CBSE Class 11 Maths Chapter 13 Statistics – NCERT Solutions Exercise 13.1 [2025–26]

Exercise 13.1

1. Calculate the mean deviation about mean for the numbers 

$\mathbf{4},\mathbf{7},\mathbf{8},\mathbf{9},\mathbf{10},\mathbf{12},\mathbf{13},\mathbf{17}$.

Ans.

We are provided the numbers

$4,7,8,9,10,12,13,17$.

Therefore, mean of the numbers,

 $ \overline{x}=\dfrac{4+7+8+9+10+12+13+17}{8} $

$ =\dfrac{80}{8} $

$ =10 $

The deviations for the given numbers with the mean $\overline{x}=10$, 

that is ${{x}_{i}}-\overline{x}$, for $i=1,2,...,8$ are 

$-6,-3,-2,-1,\,\,0,\,\,2,\,\,3,\,\,7$.

Also, the absolute values of the obtained deviations, that is $\left| {{x}_{i}}-x \right|$ are

$6,3,2,1,0,2,3,7$.

Thus, the mean deviation about the mean $\overline{x}$ for the given numbers is

$ M.D.\left( \overline{x} \right)=\dfrac{\sum\limits_{i=1}^{8}{\left| {{x}_{i}}-overline{x} \right|}}{8}\\ $

 $ =\dfrac{6+3+2+1+0+2+3+7}{8}\\ $

$ =\dfrac{24}{8}\\ $

$ =3\\ $

2. Calculate the mean deviation about the mean for the following numbers.

$\mathbf{38},\mathbf{70},\mathbf{48},\mathbf{40},\mathbf{42},\mathbf{55},\mathbf{63},\mathbf{46},\mathbf{54},\mathbf{44}$

Ans.

We are provided that numbers

$\text{38,70,48,40,42,55,63,46,54,44}$.

Therefore, the mean of the numbers,

$ \overline{x}=\dfrac{38+70+48+40+42+55+63+46+54+44}{10}\\ $

$ =\dfrac{500}{10}\\ $

$ =50\\ $

Now, the deviations for the given numbers with the mean $\overline{x}=50$, that is

${{x}_{i}}-x$, for $i=1,2,3,...,10$ are 

$-12,20,-2,-10,-8,5,13,-4,4,-6$.

So, the absolute values of the obtained deviations for the given numbers, that is $\left| {{x}_{i}}-x \right|$ are

$12,20,2,10,8,5,13,4,4,6$.

Thus, the mean deviation about the mean $\overline{x}$ for the given numbers is

$M.D.\left( \overline{x} \right)=\dfrac{\sum\limits_{i=1}^{10}{\left| {{x}_{i}}-x \right|}}{10}$

$=\dfrac{12+20+2+10+8+5+13+4+4+6}{10}$

$=\dfrac{84}{10}$

$=8.4$

3. Calculate the mean deviation about the median for the following numbers.

$\mathbf{13},\mathbf{17},\mathbf{16},\mathbf{14},\mathbf{11},\mathbf{13},\mathbf{10},\mathbf{16},\mathbf{11},\mathbf{18},\mathbf{12},\mathbf{17}$.

Ans.

We are provided the numbers

$\text{13,17,16,14,11,13,10,16,11,18,12,17}$.

There is total $12$ numbers, and it is even.

Now, arranging the numbers in the ascending order, gives

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

It is known that, 

Median, $M=\dfrac{{{\left( \dfrac{n}{2} \right)}^{th}}\,\,number+{{\left( \dfrac{n}{2}+1 \right)}^{th}}number}{2}$, when $n$ is even.

Therefore, the median

$M=\dfrac{{{\left( \dfrac{12}{2} \right)}^{th}}number+{{\left( \dfrac{12}{2}+1 \right)}^{th}}number}{2}$

$=\dfrac{{{6}^{th}}\,number+{{7}^{th}}\,\,number}{2}$

$=\dfrac{13+14}{2}$

$=\dfrac{27}{2}$

$=13.5$

Now, the deviations about the median $M$ for the given numbers, that is ${{x}_{i}}-M$, for $i=1,2,3,...,12$ are

$-3.5,-2.5,-2.5,-1.5,-0.5,-0.5,\,0.5,\,2.5,\,3.5,\,3.5,\,3.5,\,4.5$

So, the absolute values of the resulted deviations, that is $\left| {{x}_{i}}-M \right|$, for $i=1,2,3,...,12$ are

$3.5,\,2.5,\,2.5,\,1.5,\,0.5,\,0.5,\,0.5,\,2.5,\,2.5,\,3.5,\,3.5,\,4.5\,$.

Thus, the mean deviation about the median $M$ for the given numbers,

$M.D.\left( M \right)=\dfrac{\sum\limits_{i=1}^{12}{\left| {{x}_{i}}-M \right|}}{12}$

$=\dfrac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+4.5}{12}$

$=\dfrac{28}{12}$

$=2.33$

4. Calculate the mean deviation about the median for the following numbers.

$\mathbf{36},\,\mathbf{72},\,\mathbf{46},\,\mathbf{42},\,\mathbf{60},\,\mathbf{45},\,\mathbf{53},\,\mathbf{46},\,\mathbf{51},\,\mathbf{49}$.

Ans.

We are provided the numbers

$\text{36,}\,\text{72,}\,\text{46,}\,\text{42,}\,\text{60,}\,\text{45,}\,\text{53,}\,\text{46,}\,\text{51,}\,\text{49}$.

There is total $10$ numbers, and it is even.

Now, arranging the numbers in the ascending order, gives

$36,\,42,\,45,\,46,\,46,\,49,\,51,\,53,\,60,\,72$.

It is known that, 

Median, $M=\dfrac{{{\left( \dfrac{n}{2} \right)}^{th}}\,\,number+{{\left( \dfrac{n}{2}+1 \right)}^{th}}number}{2}$, when $n$ is even.

Therefore, the median

$M=\dfrac{{{\left( \dfrac{10}{2} \right)}^{th}}number+{{\left( \dfrac{10}{2}+1 \right)}^{th}}number}{2}$

$=\dfrac{{{5}^{th}}\,number+{{6}^{th}}\,\,number}{2}$

$=\dfrac{46+49}{2}$

$=\dfrac{95}{2}$

$=47.5$

Now, the deviations about the median $M$ for the given numbers, that is ${{x}_{i}}-M$, for $i=1,2,3,...,10$ are

$-11.5,\,-5.5,\,-2.5,\,-1.5,\,-1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$

So, the absolute values of the resulted deviations, that is $\left| {{x}_{i}}-M \right|$, for $i=1,2,3,...,10$ are

$11.5,\,5.5,\,2.5,\,1.5,\,1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$.

Thus, the mean deviation about the median $M$ for the given numbers,

$M.D.\left( M \right)=\dfrac{\sum\limits_{i=1}^{10}{\left| {{x}_{i}}-M \right|}}{10}$

$=\dfrac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10}$

$=\dfrac{70}{10}$

$=7$.

5. Use the following data to calculate the mean deviation about the mean.

Ans.

Consider the following table of data.

Thus, from the above table we have,

$N=\sum\limits_{i=1}^{5}{{{f}_{i}}}=25$,

$\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}=350}$

Therefore, the mean, 

$ \overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}}\\ $

$  =\dfrac{1}{25}\times 350\\ $

$ =14\\ $

Hence, the mean deviation about the mean is given by

$ M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}\\ $

$ =\dfrac{1}{25}\times 158\\ $

$  =6.32\\ $

6. Use the following data to calculate the mean deviation about the mean.

Ans.

Consider the following table of data.

Thus, from the above table we have,

$N=\sum\limits_{i=1}^{5}{{{f}_{i}}}=80$,

$\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}=}4000$

Therefore, the mean, 

$ \overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}}\\ $

$ =\dfrac{1}{80}\times 4000\\ $

 $ =50\\ $

Hence, the mean deviation about the mean is given by

$ M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}\\ $

$ =\dfrac{1}{80}\times 1280\\ $

$ =16\\ $.

7. Use the following data to calculate the mean deviation about the median.

Ans.

Notice that the given data is already in ascending order.

Consider the following table that shows the cumulative frequencies of the provided data.

Thus, from the above table we have,

$N=26$, that is an even number.

Therefore, the median can be obtained by calculating the mean of the ${{13}^{th}}$ and ${{14}^{th}}$ observations and both the observations belongs to the cumulative frequency $14$.

Therefore,

Median,

 $ M=\dfrac{{{13}^{th}}\,\,observation+{{14}^{th}}\,observation}{2}\\ $

$ =\dfrac{7+7}{2}\\ $

$ =7\\ $

Now, the following table shows the absolute values of the deviations about the median, that is $\left| {{x}_{i}}-M \right|$.

Thus, from the above table, we have

$\sum\limits_{i=1}^{6}{{{f}_{i}}}=26$ and $\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-M \right|}=84$.

Hence, the required mean deviation about the median,

$ M.D\left( M \right)=\dfrac{1}{N}\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-M \right|}\\ $

$ =\dfrac{1}{26}\times 84\\ $

$ =3.23\\ $

8. Use the following data to calculate the mean deviation about the median.

Ans.

Notice that the given data is already in ascending order.

Consider the following table that shows the cumulative frequencies of the provided data.

Thus, from the above table we have,

$N=29$, that is an odd number.

Therefore, the median will be the $\dfrac{29+1}{2}=\dfrac{30}{2}={{15}^{th}}$ observation.

Notice that, the ${{15}^{th}}$ observation belongs to the cumulative frequency $21$. So, the median is the corresponding observation, that is $30$.

Now, the following table shows the absolute values of the deviations about the median, that is $\left| {{x}_{i}}-M \right|$.

Thus, from the above table, we have

$\sum\limits_{i=1}^{5}{{{f}_{i}}}=29$ and $\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-M \right|}=148$.

Hence, the required mean deviation about the median,

  $ M.D\left( M \right)=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-M \right|} $

 $ =\dfrac{1}{29}\times 148 $

 $ =5.1 $

9. Use the following data to calculate the mean deviation about the mean.

Ans.

Consider the following table of data.

Thus, from the above table we have,

\[N=\sum\limits_{i=1}^{8}{{{f}_{i}}}=50\],

$\sum\limits_{i=1}^{8}{{{f}_{i}}{{x}_{i}}=}17900$

Therefore, the mean,  

$ \overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{8}{{{f}_{i}}{{x}_{i}}} $

 $ =\dfrac{1}{50}\times 17900 $

 $ =358 $

Hence, the mean deviation about the mean is given by

$ M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{8}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}\\ $

$ =\dfrac{1}{50}\times 7896\\ $

$=157.92\\ $

10. Use the following data to calculate the mean deviation about the mean.

Ans.

Consider the following table of data.

Thus, from the above table we have,

\[N=\sum\limits_{i=1}^{6}{{{f}_{i}}}=100\],

$\sum\limits_{i=1}^{6}{{{f}_{i}}{{x}_{i}}=}12530$

Therefore, the mean,

 $\overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{6}{{{f}_{i}}{{x}_{i}}} \\ $ 

 $=\dfrac{1}{100}\times 12530 \\ $ 

 $=125.3 \\ $ 

Hence, the mean deviation about the mean is given by
  $M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|} \\ $ 

 $=\dfrac{1}{100}\times 1128.8 \\ $ 

 $=11.28 \\ $ 

.

11. Use the following data including ages of $\mathbf{100}$ men to find the mean deviation about the median.

Ans.

Observe that the provided data is not continuous. So, to convert these discrete data to continuous data, subtract $0.5$ from the lower limit and add $0.5$ to the upper limit of the ages.

So, consider the following table of data.

Therefore, here $N=100$ and it is an even number.

So, $\dfrac{N}{2}=\dfrac{100}{2}=50$. That is, the cumulative frequency greater than $50$ is $63$. Thus, the median class $=$$35.5-40.5$.

Now, we known that,

Median, $M=l+\dfrac{\dfrac{N}{2}-C}{f}\times h$

We have $l=35.5$, $C=37$, $f=26$, $h=5$. Also, $N=100$.

Therefore,

$M=35.5+\dfrac{50-37}{26}\times 5=35.5+\dfrac{13\times 5}{26}=35.5+2.5=38$.

Hence, the mean deviation about the median for the given data is

$M.D.\left( M \right)=\dfrac{1}{N}\sum\limits_{i=1}^{8}{{{f}_{i}}\left| {{x}_{i}}-M \right|} \\ $ 

 $=\dfrac{1}{100}\times 735 \\ $ 

 $=7.35 \\ $ 

NCERT Solutions for Class 11 Maths Chapter 13 Statistics Exercise 13.1

Opting for the NCERT solutions for Ex 13.1 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.1 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 13 Exercise 13.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 13 Exercise 13.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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Important Topics Covered in Exercise 13.1 of Class 11 Maths NCERT Solutions

Exercise 13.1 of Class 11 Maths NCERT Solutions is based on the mean deviation which is a measure of dispersion. There are four main types of measures of dispersion which are range, standard deviation, and quartile deviation. The range of any given data gives us an overall idea of variability but does not tell us about the dispersion of the data from a measure of central tendency. For this reason, we use the standard deviation and mean. These measures can be performed on both grouped data and ungrouped data. Grouped data can be categorised into two types that are discrete frequency distribution and continuous frequency distribution. 

This exercise consists of questions on finding the mean deviation about the mean and mean deviation about the median. The solution provided in this exercise will be helpful for the students to have a better understanding of the concepts. It will be also helpful to build a solid base for learning advanced topics in Statistics.

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