NCERT Solutions for Class 11 Maths Chapter 13 Statistics

Question:1. Find the mean deviation about the mean for the data. $4,7,8,9,10,12,13,17$

Answer:

Mean ($\bar{x}$) of the given data:

$\bar{x}=\frac{1}{8}\sum _{i=1}^8{x}_i=\frac{4+7+8+9+10+12+13+17}{8}=10$

The respective absolute values of the deviations from mean, $|x_i-\bar{x}|$ are
6, 3, 2, 1, 0, 2, 3, 7.

$\therefore$ $\sum _{i=1}^8|x_i-10|=24$

$\therefore$ $M.D.(\bar{x})=\frac{1}{n}\sum _{i=1}^n|x_i-\bar{x}|=\frac{24}{8}=3$

Hence, the mean deviation from the mean is 3.

Question:2. Find the mean deviation about the mean for the data. $38,70,48,40,42,55,63,46,54,44$

Answer:

Mean ( $\bar{x}$ ) of the given data:

$$\begin{gathered} \bar{x}=\frac{1}{8}\sum _{i=1}^8{x}_i=\frac{38+70+48+40+42+55+63+46+54+44}{10} \\ =\frac{500}{10}=50 \end{gathered}$$

The respective absolute values of the deviations from mean, $|x_i-\bar{x}|$ are
12, 20, 2, 10, 8, 5, 13, 4, 4, 6.

$\therefore$ $\sum _{i=1}^8|x_i-50|=84$

$\therefore$ $M.D.(\bar{x})=\frac{1}{n}\sum _{i=1}^n|x_i-\bar{x}|=\frac{84}{10}=8.4$

Hence, the mean deviation from the mean is 8.4.

Question 3. Find the mean deviation about the median. $13,17,16,14,11,13,10,16,11,18,12,17$

Answer:

Number of observations, n = 12, which is even.

Arranging the values in ascending order:

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

Now, Median (M)

$$\begin{gathered} =\frac{(\frac{12}{2}{)}^{th}\text{observation}+(\frac{12}{2}+1{)}^{th}\text{observation}}{2} \\ =\frac{13+14}{2}=\frac{27}{2}=13.5 \end{gathered}$$

The respective absolute values of the deviations from the median, $|x_i-M|$ are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

$\therefore$ $\sum _{i=1}^8|x_i-13.5|=28$

$\therefore$ $M.D.(M)=\frac{1}{12}\sum _{i=1}^n|x_i-M|$

$=\frac{28}{12}=2.33$

Hence, the mean deviation from the median is 2.33.

Question:4. Find the mean deviation about the median. $36,72,46,42,60,45,53,46,51,49$

Answer:

Number of observations, n = 10, which is even.

Arranging the values in ascending order:

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Now, Median (M)

$$\begin{gathered} =\frac{(\frac{10}{2}{)}^{th}\text{observation}+(\frac{10}{2}+1{)}^{th}\text{observation}}{2} \\ =\frac{46+49}{2}=\frac{95}{2}=47.5 \end{gathered}$$

The respective absolute values of the deviations from the median, $|x_i-M|$ are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

$\therefore$ $\sum _{i=1}^8|x_i-47.5|=70$

$\therefore$ $M.D.(M)=\frac{1}{10}\sum _{i=1}^n|x_i-M|$

$=\frac{70}{10}=7$

Hence, the mean deviation from the median is 7.

Question:5. Find the mean deviation from the mean.

${x}_{i}510152025$
$f_{i}74635$

Answer:

$N=\sum _{i=1}^5{f}_i=25;\sum _{i=1}^5{f}_i{x}_i=350$

$\bar{x}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i=\frac{350}{25}=14$

Now, we calculate the absolute values of the deviations from the an, $|x_i-\bar{x}|$ and

$\sum f_i|x_i-\bar{x}|$ = 158

$\therefore$ $M.D.(\bar{x})=\frac{1}{25}\sum _{i=1}^n{f}_i|x_i-\bar{x}|$

$=\frac{158}{25}=6.32$

Hence, the mean deviation from the mean is 6.32.

Question:6. Find the mean deviation from the mean.

${x}_{i}1030507090$
$f_{i}42428168$

Answer:

$N=\sum _{i=1}^5{f}_i=80;\sum _{i=1}^5{f}_i{x}_i=4000$

$\bar{x}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i=\frac{4000}{80}=50$

Now, we calculate the absolute values of the deviations from the an, $|x_i-\bar{x}|$ and

$\sum f_i|x_i-\bar{x}|$ = 1280

$\therefore$ $M.D.(\bar{x})=\frac{1}{80}\sum _{i=1}^5{f}_i|x_i-\bar{x}|$

$=\frac{1280}{80}=16$

Hence, the mean deviation from the mean is 16.

Question 7. Find the mean deviation about the median.

Answer:

Now, N = 26 which is even.

Median is the mean of ${13}^{th}$ and ${14}^{th}$ observations.

Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Therefore, Median, M $=\frac{{13}^{th}observation+{14}^{th}observation}{2}=\frac{7+7}{2}=\frac{14}{2}=7$

Now, we calculate the absolute values of the deviations from the median, $|x_i-M|$ and

$\sum f_i|x_i-M|$ = 84

$\therefore$ $M.D.(M)=\frac{1}{26}\sum _{i=1}^6|x_i-M|$

$=\frac{84}{26}=3.23$

Hence, the mean deviation from the median is 3.23.

Question:8. Find the mean deviation about the median.

Answer:

Now, N = 30, which is even.

Median is the mean of ${15}^{th}$ and ${16}^{th}$ observations.

Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.

Therefore, Median, M $=\frac{{15}^{th}\text{observation}+{16}^{th}\text{observation}}{2}=\frac{30+30}{2}=30$

Now, we calculate the absolute values of the deviations from the median, $|x_i-M|$ and

$\sum f_i|x_i-M|$ = 149.5

$\therefore$ $M.D.(M)=\frac{1}{29}\sum _{i=1}^5|x_i-M|$

$=\frac{149.5}{29}=5.1$

Hence, the mean deviation from the median is 5.1.

Question 9. Find the mean deviation from the mean.

Answer:

$N=\sum _{i=1}^8{f}_i=50;\sum _{i=1}^8{f}_i{x}_i=17900$

$\bar{x}=\frac{1}{N}\sum _{i=1}^8{f}_i{x}_i=\frac{17900}{50}=358$

Now, we calculate the absolute values of the deviations from mean n, $|x_i-\bar{x}|$ and

$\sum f_i|x_i-\bar{x}|$ = 7896

$\therefore$ $M.D.(\bar{x})=\frac{1}{50}\sum _{i=1}^8{f}_i|x_i-\bar{x}|$

$=\frac{7896}{50}=157.92$

Hence, the mean deviation from the mean is 157.92.

Question 10. Find the mean deviation from the mean.

Answer:

$N=\sum _{i=1}^6{f}_i=100;\sum _{i=1}^6{f}_i{x}_i=12530$

$\bar{x}=\frac{1}{N}\sum _{i=1}^6{f}_i{x}_i=\frac{12530}{100}=125.3$

Now, we calculate the absolute values of the deviations from the mean, $|x_i-\bar{x}|$ and

$\sum f_i|x_i-\bar{x}|$ = 1128.8

$\therefore$ $M.D.(\bar{x})=\frac{1}{100}\sum _{i=1}^6{f}_i|x_i-\bar{x}|$

$=\frac{1128.8}{100}=11.29$

Hence, the mean deviation from the mean is 11.29

Question 11. Find the mean deviation about the median for the following data :

Answer:

Now, N = 50, which is even.

The class interval containing ${\left(\frac{N}{2}\right)}^{th}$ or ${25}^{th}$ item is 20-30. Therefore, 20-30 is the median class.

We know,

Median $=l+\frac{\frac{N}{2}-C}{f}\times h$

Here, l = 20, C = 14, f = 14, h = 10 and N = 50

Therefore, Median $=20+\frac{25-14}{14}\times 10=20+7.85=27.85$

Now, we calculate the absolute values of the deviations from the median, $|x_i-M|$ and

$\sum f_i|x_i-M|$ = 517.1

$\therefore$ $M.D.(M)=\frac{1}{50}\sum _{i=1}^6{f}_i|x_i-M|$

$=\frac{517.1}{50}=10.34$

Hence, the mean deviation from the median is 10.34

Question:12. Calculate the mean deviation about median age for the age distribution of $100$ persons given below:

[Hint: Convert the given data into continuous frequency distribution by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval]

Answer:

Now, N = 100, which is even.

The class interval containing ${\left(\frac{N}{2}\right)}^{th}$ or ${50}^{th}$ item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

We know,

Median $=l+\frac{\frac{N}{2}-C}{f}\times h$

Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100

Therefore, Median $=35.5+\frac{50-37}{26}\times 5=35.5+2.5=38$

Now, we calculate the absolute values of the deviations from the median, $|x_i-M|$ and

$\sum f_i|x_i-M|$ = 735

$\therefore$ $M.D.(M)=\frac{1}{100}\sum _{i=1}^8{f}_i|x_i-M|$

$=\frac{735}{100}=7.35$

Hence, the mean deviation from the median is 7.35.

Question 1. Find the mean and variance for each of the data.

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

Answer:

Mean ( $\bar{x}$ ) of the given data:

$\bar{x}=\frac{1}{8}\sum _{i=1}^8{x}_i=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9$

The respective values of the deviations from mean, $(x_i-\bar{x})$ are

-3, -2, 1, 3, 4, -5, -1, 3

$\therefore$ $\sum _{i=1}^8(x_i-\bar{x}{)}^2=74$

$\therefore$ Variance, ${\sigma }^2=\frac{1}{n}\sum _{i=1}^n(x_i-\bar{x}{)}^2$

So, $\frac{1}{8}\sum _{i=1}^8(x_i-\bar{x}{)}^2=\frac{74}{8}=9.25$

Hence, Mean = 9 and Variance = 9.25

Question 2. Find the mean and variance for each of the data:

First natural numbers.

Answer:

Mean ( $\bar{x}$ ) of first n natural numbers:

$\bar{x}=\frac{1}{n}\sum _{i=1}^n{x}_i=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}$

We know, Variance ${\sigma }^2=\frac{1}{n}\sum _{i=1}^n(x_i-\bar{x}{)}^2$

${\sigma }^2=\frac{1}{n}\sum _{i=1}^n{(x_i-\frac{n+1}{2})}^2$

We know that $(a-b{)}^2=a^2-2ab+b^2$

$$\begin{gathered} \therefore n{\sigma }^2=\sum _{i=1}^n{x}_i^2+\sum _{i=1}^n(\frac{n+1}{2}{)}^2-2\sum _{i=1}^n{x}_i\frac{n+1}{2} \\ =\frac{n(n+1)(2n+1)}{6}+\frac{(n+1{)}^2}{4}\times n-2.\frac{(n+1)}{2}.\frac{n(n+1)}{2} \end{gathered}$$

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

$$\begin{gathered} \Rightarrow {\sigma }^2=\frac{(n+1)(2n+1)}{6}-\frac{(n+1{)}^2}{4} \\ =(n+1)\left[\frac{4n+2-3n-3}{12}\right] \\ =(n+1).\frac{(n-1)}{12} \\ =\frac{n^2-1}{12} \end{gathered}$$

Hence, Mean = $\frac{n+1}{2}$ and Variance = $\frac{n^2-1}{12}$

Question 3. Find the mean and variance for each of the data:

First 10 multiples of 3.

Answer:

The first 10 multiples of 3 are:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Mean ( $\bar{x}$ ) of the above values:

$$\begin{gathered} \bar{x}=\frac{1}{10}\sum _{i=1}^{10}{x}_i=\frac{3+6+9+12+15+18+21+24+27+30}{10} \\ =3.\frac{\frac{10(10+1)}{2}}{10}=16.5 \end{gathered}$$