NCERT Solutions Class 11 Maths Chapter 13 Statistics

Exercise 13.1: In this exercise, students will learn about measures of central tendency, including mean, median, and mode, and their properties. They will also practice finding the measures of central tendency for grouped data and ungrouped data.

Exercise 13.2: This exercise focuses on measures of dispersion, including range, quartile deviation, mean deviation, and standard deviation. Students will learn about the properties of these measures and practice finding them for grouped data and ungrouped data.

Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of statistics to solve various problems and answer questions. They will also practice finding measures of central tendency and dispersion for grouped data and ungrouped data, as well as organizing data using frequency distribution tables.

Access NCERT Solutions for Class 11 Maths Chapter 13 – Statistics

Exercise 13.1

1. Find the mean deviation about the mean for the data \[{\text{4,}}\,{\text{7,}}\,{\text{8,}}\,{\text{9,}}\,{\text{10,}}\,{\text{12,}}\,{\text{13,}}\,{\text{17}}\].

Ans: Consider the given data, which is, \[4,\,7,\,8,\,9,\,10,\,12,\,13,\,17\].

Therefore, the mean of the data is,

$\overline x  = \dfrac{{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}}{8} = \dfrac{{80}}{8} = 10$ 

Observe that the deviations of the respective observations from the mean $\left( {\overline x } \right)$, that is, ${x_i} - \overline x $ can be calculated as, $ - 6,\, - 3, - 2,\, - 1,\,0,\,2,\,3,\,7$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - \overline x } \right|$ are $6,\,3,2,\,1,\,0,\,2,\,3,\,7$.

Now, the mean deviation about the mean is,

\[\therefore M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^8 {\left| {{x_i} - \overline x } \right|} }}{8} = \dfrac{{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}}{8} = \dfrac{{24}}{8} = 3\] 

2. Find the mean deviation about the mean for the data \[{\text{38,}}\,{\text{70,}}\,{\text{48,40,}}\,{\text{42,}}\,{\text{55,}}\,{\text{63,}}\,{\text{46,}}\,{\text{54,}}\,{\text{44}}\].

Ans: Consider the given data, which is, \[38,\,70,\,48,40,\,42,\,55,\,63,\,46,\,54,\,44\].

Therefore, the mean of the data is,

$\overline x  = \dfrac{{38\, + 70 + \,48 + 40 + \,42 + \,55 + \,63 + \,46 + \,54 + \,44}}{{10}} = \dfrac{{500}}{{10}} = 50$ 

Observe that the deviations of the respective observations from the mean $\left( {\overline x } \right)$, that is, ${x_i} - \overline x $ can be calculated as, $ - 12,\,20, - 2,\, - 10,\, - 8,\,5,\,13,\, - 4,\,4,\, - 6$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - \overline x } \right|$ are $12,\,20,2,\,10,\,8,\,5,\,13,\,4,\,4,\,6$.

Now, the mean deviation about the mean is,

\[\therefore M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^{10} {\left| {{x_i} - \overline x } \right|} }}{{10}} = \dfrac{{12 + \,20 + 2 + \,10 + \,8 + \,5 + \,13 + \,4 + \,4 + \,6}}{{10}} = \dfrac{{84}}{{10}} = 8.4\] 

3. Find the mean deviation about the median for the data \[{\text{13,1}}\,{\text{7,}}\,{\text{16,14,}}\,{\text{11,}}\,{\text{13,}}\,{\text{10,}}\,{\text{16,}}\,{\text{11,}}\,{\text{18,}}\,{\text{12,}}\,{\text{17}}\].

Ans: Consider the given data, which is, \[13,1\,7,\,16,14,\,11,\,13,\,10,\,16,\,11,\,18,\,12,\,17\].

Observe that the number of observations in this case is $12$, that is, even and on arranging the data in ascending order it can be obtained as, \[10,1\,1,\,11,12,\,13,\,13,\,14,\,16,\,16,\,17,\,17,\,18\]  Therefore, the median of the data is the average of the ${6^{th}}$ and the ${7^{th}}$ observations,

$\therefore M = \dfrac{{13 + 14}}{2} = \dfrac{{27}}{2} = 13.5$ 

Observe that the deviations of the respective observations from the median $\left( M \right)$, that is, ${x_i} - M$ can be calculated as, $ - 3.5,\, - 2.5, - 2.5,\, - 1.5,\, - 0.5,\, - 0.5,\,0.5,\,2.5,\,2.5,\,3.5,3.5,4.5$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - M} \right|$ are $3.5,\,2.5,2.5,\,1.5,\,0.5,\,0.5,\,0.5,\,2.5,\,2.5,\,3.5,3.5,4.5$.

Now, the mean deviation about the median is,

\[\therefore M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^{12} {\left| {{x_i} - M} \right|} }}{{12}} = \dfrac{{3.5 + \,2.5 + 2.5 + \,1.5 + \,0.5 + \,0.5 + \,0.5 + \,2.5 + \,2.5 + \,3.5 + 3.5 + 4.5}}{{12}}\]\[ \Rightarrow M.D.\left( M \right) = \dfrac{{28}}{{12}} = 2.33\] 

4. Find the mean deviation about the median for the data \[{\text{36,72,}}\,{\text{46,42,}}\,{\text{60,}}\,{\text{45,}}\,{\text{53,}}\,{\text{46,}}\,{\text{51,}}\,{\text{49}}\].

Ans: Consider the given data, which is, \[36,72,\,46,42,\,60,\,45,\,53,\,46,\,51,\,49\].

Observe that the number of observations in this case is $10$, that is, even and on arranging the data in ascending order it can be obtained as, \[36,42,\,45,46,\,46,\,49,\,51,\,53,\,60,\,72\]  Therefore, the median of the data is the average of the ${5^{th}}$ and the ${6^{th}}$ observations,

$\therefore M = \dfrac{{46 + 49}}{2} = \dfrac{{95}}{2} = 47.5$ 

Observe that the deviations of the respective observations from the median $\left( M \right)$, that is, ${x_i} - M$ can be calculated as, $ - 11.5,\, - 5.5, - 2.5,\, - 1.5,\, - 1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - M} \right|$ are $11.5,\,5.5,2.5,\,1.5,\,1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$.

Now, the mean deviation about the median is,

\[\therefore M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^{10} {\left| {{x_i} - M} \right|} }}{{10}} = \dfrac{{11.5 + \,5.5 + 2.5 + \,1.5 + \,1.5 + \,1.5 + \,3.5 + \,5.5 + \,12.5 + \,24.5}}{{10}}\]\[ \Rightarrow M.D.\left( M \right) = \dfrac{{70}}{{10}} = 7\] 

5. Find the mean deviation about the mean for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^5 {{f_i}{x_i}}  = \dfrac{1}{{25}} \times 350 = 14\] 

Therefore, the mean deviation about the mean is,

\[M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{158}}{{25}} = 6.32\]

6. Find the mean deviation about the mean for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^5 {{f_i}{x_i}}  = \dfrac{1}{{80}} \times 4000 = 50\] 

Therefore, the mean deviation about the mean is,

\[M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{1280}}{{80}} = 16\]

7. Find the mean deviation about the median for the data.

Ans: It can be clearly observed that the given observations are already in ascending order and hence on adding a column corresponding to the cumulative frequencies of the given data, the following table can be obtained as shown below,

Observe that the number of observations in this case is $26$, that is, even and therefore, the median is the mean of the ${13^{th}}$ and the ${14^{th}}$ observations. Observe that both of these observations lie in the cumulative frequency $14$, for which the corresponding observation is obtained as $7$,

$\therefore M = \dfrac{{7 + 7}}{2} = \dfrac{{14}}{2} = 7$ 

Now, the absolute values of the deviations from the median can be calculated using $\left| {{x_i} - M} \right|$ and therefore observe the table as shown below,  

Therefore, the mean deviation about the mean is,

\[M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - M} \right|} }}{N} = \dfrac{{16 + 4 + 6 + 10 + 48}}{{26}} = \dfrac{{84}}{{26}} = 3.23\]

8. Find the mean deviation about the median for the data.

Ans: It can be clearly observed that the given observations are already in ascending order and hence on adding a column corresponding to the cumulative frequencies of the given data, the following table can be obtained as shown below,

Observe that the number of observations, in this case, is $29$, which is, odd and therefore, the median is the ${15^{th}}$ observation. Observe that this observation lie in the cumulative frequency $21$, for which the corresponding observation is obtained as $30$,

$\therefore M = 30$ 

Now, the absolute values of the deviations from the median can be calculated using $\left| {{x_i} - M} \right|$ and therefore observe the table as shown below,  

Therefore, the mean deviation about the median is,

\[M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - M} \right|} }}{N} = \dfrac{{45 + 45 + 18 + 0 + 40}}{{29}} = \dfrac{{148}}{{29}} = 5.1\]

9. Find the mean deviation about the mean for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^8 {{f_i}{x_i}}  = \dfrac{1}{{50}} \times 17900 = 358\] 

Therefore, the mean deviation about the mean is,

\[M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^8 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{7896}}{{50}} = 157.92\]

10. Find the mean deviation about the mean for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^6 {{f_i}{x_i}}  = \dfrac{1}{{100}} \times 12530 = 125.3\]

Therefore, the mean deviation about the mean is,

\[M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{1128.8}}{{100}} = 11.28\]

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

Ans: Consider the given data and observe the table as shown below, which is,

The class interval containing $\frac{{{N^{th}}}}{2}$ or ${25^{th}}$ item is $20 - 30$

$\therefore $$20 - 30$ is the median class

Median = $l + \left( {\frac{{\left( {\left( {\frac{N}{2}} \right) - c} \right)}}{f}} \right) \times h$  …….(1)

Where $l = 20,\,\,c = 14,\,\,f = 14,\,\,h = 10,\,\,n = 50$

Substitute the above values in (1), we get

Median = $20 + \left( {\left( {\left( {\frac{{25 - 14}}{{14}}} \right)} \right)} \right) \times 10$

$\begin{gathered} = 20 + \left( {\left( {\frac{{11}}{{14}}} \right)} \right) \times 10 \hfill \\ \hfill \\ \end{gathered} $ $\begin{gathered} = 20 + \left( {\frac{{110}}{{14}}} \right) \hfill \\ \hfill \\ \end{gathered} $

\[ = 20 + 7.85\]

$ = 27.85$

So, \[\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - \bar x} \right|} \, = \,517.1\]

Therefore, the mean deviation about the median is,

M.D = \[\begin{gathered} \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - \bar x} \right|}  \hfill \\ \, \hfill \\ \end{gathered} \]

$ = \,\frac{1}{{50}} \times 517.1$

$ = \,10.34$

12. Calculate the mean deviation about the median age for the age distribution of ${\text{100}}$ persons.

Ans: It can be clearly observed that the given data is not continuous and therefore it needs to be converted into a continuous frequency distribution, which can be done by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval. Now, observe the table as shown below, which is,

Observe that the class interval containing the ${\left( {\dfrac{N}{2}} \right)^{th}}$ item or the ${50^{th}}$ item is $35.5 - 40.5$. Thus, $35.5 - 40.5$ is the median class.

Therefore, median of the data can be calculated as shown below,

\[\therefore M = l + \dfrac{{\dfrac{N}{2} - C}}{f} \times h\]   ( where l = 35.5, C = 37, f = 26, h = 5 and N = 100) 

\[ \Rightarrow M = 35.5 + \dfrac{{50 - 37}}{{26}} \times 5 = 35.5 + \dfrac{{13}}{{26}} \times 5 = 35.5 + 2.5 = 38\]

Therefore, the mean deviation about the median is,

\[M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^8 {{f_i}\left| {{x_i} - M} \right|} }}{N} = \dfrac{{735}}{{100}} = 7.35\]

Exercise 13.2

1. Find the mean and variance for the data \[{\text{6,}}\,{\text{7,}}\,{\text{10,}}\,{\text{12,}}\,{\text{13, 4,}}\,{\text{8,}}\,{\text{12}}\].

Ans: Consider the given data which is, \[6,\,7,\,10,\,12,\,13,{\text{ }}4,\,8,\,12\].

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{n}\sum\limits_{i = 1}^8 {{x_i}}  = \dfrac{{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}}{8} = \dfrac{{72}}{8} = 9\] 

 Now, observe the table as shown below, which is,

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^8 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n} = \dfrac{{74}}{8} = 9.25\]

2. Find the mean and variance for the first ${\text{n}}$ natural numbers.

Ans: The mean of the first $n$ natural numbers can be calculated as shown below,

\[\therefore \overline x  = \dfrac{{\dfrac{{n\left( {n + 1} \right)}}{2}}}{n} = \dfrac{{n + 1}}{2}\] 

Now, the variance can be calculated as,

\[\therefore Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}\]

\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {\left[ {{x_i} - {{\left( {\dfrac{{n + 1}}{2}} \right)}^2}} \right]} \]

\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}^2}  - \dfrac{1}{n}\sum\limits_{i = 1}^n {2\left( {\dfrac{{n + 1}}{n}} \right){x_i}}  + \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\dfrac{{n + 1}}{2}} \right)}^2}} \]

\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{\left( {n + 1} \right)(2n + 1)}}{6} - \dfrac{{{{\left( {n + 1} \right)}^2}}}{2} + \dfrac{{{{\left( {n + 1} \right)}^2}}}{4}\]

\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \left( {n + 1} \right)\left[ {\dfrac{{4n + 2 - 3n - 3}}{{12}}} \right]\]

\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{{n^2} - 1}}{{12}}\]

3. Find the mean and variance for the first ${\text{10}}$ multiples of ${\text{3}}$.

Ans: Observe that the first $10$ multiples of $3$ are, \[3,\,6,\,9,\,12,\,15,\,18,\,21,\,24,\,27,\,30\].

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{n}\sum\limits_{i = 1}^{10} {{x_i}}  = \dfrac{{3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30}}{{10}} = \dfrac{{165}}{{10}} = 16.5\] 

 Now, observe the table as shown below, which is,

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n} = \dfrac{{742.5}}{{10}} = 74.25\]

4. Find the mean and variance for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{x_i}}  = \dfrac{1}{{40}} \times 760 = 19\] 

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = \dfrac{{1736}}{{40}} = 43.4\]

5. Find the mean and variance for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{x_i}}  = \dfrac{1}{{22}} \times 2200 = 100\] 

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = \dfrac{{640}}{{22}} = 29.09\]

6. Find the mean, variance and standard deviation using the shortcut method.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h = 64 + \dfrac{0}{{100}} \times 1 = 64\] 

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 - \left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)} }^2}} \right] = \dfrac{1}{{{{\left( {100} \right)}^2}}}\left[ {100 \times 286 - 0} \right] = 2.86\]

Now the standard deviation can be calculated as shown below,

\[\therefore \sigma  = \sqrt {2.86}  = 1.69\]

7. Find the mean and variance for the following frequency distribution.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{N} \times h = 105 + \dfrac{2}{{30}} \times 30 = 107\] 

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} } \right] = \dfrac{{{{\left( {30} \right)}^2}}}{{{{\left( {30} \right)}^2}}}\left[ {30 \times 76 - {{\left( 2 \right)}^2}} \right] = 2280 - 4 = 2276\]

8. Find the mean and variance for the following frequency distribution.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{N} \times h = 25 + \dfrac{{10}}{{50}} \times 10 = 27\] 

Therefore, the variance is,

$ Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} } \right] = \dfrac{{{{\left( {10} \right)}^2}}}{{{{\left( {50} \right)}^2}}}\left[ {50 \times 68 - {{\left( {10} \right)}^2}} \right] \\$

$   \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{1}{{25}}\left[ {3400 - 100} \right] = 132  \\ $

9. Find the mean, variance and standard deviation using the shortcut method.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h = 92.5 + \dfrac{6}{{60}} \times 5 = 92.5 + 0.5 = 93\] 

Therefore, the variance is,

$Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 - \left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)} }^2}} \right] = \dfrac{{{{\left( 5 \right)}^2}}}{{{{\left( {100} \right)}^2}}}\left[ {60 \times 254 - {{\left( 6 \right)}^2}} \right]  \\$

$   \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{25}}{{3600}}\left( {15204} \right) = 105.58  \\ $

Now the standard deviation can be calculated as shown below,

\[\therefore \sigma  = \sqrt {105.58}  = 10.27\]

10. The diameters of circles (in mm) drawn in a design are given below. Find the mean, variance and standard deviation using the shortcut method.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{N} \times h = 42.5 + \dfrac{{25}}{{100}} \times 4 = 43.5\] 

Therefore, the variance is,

$  Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum\limits_{i = 1}^5 {{f_i}{y_i}^2 - \left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)} }^2}} \right] = \dfrac{{{{\left( 4 \right)}^2}}}{{{{\left( {100} \right)}^2}}}\left[ {100 \times 199 - {{\left( {25} \right)}^2}} \right]  \\$

$   \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{16}}{{10000}}\left( {19900 - 625} \right) = \dfrac{{16}}{{10000}}\left( {19275} \right) = 30.84 \\\ $

Now the standard deviation can be calculated as shown below,

\[\therefore \sigma  = \sqrt {30.84}  = 5.55\]

Miscellaneous Exercise

1. The mean and variance of eight observations are ${\text{9}}$ and ${\text{9}}{\text{.25}}$ respectively. If six of the observations are \[{\text{6,}}\,{\text{7,10,}}\,{\text{12,}}\,{\text{12}}\] and \[{\text{13}}\], find the remaining two observations.

Ans: Consider the remaining two observations to be $x$ and $y$. Thus the eight observations are \[6,\,7,10,\,12,\,12,\,13,\,x,\,y\].

Therefore, from the mean of the data it can be obtained that,

\[\overline x  = \dfrac{{6 + 7 + \,10 + 12 + \,12 + 13 + x + y}}{8} = 9\] 

\[ \Rightarrow 60 + x + y = 72\]

\[ \Rightarrow x + y = 12\] ......(i)

Again, from the variance of the data it can be obtained that,

\[{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^8 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = 9.25\]

\[\therefore \dfrac{1}{8}\left[ {{{\left( { - 3} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {x^2} + {y^2} - (2)(9)\left( {x + y} \right) + \left( {2{{\left( 9 \right)}^2}} \right)} \right] = 9.25\]

\[ \Rightarrow \dfrac{1}{8}\left[ {9 + 4 + 9 + 16 + {x^2} + {y^2} - (18)\left( {12} \right) + 162} \right] = 9.25\]   [By, using (i)]

\[ \Rightarrow \dfrac{1}{8}\left[ {{x^2} + {y^2} - 6} \right] = 9.25\]

\[ \Rightarrow {x^2} + {y^2} = 80\] ......(ii)

Observe that form equation (i), it can be obtained that,

${x^2} + {y^2} + 2xy = 144$ ......(iii)

Also, from equations (ii) and (iii), it can be obtained that,

\[2xy = 64\] ......(iv) 

Now, on subtracting equation (iv) from equation (ii), it can be obtained that,

\[{x^2} + {y^2} - 2xy = 16\] 

\[ \Rightarrow x - y =  \pm 4\] ......(v)

It can be calculated from equations (i) and (v) that, $x = 8$ and $y = 4$, when \[x - y = 4\] and $x = 4$ and $y = 8$, when \[x - y =  - 4\]. Henceforth, the remaining two observations are $4$ and $8$.

2. The mean and variance of ${\text{7}}$ observations are ${\text{8}}$ and ${\text{16}}$ respectively. If five of the observations are \[{\text{2,}}\,{\text{4,10,}}\,{\text{12}}\] and \[{\text{14}}\]. Find the remaining two observations.

Ans: Consider the remaining two observations to be $x$ and $y$. Thus the eight observations are \[2,\,4,10,\,12,\,14,\,x,\,y\].

Therefore, from the mean of the data it can be obtained that,

\[\overline x  = \dfrac{{2 + 4 + \,10 + 12 + \,14 + x + y}}{7} = 8\] 

\[ \Rightarrow 42 + x + y = 56\]

\[ \Rightarrow x + y = 14\] ......(i)

Again, from the variance of the data it can be obtained that,

\[{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^7 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = 16\]

\[\therefore \dfrac{1}{7}\left[ {{{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 6 \right)}^2} + {x^2} + {y^2} - (2)(8)\left( {x + y} \right) + \left( {2{{\left( 8 \right)}^2}} \right)} \right] = 16\]

\[ \Rightarrow \dfrac{1}{7}\left[ {36 + 16 + 4 + 16 + {x^2} + {y^2} - (16)\left( {14} \right) + 128} \right] = 16\]   [By, using (i)]

\[ \Rightarrow \dfrac{1}{7}\left[ {{x^2} + {y^2} + 12} \right] = 16\]

\[ \Rightarrow {x^2} + {y^2} = 100\] ......(ii)

Observe that form equation (i), it can be obtained that,

${x^2} + {y^2} + 2xy = 196$ ......(iii)

Also, from equations (ii) and (iii), it can be obtained that,

\[2xy = 96\] ......(iv) 

Now, on subtracting equation (iv) from equation (ii), it can be obtained that,

\[{x^2} + {y^2} - 2xy = 4\] 

\[ \Rightarrow x - y =  \pm 2\] ......(v)

It can be calculated from equations (i) and (v) that, $x = 8$ and $y = 6$, when \[x - y = 2\]and $x = 6$ and $y = 8$, when \[x - y =  - 2\]. Henceforth, the remaining two observations are $6$ and $8$.

3. The mean and standard deviation of six observations are ${\text{8}}$ and ${\text{4}}$ respectively. If each observation is multiplied by ${\text{3}}$, find the new mean and new standard deviation of the resulting observations.

Ans: Assume the observations to be \[{x_1},\,{x_2},\,{x_3},\,{x_4},\,{x_5},\,{x_6}\].

Therefore, the mean of the data is,

\[\overline x  = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8\] 

\[ \Rightarrow \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8\] ......(i)

Observe that when each of the observation is multiplied with $3$ and if we consider the resulting observations as ${y_i}$, then observe as shown below,

$\therefore {y_1} = 3{x_1}$ 

$ \Rightarrow {x_1} = \dfrac{1}{3}{y_1},\,\forall i = 1,\,2,\,3,\,....,\,6\& i \in {\mathbb{Z}^ + }$  

Now, the mean of the new data is,

\[\therefore \overline y  = \dfrac{{{y_1} + {y_2} + {y_3} + {y_4} + {y_5} + {y_6}}}{6}\] 

\[ \Rightarrow \overline y  = \dfrac{{3({x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6})}}{6}\]

\[ \Rightarrow \overline y  = 3 \times 8\]      [By using (i)]

\[ \Rightarrow \overline y  = 24\]

Therefore, the standard deviation of the data is,

\[\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}  = 4\]

\[ \Rightarrow {\left( 4 \right)^2} = \dfrac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{6}\]

\[ \Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}}  = 96\] ......(ii)

Again, it can be observed from equations (i) and (ii) that \[\overline y  = 3\overline x \] and \[\overline x  = \dfrac{{\overline y }}{3}\] and hence on substituting the values of ${x_i}$ and $\overline x $ in equation (ii) it can be clearly obtained as shown below,

 \[\therefore \sum\limits_{i = 1}^6 {{{\left( {\dfrac{{{y_i}}}{3} - \dfrac{{\overline y }}{3}} \right)}^2}}  = 96\]

\[ \Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{y_i} - \overline y } \right)}^2}}  = 864\]

Henceforth, the standard deviation of the new data can be calculated as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{{864}}{6}}  = \sqrt {144}  = 12\]

4. Given that $\overline {\text{x}} $ is the mean and $\sigma^{2}$ is the variance of ${\text{n}}$ observations ${{\text{x}}_{\text{1}}}{\text{,}}\,{{\text{x}}_{\text{2}}}{\text{,}}\,......{\text{,}}\,{{\text{x}}_{\text{n}}}$.Prove that the mean and variance of observations ${\text{a}}{{\text{x}}_{\text{1}}}{\text{,}}\,{\text{a}}{{\text{x}}_{\text{2}}}{\text{,}}\,{\text{a}}{{\text{x}}_{\text{3}}}{\text{, }}.....{\text{,}}\,{\text{a}}{{\text{x}}_{\text{n}}}$ are ${\text{a}}\overline {\text{x}} $ and ${{\text{a}}^{\text{2}}}{{\sigma}}^{\text{2}}$, respectively $\left( {{\text{a}} \ne {\text{0}}} \right)$.

Ans: Observe that the given observations are ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.

\[\therefore {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{y_i}{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}\]......(i)

Observe that when each of the observation is multiplied with $a$ and if we consider the resulting observations as ${y_i}$, then observe as shown below,

$\therefore {y_i} = a{x_i}$ 

$ \Rightarrow {x_i} = \dfrac{1}{a}{y_i},\,\forall i = 1,\,2,\,3,\,....,\,n\& i \in {\mathbb{Z}^ + }$  

Now, the mean of the new data is,

$\therefore \overline y  = \dfrac{1}{n}\sum\limits_{i = 1}^n {{y_i}}  = \dfrac{1}{n}\sum\limits_{i = 1}^n {a{x_i}}  = \dfrac{a}{n}\sum\limits_{i = 1}^n {{x_i}}  = a\overline x \,\,\,\,\,\,\,\left[ {\because \overline x  = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}} } \right]$ 

Again on substituting the values of ${x_i}$ and $\overline x $ in equation (i) it can be clearly obtained as shown below,

 \[\therefore {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {\dfrac{{{y_i}}}{a} - \dfrac{{\overline y }}{a}} \right)}^2}} }}{n}\]

\[ \Rightarrow {a^2}{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{y_i} - \overline y } \right)}^2}} }}{n}\]

Henceforth, it can be clearly proved that the mean and variance of the new data is $a\overline x $ and ${a^2}{\sigma ^2}$, respectively.

5. The mean and standard deviation of ${\text{20}}$ observations are found to be ${\text{10}}$ and ${\text{2}}$, respectively. On rechecking it was found that an observation ${\text{8}}$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If wrong item is omitted 

Ans: Observe that the incorrect number of observations, incorrect mean and the incorrect standard deviation are $20$, $10$ and $2$, respectively. ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{n} = 10\] 

\[ \Rightarrow \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{{20}} = 10\] 

\[ \Rightarrow \sum\limits_{i = 1}^{20} {{x_i}}  = 200\]

Now, the correct mean is,

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{19} {{x_i}} }}{n}\] 

\[ \Rightarrow \overline x  = \dfrac{{192}}{{19}} = 10.1\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{19} {{x_i}}  = 192} \right]\]

Again observe as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} }  = 2\]

\[ \Rightarrow 4 = \dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} \]

\[ \Rightarrow 2080 = \sum\limits_{i = 1}^{20} {{x_i}^2} \]

Therefore, the correct standard deviation of the data is,

\[\sum\limits_{i = 1}^{20} {{x_i}^2}  - {\left( 8 \right)^2}\]

\[ \Rightarrow 2080 - 64\]

\[ \Rightarrow 2016\]

\[\therefore \sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^{19} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}} \] 

\[ \Rightarrow \sigma  = \sqrt {\dfrac{{2016}}{{19}} - {{\left( {10.1} \right)}^2}} \]

\[ \Rightarrow \sigma  = \sqrt {1061.1 - 102.1}  = \sqrt {4.09}  = 2.02\]

(ii) If it is replaced by ${\text{12}}$.  

Ans: Observe that the incorrect sum of observations is $200$.

\[\therefore \sum\limits_{i = 1}^{20} {{x_i}}  = 200 - 8 + 12 = 204\]

Now, the correct mean is,

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{n}\] 

\[ \Rightarrow \overline x  = \dfrac{{204}}{{20}} = 10.2\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{20} {{x_i}}  = 204} \right]\]

Again observe as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} }  = 2\]

\[ \Rightarrow 4 = \dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} \]

\[ \Rightarrow 2080 = \sum\limits_{i = 1}^{20} {{x_i}^2} \]

Therefore, the correct standard deviation of the data is,

\[\sum\limits_{i = 1}^{20} {{x_i}^2}  - {\left( 8 \right)^2} + {\left( {12} \right)^2}\]

\[ \Rightarrow 2080 - 64 + 144\]

\[ \Rightarrow 2160\]

\[\therefore \sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^{20} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}} \] 

\[ \Rightarrow \sigma  = \sqrt {\dfrac{{2160}}{{20}} - {{\left( {10.2} \right)}^2}} \]

\[ \Rightarrow \sigma  = \sqrt {108 - 104.04}  = \sqrt {3.96}  = 1.98\]

6. The mean and standard deviation of a group of ${\text{100}}$ observations were found to be ${\text{20}}$ and ${\text{3}}$, respectively. Later on it was found that ${\text{3}}$ observations are incorrect which were recorded as ${\text{21,}}\,{\text{21}}$ and ${\text{18}}$. Find the mean and standard deviation if the incorrect  observations are omitted.

Ans: Observe that the number of observations, incorrect mean and the incorrect standard deviation are $100$, $20$ and $3$, respectively. ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{100} {{x_i}} }}{n} = 20\] 

\[ \Rightarrow \dfrac{{\sum\limits_{i = 1}^{100} {{x_i}} }}{{100}} = 20\] 

\[ \Rightarrow \sum\limits_{i = 1}^{100} {{x_i}}  = 2000\]

Now, the correct mean is,

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{97} {{x_i}} }}{n}\] 

\[ \Rightarrow \overline x  = \dfrac{{2000 - 60}}{{97}} = 20\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{97} {{x_i}}  = 1940} \right]\]

Again observe as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} }  = 3\]

\[ \Rightarrow 9 = \dfrac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} \]

\[ \Rightarrow 40900 = \sum\limits_{i = 1}^{100} {{x_i}^2} \]

Therefore, the correct standard deviation of the data is,

\[\sum\limits_{i = 1}^{100} {{x_i}^2}  - (2){\left( {21} \right)^2} - {(18)^2}\]

\[ \Rightarrow 40900 - 1206\]

\[ \Rightarrow 39694\]

\[\therefore \sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^{97} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}} \] 

\[ \Rightarrow \sigma  = \sqrt {\dfrac{{39694}}{{97}} - {{\left( {20} \right)}^2}} \]

\[ \Rightarrow \sigma  = \sqrt {409.22 - 400}  = \sqrt {9.22}  = 3.04\].

Class 11 Maths NCERT Solutions Chapter 13 - Summary

• Statistics: It is defined as the process of collection and classification of data, interpretation, and presentation of the data, and analysis of data. Statistics also is defined as concluding the sample data that is collected using experiments. Statistics is applied in various fields such as sociology, psychology, geology, probability, and so on.

• Measures of Dispersion in Statistics: The dispersion in the data is measured based on the observations data and measure of central tendency type. 

• There are different types to represent the measures of dispersion. They are Range, Mean deviation, Quartile deviation, and Standard deviation.

• Range in Statistics: The range is the difference between the maximum value and the minimum value in the given data set.

The formula of Range = Maximum Value – Minimum Value

• Mean Deviation in Statistics: The term “mean deviation” is defined as the difference between the observed value of a data point and the expected value.

• Variance and Standard Deviation: These are the two important measurements in statistics. Variance is a measure of how data values vary from the mean, on the other hand the standard deviation is the measure of the distribution of statistical data. The variance and the standard deviation are measured in different units.

• Analysis of Frequency Distributions: A frequency distribution represents the frequency of items of a data set in a graphical format or a tabular format. With the help of a frequency distribution, we will get a visual display of the frequency of data items which represents the number of times they repeated.

Measures of dispersion Range, Quartile deviation, mean deviation, variance, and standard deviation are measures of dispersion. Range $=$ Maximum Value - Minimum Value

• Mean deviation for ungrouped data

$$ \begin{aligned} & M D .(\bar{x})=\frac{\sum f_i\left(x_i-\bar{x}\right)}{N}, \\ & M D .(M)=\frac{\sum f_i\left(x_i-M\right)}{N}, \end{aligned} $$

where $N=\sum f_i$

• Variance and standard deviation for ungrouped data

$$ \begin{aligned} & \sigma^2=\frac{1}{n} \sum\left(x_l-\bar{x}\right)^2, \\ & \sigma=\sqrt{\frac{1}{n} \sum\left(x_l-\bar{x}\right)^2}, \end{aligned} $$

• Variance and standard deviation of a discrete frequency distribution

$$ \begin{aligned} & \sigma^2=\frac{1}{N} \sum f_i\left(x_i-\bar{x}\right)^2, \\ & \sigma=\sqrt{\frac{1}{N} \sum f_i\left(x_i-\bar{x}\right)^2} \end{aligned} $$

• Variance and standard deviation of a continuous frequency distribution

$$ \begin{aligned} & \sigma^2=\frac{1}{N} \sum f_i\left(x_i-\bar{x}\right)^2, \\ & \sigma=\frac{1}{N} \sqrt{N \sum f_i x_i^2-\left(\sum f_i x_i\right)^2} \end{aligned} $$

• Shortcut method to find variance and standard deviation.

$$ \begin{aligned} & \sigma^2=\frac{h^2}{N^2}\left[N \sum f_i y_i^2-\left(\sum f_i y_i\right)^2\right], \\ & \sigma=\frac{h}{N} \sqrt{N \sum f_i x_i^2-\left(\sum f_i x_i\right)^2} \text { where } \frac{x_i-A}{h} \end{aligned} $$

Coefficient of variation(C.V.) $=\frac{\sigma}{\bar{x}} \times 100, \bar{x} \neq 0$.

Class 11 Maths NCERT Solutions Chapter 13

Class 11 Maths Ch 15 NCERT Solutions shows the sums on a step-by-step basis. The concepts touched up in the chapter and the example sums are indicated below:

Mean Deviation About Mean

1. Ungrouped

Example: Finding mean deviation about the mean of 6, 7, 10, 12, 13, 12, 8, 4

2. Discrete Frequency

Example: Finding mean deviation about the mean of the below data:

3. Continuous Frequency Distribution

Example: Finding the mean deviation about the mean of the below data:

• Mean Deviation About Median

1. Ungrouped

Example: Finding the mean deviation about the median of 3, 5, 9, 3, 10, 12, 4, 18, 7, 19, 21

2. Discrete Frequency

Example: Finding mean deviation about the median of the below data:

3. Continuous Frequency Distribution

Example: Calculation of the mean deviation about median of the below data:

• Standard Deviation and Variance

1. Ungrouped Data

Example: Finding the variance of 6, 8, 10, 12, 14, 16, 18, 20, 22, 24

2. Discrete Frequency

Example: Finding the variance and standard deviation of the below data:

3. Continuous Frequency

Example: Calculation of the mean, variance and standard deviation of the following distribution: 

• Coefficient of Variation

Example: Determine which group is more variable from the below data:

• Indirect Questions

1. Multiplication of Observation

Example: Given that the mean and standard deviation of six observations are 8 and 4 respectively, if each observation is multiplied by 3, what would be the new mean and new standard deviation of the resultant observations?

2. Finding Remaining Observations

Example: The variance and mean of 7 observations are 16 and 8 respectively. If 5 of the observations amount to 2, 4, 10, 12, 14, what are the remaining 2 observations?

3. Incorrect Observation

Example: The standard deviation and mean of 100 observations are 5.1 and 40 respectively. However, a mistake was detected in case of one observation where instead of 40, 50 was taken. What would be the correct standard deviation and mean?

Overview of Deleted Syllabus for CBSE Class 11 Maths Statistics

Class 11 Maths Chapter 13: Exercises Breakdown

Conclusion

Class 11 Maths Statistics Solutions, we have thoroughly examined the principles and methods of statistics, including measures of dispersion, mean deviation, variance, and standard deviation. Understanding these concepts is crucial for interpreting and analyzing data effectively. By mastering these tools, Students can calculate and interpret measures of central tendency and dispersion, applying them to real-world data for meaningful insights. This chapter is significant in exams, with an average of 3-4 questions being asked in previous years.

Other Study Material for CBSE Class 11 Maths Chapter 13

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Important Related Links for CBSE Class 11 Maths

• CBSE Class 11 NCERT Books for Maths

• Important Questions for CBSE Class 11 Maths

• CBSE Class 11 NCERT RS Aggarwal

• CBSE Class 11 NCERT RD Sharma