	4	$7 \times 4 = 28$
85 - 90	7	87.5	-1	$7 \times (-1) = -7$	1	$7 \times 1 = 7$
90 - 95	15	92.5	0	$15 \times 0 = 0$	0	$15 \times 0 = 0$
95 - 100	9	97.5	1	$9 \times 1 = 9$	1	$9 \times 1 = 9$
100 - 105	6	102.5	2	$6 \times 2 = 12$	4	$6 \times 4 = 24$
105 - 110	6	107.5	3	$6 \times 3 = 18$	9	$6 \times 9 = 54$
110 - 115	3	112.5	4	$3 \times 4 = 12$	16	$3 \times 16 = 48$
Total	$\sum f_i = 60$			$\sum f_i u_i = 6$		$\sum f_i u_i^2 = 254$

From the table:

Total frequency, $\sum f_i = 60$.

Sum of $f_i u_i$, $\sum f_i u_i = 6$.

Sum of $f_i u_i^2$, $\sum f_i u_i^2 = 254$.

Assumed mean, $A = 92.5$.

Class width, $h = 5$.

---

Calculate the mean:

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

$\bar{x} = 92.5 + 5 \left(\frac{6}{60}\right)$

$\bar{x} = 92.5 + 5 \left(\frac{1}{10}\right)$

$\bar{x} = 92.5 + 0.5 = 93$

The mean is $\mathbf{93}$.

---

Calculate the variance:

$\sigma^2 = h^2 \left[\frac{\sum f_i u_i^2}{\sum f_i} - \left(\frac{\sum f_i u_i}{\sum f_i}\right)^2\right]$

$\sigma^2 = 5^2 \left[\frac{254}{60} - \left(\frac{6}{60}\right)^2\right]$

$\sigma^2 = 25 \left[\frac{254}{60} - \left(\frac{1}{10}\right)^2\right]$

$\sigma^2 = 25 \left[\frac{254}{60} - \frac{1}{100}\right]$

Find a common denominator for 60 and 100, which is 300.

$\sigma^2 = 25 \left[\frac{254 \times 5}{300} - \frac{1 \times 3}{300}\right]$

$\sigma^2 = 25 \left[\frac{1270 - 3}{300}\right]$

$\sigma^2 = 25 \left[\frac{1267}{300}\right]$

$\sigma^2 = \frac{25 \times 1267}{300} = \frac{\cancel{25} \times 1267}{\cancel{300}_{12}}$

$\sigma^2 = \frac{1267}{12}$

The variance is $\mathbf{\frac{1267}{12}}$.

As a decimal, $\sigma^2 \approx 105.5833$

The variance is approximately $\mathbf{105.58}$.

---

Calculate the standard deviation:

$\sigma = \sqrt{\sigma^2} = \sqrt{\frac{1267}{12}}$

Using a calculator, $\sqrt{\frac{1267}{12}} \approx \sqrt{105.5833} \approx 10.275$

The standard deviation is approximately $\mathbf{10.28}$.

Given Data:

The given data is a grouped frequency distribution with discontinuous classes:

Diameters (Classes)	33 - 36	37 - 40	41 - 44	45 - 48	49 - 52
No. of circles (Frequency, $f_i$)	15	17	21	22	25

---

To Find:

The mean and standard deviation of the diameters.

---

Solution:

First, we make the data continuous by adjusting the class boundaries. The adjustment factor is $\frac{37-36}{2} = 0.5$. We subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.

The new continuous classes are: 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5-48.5, 48.5-52.5.

Let $n = \sum f_i$ be the total number of circles.

We will use the Step-Deviation Method to calculate the mean and standard deviation.

First, find the mid-point ($x_i$) of each continuous class interval.

Let the assumed mean be $A$. We choose $A=50.5$ (the mid-point of the class 48.5 - 52.5, which has the highest frequency).

The class width is $h = 36.5 - 32.5 = 4$.

Let $u_i = \frac{x_i - A}{h} = \frac{x_i - 50.5}{4}$.

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

The formula for the variance is $\sigma^2 = h^2 \left[\frac{\sum f_i u_i^2}{\sum f_i} - \left(\frac{\sum f_i u_i}{\sum f_i}\right)^2\right]$.

The standard deviation is $\sigma = \sqrt{\sigma^2}$.

Let's create a table to calculate the necessary values:

Continuous Classes	Frequency ($f_i$)	Mid-point ($x_i$)	$u_i = \frac{x_i - 50.5}{4}$	$f_i u_i$	$u_i^2$	$f_i u_i^2$
32.5 - 36.5	15	34.5	-4	$15 \times (-4) = -60$	16	$15 \times 16 = 240$
36.5 - 40.5	17	38.5	-3	$17 \times (-3) = -51$	9	$17 \times 9 = 153$
40.5 - 44.5	21	42.5	-2	$21 \times (-2) = -42$	4	$21 \times 4 = 84$
44.5 - 48.5	22	46.5	-1	$22 \times (-1) = -22$	1	$22 \times 1 = 22$
48.5 - 52.5	25	50.5	0	$25 \times 0 = 0$	0	$25 \times 0 = 0$
Total	$\sum f_i = 100$			$\sum f_i u_i = -175$		$\sum f_i u_i^2 = 499$

From the table:

Total frequency, $\sum f_i = 100$.

Sum of $f_i u_i$, $\sum f_i u_i = -175$.

Sum of $f_i u_i^2$, $\sum f_i u_i^2 = 499$.

Assumed mean, $A = 50.5$.

Class width, $h = 4$.

---

Calculate the mean:

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

$\bar{x} = 50.5 + 4 \left(\frac{-175}{100}\right)$

$\bar{x} = 50.5 + 4 (-1.75)$

$\bar{x} = 50.5 - 7 = 43.5$

The mean diameter is $\mathbf{43.5}$ mm.

---

Calculate the variance:

$\sigma^2 = h^2 \left[\frac{\sum f_i u_i^2}{\sum f_i} - \left(\frac{\sum f_i u_i}{\sum f_i}\right)^2\right]$

$\sigma^2 = 4^2 \left[\frac{499}{100} - \left(\frac{-175}{100}\right)^2\right]$

$\sigma^2 = 16 \left[4.99 - (-1.75)^2\right]$

$\sigma^2 = 16 \left[4.99 - 3.0625\right]$

$\sigma^2 = 16 \left[1.9275\right]$

$\sigma^2 = 30.84$

The variance is $\mathbf{30.84}$ $mm^2$.

---

Calculate the standard deviation:

$\sigma = \sqrt{\sigma^2} = \sqrt{30.84}$

Using a calculator, $\sqrt{30.84} \approx 5.553377$

The standard deviation is approximately $\mathbf{5.55}$ mm.

Given:

Number of observations, $n = 20$.

Variance of the original observations, $\sigma_{old}^2 = 5$.

Each observation is multiplied by 2.

---

To Find:

The new variance of the resulting observations.

---

Solution:

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

The mean of the original observations is $\bar{x} = \frac{1}{20} \sum\limits_{i=1}^{20} x_i$.

The variance of the original observations is given by:

Let the new observations be $y_i$. Each observation is multiplied by 2, so $y_i = 2x_i$ for $i = 1, 2, \dots, 20$.

The mean of the new observations is $\bar{y} = \frac{1}{20} \sum\limits_{i=1}^{20} y_i = \frac{1}{20} \sum\limits_{i=1}^{20} (2x_i) = \frac{2}{20} \sum\limits_{i=1}^{20} x_i = 2 \left(\frac{1}{20} \sum\limits_{i=1}^{20} x_i\right)$.

The variance of the new observations is $\sigma_{new}^2 = \frac{1}{20} \sum\limits_{i=1}^{20} (y_i - \bar{y})^2$.

Substitute $y_i = 2x_i$ and $\bar{y} = 2\bar{x}$:

$\sigma_{new}^2 = \frac{1}{20} \sum\limits_{i=1}^{20} (2x_i - 2\bar{x})^2$

$\sigma_{new}^2 = \frac{1}{20} \sum\limits_{i=1}^{20} (2(x_i - \bar{x}))^2$

$\sigma_{new}^2 = \frac{1}{20} \sum\limits_{i=1}^{20} 4(x_i - \bar{x})^2$

$\sigma_{new}^2 = 4 \left(\frac{1}{20} \sum\limits_{i=1}^{20} (x_i - \bar{x})^2\right)$

We know that $\frac{1}{20} \sum\limits_{i=1}^{20} (x_i - \bar{x})^2$ is the old variance, which is 5.

Substitute the value of $\sigma_{old}^2$:

$\sigma_{new}^2 = 4 \times 5$

$\sigma_{new}^2 = 20$

Alternatively, the property of variance states that if each observation in a data set is multiplied by a constant $c$, the new variance is $c^2$ times the original variance.

New Variance = $( \text{Multiplication Factor} )^2 \times \text{Old Variance}$

New Variance = $(2)^2 \times 5$

New Variance = $4 \times 5 = 20$

The new variance of the resulting observations is $\mathbf{20}$.

Given:

Number of observations, $n = 5$.

Mean of the observations, $\bar{x} = 4.4$.

Variance of the observations, $\sigma^2 = 8.24$.

Three of the observations are 1, 2, and 6.

---

To Find:

The other two observations.

---

Solution:

Let the five observations be $x_1, x_2, x_3, x_4, x_5$. We are given $x_1=1$, $x_2=2$, $x_3=6$.

Let the other two observations be $a$ and $b$. The observations are $1, 2, 6, a, b$.

The mean is given by the formula $\bar{x} = \frac{\sum x_i}{n}$.

The sum of the observations is $\sum x_i = 1 + 2 + 6 + a + b = 9 + a + b$.

Using the given mean:

$4.4 = \frac{9 + a + b}{5}$

$4.4 \times 5 = 9 + a + b$

$22 = 9 + a + b$

---

The variance is given by the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

The sum of the squares of the observations is:

$\sum x_i^2 = 1^2 + 2^2 + 6^2 + a^2 + b^2 = 1 + 4 + 36 + a^2 + b^2 = 41 + a^2 + b^2$.

The square of the mean is $(\bar{x})^2 = (4.4)^2 = 19.36$.

Using the given variance:

$8.24 = \frac{41 + a^2 + b^2}{5} - 19.36$

$8.24 + 19.36 = \frac{41 + a^2 + b^2}{5}$

$27.6 = \frac{41 + a^2 + b^2}{5}$

$27.6 \times 5 = 41 + a^2 + b^2$

$138 = 41 + a^2 + b^2$

---

Now we have a system of two equations with two variables:

From equation (i), we can express $b$ in terms of $a$:

$b = 13 - a$

Substitute this into equation (ii):

$a^2 + (13 - a)^2 = 97$

Expand the term $(13 - a)^2$:

$a^2 + (169 - 26a + a^2) = 97$

Combine like terms:

$2a^2 - 26a + 169 = 97$

Move all terms to one side to form a quadratic equation:

$2a^2 - 26a + 169 - 97 = 0$

$2a^2 - 26a + 72 = 0$

Divide the entire equation by 2:

$a^2 - 13a + 36 = 0$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to 36 and add up to -13. These numbers are -4 and -9.

$(a - 4)(a - 9) = 0$

This gives two possible values for $a$:

$a - 4 = 0 \implies a = 4$

$a - 9 = 0 \implies a = 9$

Case 1: If $a = 4$, substitute this into equation (i):

$4 + b = 13 \implies b = 13 - 4 = 9$

Case 2: If $a = 9$, substitute this into equation (i):

$9 + b = 13 \implies b = 13 - 9 = 4$

In both cases, the other two observations are 4 and 9.

The other two observations are $\mathbf{4}$ and $\mathbf{9}$.

Given:

A set of $n$ observations: $x_1, x_2, \dots, x_n$.

A new set of observations $y_1, y_2, \dots, y_n$, where $y_i = x_i + a$ for each $i$, and $a$ is a constant (positive or negative number).

---

To Show:

The variance of the new observations is equal to the variance of the original observations (i.e., the variance remains unchanged).

---

Solution:

Let $\bar{x}$ be the mean of the original observations $x_1, x_2, \dots, x_n$.

By definition, the mean is:

$\bar{x} = \frac{1}{n} \sum\limits_{i=1}^n x_i$

The variance of the original observations, denoted by $\sigma_x^2$, is given by:

Now, consider the new observations $y_i = x_i + a$.

Let $\bar{y}$ be the mean of the new observations $y_1, y_2, \dots, y_n$.

The mean of the new observations is:

$\bar{y} = \frac{1}{n} \sum\limits_{i=1}^n y_i$

Substitute $y_i = x_i + a$ into the formula for $\bar{y}$:

$\bar{y} = \frac{1}{n} \sum\limits_{i=1}^n (x_i + a)$

Using the property of summation, $\sum (u_i + v_i) = \sum u_i + \sum v_i$ and $\sum c = nc$:

$\bar{y} = \frac{1}{n} \left(\sum\limits_{i=1}^n x_i + \sum\limits_{i=1}^n a\right)$

$\bar{y} = \frac{1}{n} \left(\sum\limits_{i=1}^n x_i + na\right)$

$\bar{y} = \frac{1}{n} \sum\limits_{i=1}^n x_i + \frac{na}{n}$

$\bar{y} = \bar{x} + a$

Now, let's find the variance of the new observations, denoted by $\sigma_y^2$.

By definition, the variance of the new observations is:

$\sigma_y^2 = \frac{1}{n} \sum\limits_{i=1}^n (y_i - \bar{y})^2$

Substitute $y_i = x_i + a$ and $\bar{y} = \bar{x} + a$ into the formula for $\sigma_y^2$:

$\sigma_y^2 = \frac{1}{n} \sum\limits_{i=1}^n ((x_i + a) - (\bar{x} + a))^2$

Simplify the term inside the parentheses:

$(x_i + a) - (\bar{x} + a) = x_i + a - \bar{x} - a = x_i - \bar{x}$

So, the formula for $\sigma_y^2$ becomes:

$\sigma_y^2 = \frac{1}{n} \sum\limits_{i=1}^n (x_i - \bar{x})^2$

Comparing this expression with the formula for $\sigma_x^2$ in equation (i), we see that:

This shows that the variance of the new observations ($y_i$) is equal to the variance of the original observations ($x_i$).

Thus, if each observation is increased by a constant 'a', the variance remains unchanged.

Given:

Number of observations, $n = 100$.

Incorrect Mean, $\bar{x}_{incorrect} = 40$.

Incorrect Standard Deviation, $\sigma_{incorrect} = 5.1$.

Incorrect observation taken: 50.

Correct observation: 40.

---

To Find:

The correct mean and standard deviation.

---

Solution:

Let the original (incorrect) observations be $x_1, x_2, \dots, x_{100}$. One of these observations was incorrectly taken as 50 instead of 40.

We know the formula for the mean: $\bar{x} = \frac{\sum x_i}{n}$.

Using the incorrect mean, we can find the incorrect sum of observations:

$\bar{x}_{incorrect} = \frac{\sum x_{i, incorrect}}{n}$

$40 = \frac{\sum x_{i, incorrect}}{100}$

The correct sum of observations is obtained by subtracting the incorrect observation and adding the correct observation:

$\sum x_{i, correct} = \sum x_{i, incorrect} - (\text{Incorrect Value}) + (\text{Correct Value})$

$\sum x_{i, correct} = 4000 - 50 + 40 = 3990$

---

Now, calculate the correct mean:

$\bar{x}_{correct} = \frac{\sum x_{i, correct}}{n}$

$\bar{x}_{correct} = \frac{3990}{100} = 39.9$

The correct mean is $\mathbf{39.9}$.

---

Next, we need to find the correct standard deviation. We know the variance is related to the sum of squared deviations or the sum of squares of the observations.

The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

Using the incorrect standard deviation, we can find the incorrect variance:

$\sigma_{incorrect}^2 = (5.1)^2 = 26.01$

Using the variance formula with incorrect values:

$\sigma_{incorrect}^2 = \frac{\sum x_{i, incorrect}^2}{n} - (\bar{x}_{incorrect})^2$

$26.01 = \frac{\sum x_{i, incorrect}^2}{100} - (40)^2$

$26.01 = \frac{\sum x_{i, incorrect}^2}{100} - 1600$

$26.01 + 1600 = \frac{\sum x_{i, incorrect}^2}{100}$

$1626.01 = \frac{\sum x_{i, incorrect}^2}{100}$

The correct sum of squares of observations is obtained by subtracting the square of the incorrect observation and adding the square of the correct observation:

$\sum x_{i, correct}^2 = \sum x_{i, incorrect}^2 - (\text{Incorrect Value})^2 + (\text{Correct Value})^2$

$\sum x_{i, correct}^2 = 162601 - (50)^2 + (40)^2$

$\sum x_{i, correct}^2 = 162601 - 2500 + 1600$

$\sum x_{i, correct}^2 = 161701$

---

Now, calculate the correct variance:

$\sigma_{correct}^2 = \frac{\sum x_{i, correct}^2}{n} - (\bar{x}_{correct})^2$

$\sigma_{correct}^2 = \frac{161701}{100} - (39.9)^2$

$\sigma_{correct}^2 = 1617.01 - 1592.01$

$\sigma_{correct}^2 = 25$

The correct variance is $\mathbf{25}$.

---

Finally, calculate the correct standard deviation:

$\sigma_{correct} = \sqrt{\sigma_{correct}^2} = \sqrt{25} = 5$

The correct standard deviation is $\mathbf{5}$.

Given:

Number of observations, $n = 8$.

Mean of the observations, $\bar{x} = 9$.

Variance of the observations, $\sigma^2 = 9.25$.

Six of the observations are 6, 7, 10, 12, 12, and 13.

---

To Find:

The remaining two observations.

---

Solution:

Let the remaining two observations be $a$ and $b$.

The eight observations are 6, 7, 10, 12, 12, 13, $a$, $b$.

The sum of the observations is $\sum x_i = 6 + 7 + 10 + 12 + 12 + 13 + a + b = 60 + a + b$.

The mean is given by the formula $\bar{x} = \frac{\sum x_i}{n}$.

Using the given mean:

$9 = \frac{60 + a + b}{8}$

$9 \times 8 = 60 + a + b$

$72 = 60 + a + b$

---

The variance is given by the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

The sum of the squares of the known observations is:

$6^2 + 7^2 + 10^2 + 12^2 + 12^2 + 13^2 = 36 + 49 + 100 + 144 + 144 + 169 = 642$.

The sum of the squares of all observations is $\sum x_i^2 = 642 + a^2 + b^2$.

Using the given variance and mean:

$9.25 = \frac{642 + a^2 + b^2}{8} - (9)^2$

$9.25 = \frac{642 + a^2 + b^2}{8} - 81$

$9.25 + 81 = \frac{642 + a^2 + b^2}{8}$

$90.25 = \frac{642 + a^2 + b^2}{8}$

$90.25 \times 8 = 642 + a^2 + b^2$

$722 = 642 + a^2 + b^2$

---

Now we have a system of two equations with two variables $a$ and $b$:

From equation (i), we can express $b$ in terms of $a$:

$b = 12 - a$

Substitute this into equation (ii):

$a^2 + (12 - a)^2 = 80$

Expand the term $(12 - a)^2$ using $(x-y)^2 = x^2 - 2xy + y^2$:

$a^2 + (144 - 24a + a^2) = 80$

Combine like terms:

$2a^2 - 24a + 144 = 80$

Move all terms to one side to form a quadratic equation:

$2a^2 - 24a + 144 - 80 = 0$

$2a^2 - 24a + 64 = 0$

Divide the entire equation by 2:

$a^2 - 12a + 32 = 0$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to 32 and add up to -12. These numbers are -4 and -8.

$(a - 4)(a - 8) = 0$

This gives two possible values for $a$:

$a - 4 = 0 \implies a = 4$

$a - 8 = 0 \implies a = 8$

Case 1: If $a = 4$, substitute this into equation (i):

$4 + b = 12 \implies b = 12 - 4 = 8$

Case 2: If $a = 8$, substitute this into equation (i):

$8 + b = 12 \implies b = 12 - 8 = 4$

In both cases, the other two observations are 4 and 8.

The remaining two observations are $\mathbf{4}$ and $\mathbf{8}$.

Given:

Number of observations, $n = 7$.

Mean of the observations, $\bar{x} = 8$.

Variance of the observations, $\sigma^2 = 16$.

Five of the observations are 2, 4, 10, 12, and 14.

---

To Find:

The remaining two observations.

---

Solution:

Let the remaining two observations be $a$ and $b$.

The seven observations are 2, 4, 10, 12, 14, $a$, $b$.

The sum of the observations is $\sum x_i = 2 + 4 + 10 + 12 + 14 + a + b = 42 + a + b$.

The mean is given by the formula $\bar{x} = \frac{\sum x_i}{n}$.

Using the given mean:

$8 = \frac{42 + a + b}{7}$

$8 \times 7 = 42 + a + b$

$56 = 42 + a + b$

---

The variance is given by the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

The sum of the squares of the known observations is:

$2^2 + 4^2 + 10^2 + 12^2 + 14^2 = 4 + 16 + 100 + 144 + 196 = 460$.

The sum of the squares of all observations is $\sum x_i^2 = 460 + a^2 + b^2$.

Using the given variance and mean:

$16 = \frac{460 + a^2 + b^2}{7} - (8)^2$

$16 = \frac{460 + a^2 + b^2}{7} - 64$

$16 + 64 = \frac{460 + a^2 + b^2}{7}$

$80 = \frac{460 + a^2 + b^2}{7}$

$80 \times 7 = 460 + a^2 + b^2$

$560 = 460 + a^2 + b^2$

---

Now we have a system of two equations with two variables $a$ and $b$:

From equation (i), we can express $b$ in terms of $a$:

$b = 14 - a$

Substitute this into equation (ii):

$a^2 + (14 - a)^2 = 100$

Expand the term $(14 - a)^2$ using $(x-y)^2 = x^2 - 2xy + y^2$:

$a^2 + (14^2 - 2 \times 14 \times a + a^2) = 100$

$a^2 + 196 - 28a + a^2 = 100$

Combine like terms:

$2a^2 - 28a + 196 = 100$

Move all terms to one side to form a quadratic equation:

$2a^2 - 28a + 196 - 100 = 0$

$2a^2 - 28a + 96 = 0$

Divide the entire equation by 2:

$a^2 - 14a + 48 = 0$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to 48 and add up to -14. These numbers are -6 and -8.

$(a - 6)(a - 8) = 0$

This gives two possible values for $a$:

$a - 6 = 0 \implies a = 6$

$a - 8 = 0 \implies a = 8$

Case 1: If $a = 6$, substitute this into equation (i):

$6 + b = 14 \implies b = 14 - 6 = 8$

Case 2: If $a = 8$, substitute this into equation (i):

$8 + b = 14 \implies b = 14 - 8 = 6$

In both cases, the other two observations are 6 and 8.

The remaining two observations are $\mathbf{6}$ and $\mathbf{8}$.

Given:

Number of observations, $n = 6$.

Original Mean, $\bar{x}_{old} = 8$.

Original Standard Deviation, $\sigma_{old} = 4$.

Each observation is multiplied by a constant, $c = 3$.

---

To Find:

The new mean and new standard deviation of the resulting observations.

---

Solution:

Let the original observations be $x_1, x_2, \dots, x_6$.

The original mean is $\bar{x}_{old} = \frac{1}{6} \sum\limits_{i=1}^6 x_i = 8$.

The original standard deviation is $\sigma_{old} = 4$.

The original variance is $\sigma_{old}^2 = 4^2 = 16$.

The original variance is also given by $\sigma_{old}^2 = \frac{1}{6} \sum\limits_{i=1}^6 (x_i - \bar{x}_{old})^2 = 16$.

The new observations are $y_i$, where $y_i = 3x_i$ for $i = 1, 2, \dots, 6$.

---

Calculate the new mean ($\bar{x}_{new}$):

The new mean is $\bar{x}_{new} = \frac{1}{6} \sum\limits_{i=1}^6 y_i$.

Substitute $y_i = 3x_i$:

$\bar{x}_{new} = \frac{1}{6} \sum\limits_{i=1}^6 (3x_i)$

$\bar{x}_{new} = \frac{3}{6} \sum\limits_{i=1}^6 x_i = 3 \left(\frac{1}{6} \sum\limits_{i=1}^6 x_i\right)$

$\bar{x}_{new} = 3 \times \bar{x}_{old}$

$\bar{x}_{new} = 3 \times 8 = 24$

The new mean is $\mathbf{24}$.

---

Calculate the new variance ($\sigma_{new}^2$):

The new variance is $\sigma_{new}^2 = \frac{1}{6} \sum\limits_{i=1}^6 (y_i - \bar{x}_{new})^2$.

Substitute $y_i = 3x_i$ and $\bar{x}_{new} = 3\bar{x}_{old}$:

$\sigma_{new}^2 = \frac{1}{6} \sum\limits_{i=1}^6 (3x_i - 3\bar{x}_{old})^2$

$\sigma_{new}^2 = \frac{1}{6} \sum\limits_{i=1}^6 (3(x_i - \bar{x}_{old}))^2$

$\sigma_{new}^2 = \frac{1}{6} \sum\limits_{i=1}^6 9(x_i - \bar{x}_{old})^2$

$\sigma_{new}^2 = 9 \left(\frac{1}{6} \sum\limits_{i=1}^6 (x_i - \bar{x}_{old})^2\right)$

$\sigma_{new}^2 = 9 \times \sigma_{old}^2$

$\sigma_{new}^2 = 9 \times 16 = 144$

The new variance is $\mathbf{144}$.

---

Calculate the new standard deviation ($\sigma_{new}$):

$\sigma_{new} = \sqrt{\sigma_{new}^2} = \sqrt{144} = 12$

Alternatively, the property of standard deviation states that if each observation is multiplied by a constant $c$, the new standard deviation is $|c|$ times the original standard deviation.

New Standard Deviation = $|3| \times$ Original Standard Deviation

New Standard Deviation = $3 \times 4 = 12$

The new standard deviation is $\mathbf{12}$.

Given:

A set of $n$ observations: $x_1, x_2, \dots, x_n$.

Mean of these observations is $\bar{x}$.

Variance of these observations is $\sigma^2$.

A new set of observations: $ax_1, ax_2, \dots, ax_n$, where $a$ is a non-zero constant.

---

To Prove:

The mean of the new observations is $a\bar{x}$.

The variance of the new observations is $a^2\sigma^2$.

---

Proof:

Let the original observations be $x_1, x_2, \dots, x_n$.

The mean of the original observations is defined as:

The variance of the original observations is defined as:

Now, consider the new observations $y_i = ax_i$ for $i = 1, 2, \dots, n$, where $a \neq 0$.

Let $\bar{y}$ be the mean of the new observations $y_1, y_2, \dots, y_n$.

The mean of the new observations is:

$\bar{y} = \frac{1}{n} \sum\limits_{i=1}^n y_i$

Substitute $y_i = ax_i$ into the formula for $\bar{y}$:

$\bar{y} = \frac{1}{n} \sum\limits_{i=1}^n (ax_i)$

Using the property of summation, $\sum (c \cdot u_i) = c \cdot \sum u_i$:

$\bar{y} = \frac{a}{n} \sum\limits_{i=1}^n x_i$

From equation (i), we know that $\frac{1}{n} \sum\limits_{i=1}^n x_i = \bar{x}$.

Therefore,

This proves that the mean of the new observations is $a\bar{x}$.

---

Next, let's find the variance of the new observations, denoted by $\sigma_{new}^2$.

By definition, the variance of the new observations is:

$\sigma_{new}^2 = \frac{1}{n} \sum\limits_{i=1}^n (y_i - \bar{y})^2$

Substitute $y_i = ax_i$ and $\bar{y} = a\bar{x}$ into the formula for $\sigma_{new}^2$:

$\sigma_{new}^2 = \frac{1}{n} \sum\limits_{i=1}^n (ax_i - a\bar{x})^2$

Factor out $a$ from the term inside the parentheses:

$\sigma_{new}^2 = \frac{1}{n} \sum\limits_{i=1}^n (a(x_i - \bar{x}))^2$

Using the property $(c \cdot u)^2 = c^2 \cdot u^2$:

$\sigma_{new}^2 = \frac{1}{n} \sum\limits_{i=1}^n a^2(x_i - \bar{x})^2$

Using the property of summation, $\sum (c \cdot v_i) = c \cdot \sum v_i$:

$\sigma_{new}^2 = \frac{a^2}{n} \sum\limits_{i=1}^n (x_i - \bar{x})^2$

From equation (ii), we know that $\frac{1}{n} \sum\limits_{i=1}^n (x_i - \bar{x})^2 = \sigma^2$.

Therefore,

This proves that the variance of the new observations is $a^2\sigma^2$.

Given (Incorrect Data):

Number of observations, $n_{old} = 20$.

Incorrect Mean, $\bar{x}_{old} = 10$.

Incorrect Standard Deviation, $\sigma_{old} = 2$.

Incorrect observation: 8.

---

Calculations based on Incorrect Data:

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

Using the incorrect mean, we find the incorrect sum of observations:

$\bar{x}_{old} = \frac{\sum x_{i, old}}{n_{old}}$

The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

The incorrect variance is $\sigma_{old}^2 = (\sigma_{old})^2 = (2)^2 = 4$.

Using the incorrect variance and mean, we find the incorrect sum of squares of observations:

$\sigma_{old}^2 = \frac{\sum x_{i, old}^2}{n_{old}} - (\bar{x}_{old})^2$

---

Case (i): If wrong item is omitted.

To Find:

The correct mean and standard deviation when the incorrect observation (8) is omitted.

Solution (Case i):

The number of observations after omitting one is $n_{new} = n_{old} - 1 = 20 - 1 = 19$.

The correct sum of observations is obtained by subtracting the incorrect observation from the incorrect sum:

$\sum x_{i, correct} = \sum x_{i, old} - (\text{Incorrect Value})$

Calculate the correct mean:

$\bar{x}_{correct} = \frac{\sum x_{i, correct}}{n_{new}}$

The correct mean is $\mathbf{\frac{192}{19}}$.

As a decimal, $\bar{x}_{correct} \approx 10.105$.

The correct sum of squares of observations is obtained by subtracting the square of the incorrect observation from the incorrect sum of squares:

$\sum x_{i, correct}^2 = \sum x_{i, old}^2 - (\text{Incorrect Value})^2$

Calculate the correct variance ($\sigma_{correct}^2$):

$\sigma_{correct}^2 = \frac{\sum x_{i, correct}^2}{n_{new}} - (\bar{x}_{correct})^2$

The correct variance is $\mathbf{\frac{1440}{361}}$.

Calculate the correct standard deviation ($\sigma_{correct}$):

$\sigma_{correct} = \sqrt{\sigma_{correct}^2} = \sqrt{\frac{1440}{361}}$

The correct standard deviation is $\mathbf{\frac{12\sqrt{10}}{19}}$.

As a decimal, $\sigma_{correct} \approx 1.997$.

---

Case (ii): If it is replaced by 12.

To Find:

The correct mean and standard deviation when the incorrect observation (8) is replaced by a correct observation (12).

Solution (Case ii):

The number of observations remains the same: $n_{new} = n_{old} = 20$.

The correct sum of observations is obtained by subtracting the incorrect observation and adding the correct observation:

$\sum x_{i, correct} = \sum x_{i, old} - (\text{Incorrect Value}) + (\text{Correct Value})$

Calculate the correct mean:

$\bar{x}_{correct} = \frac{\sum x_{i, correct}}{n_{new}}$

The correct mean is $\mathbf{10.2}$.

The correct sum of squares of observations is obtained by subtracting the square of the incorrect observation and adding the square of the correct observation:

$\sum x_{i, correct}^2 = \sum x_{i, old}^2 - (\text{Incorrect Value})^2 + (\text{Correct Value})^2$

Calculate the correct variance ($\sigma_{correct}^2$):

$\sigma_{correct}^2 = \frac{\sum x_{i, correct}^2}{n_{new}} - (\bar{x}_{correct})^2$

The correct variance is $\mathbf{3.96}$.

Calculate the correct standard deviation ($\sigma_{correct}$):

$\sigma_{correct} = \sqrt{\sigma_{correct}^2} = \sqrt{3.96}$

Using a calculator, $\sqrt{3.96} \approx 1.98997$.

The correct standard deviation is approximately $\mathbf{1.99}$.

Given (Incorrect Data):

Initial number of observations, $n_{old} = 100$.

Incorrect Mean, $\bar{x}_{old} = 20$.

Incorrect Standard Deviation, $\sigma_{old} = 3$.

Incorrect observations: 21, 21, and 18.

---

To Find:

The mean and standard deviation if the incorrect observations are omitted.

---

Calculations based on Incorrect Data:

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

Using the incorrect mean, we find the incorrect sum of observations:

$\bar{x}_{old} = \frac{\sum x_{i, old}}{n_{old}}$

The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

The incorrect variance is $\sigma_{old}^2 = (\sigma_{old})^2 = (3)^2 = 9$.

Using the incorrect variance and mean, we find the incorrect sum of squares of observations:

$\sigma_{old}^2 = \frac{\sum x_{i, old}^2}{n_{old}} - (\bar{x}_{old})^2$

---

Calculations after Omitting Incorrect Observations:

The number of incorrect observations is 3 (21, 21, 18).

The new number of observations is $n_{new} = n_{old} - 3 = 100 - 3 = 97$.

The sum of the incorrect observations is $21 + 21 + 18 = 60$.

The correct sum of observations is obtained by subtracting the sum of incorrect observations from the incorrect sum:

$\sum x_{i, correct} = \sum x_{i, old} - (\text{Sum of Incorrect Values})$

Calculate the correct mean:

$\bar{x}_{correct} = \frac{\sum x_{i, correct}}{n_{new}}$

The correct mean is $\mathbf{\frac{1940}{97}}$.

As a decimal, $\bar{x}_{correct} \approx 20$.

The sum of the squares of the incorrect observations is $21^2 + 21^2 + 18^2 = 441 + 441 + 324 = 1206$.

The correct sum of squares of observations is obtained by subtracting the sum of the squares of the incorrect observations from the incorrect sum of squares:

$\sum x_{i, correct}^2 = \sum x_{i, old}^2 - (\text{Sum of Squares of Incorrect Values})$

Calculate the correct variance ($\sigma_{correct}^2$):

$\sigma_{correct}^2 = \frac{\sum x_{i, correct}^2}{n_{new}} - (\bar{x}_{correct})^2$

Performing the division: $86718 \div 9409 \approx 9.216$

The correct variance is $\mathbf{\frac{86718}{9409}} \approx \mathbf{9.22}$.

---

Calculate the correct standard deviation ($\sigma_{correct}$):

$\sigma_{correct} = \sqrt{\sigma_{correct}^2} = \sqrt{\frac{86718}{9409}}$

$\sigma_{correct} \approx \sqrt{9.216} \approx 3.0358$

The correct standard deviation is approximately $\mathbf{3.04}$.