Examples (Revised) - Chapter 15 - Statistics - Ncert Solutions class 11 - Maths

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Chapter 13 - Statistics | NCERT Solutions for Class 11 Maths

Example 1

Find the mean deviation about the mean for the following data:
$
6,7,10,12,13,4,8,12
$

Solution

We proceed step-wise and get the following:
Step 1 Mean of the given data is
$
\bar{x}=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9
$

Step 2 The deviations of the respective observations from the mean $\bar{x}$, i.e., $x_i-\bar{x}$ are
$
\begin{aligned}
& 6-9,7-9,10-9,12-9,13-9,4-9,8-9,12-9 \text {, } \\
& \text { or } 3,-2,1,3,4,-5,-1,3
\end{aligned}
$

Step 3 The absolute values of the deviations, i.e., $\left|x_i-\bar{x}\right|$ are
$
3,2,1,3,4,5,1,3
$

Step 4 The required mean deviation about the mean is
$
\begin{aligned}
\text { M.D. }(\bar{x}) & =\frac{\sum_{i=1}^8\left|x_i-\bar{x}\right|}{8} \\
& =\frac{3+2+1+3+4+5+1+3}{8}=\frac{22}{8}=2.75
\end{aligned}
$
$\square$ Note Instead of carrying out the steps every time, we can carry on calculation, step-wise without referring to steps.

Example 2

Find the mean deviation about the mean for the following data :
$
12,3,18,17,4,9,17,19,20,15,8,17,2,3,16,11,3,1,0,5
$

Solution

We have to first find the mean $(\bar{x})$ of the given data
$
\bar{x}=\frac{1}{20} \sum_{i=1}^{20} x_i=\frac{200}{20}=10
$

The respective absolute values of the deviations from mean, i.e., $\left|x_i-\bar{x}\right|$ are
$
2,7,8,7,6,1,7,9,10,5,2,7,8,7,6,1,7,9,10,5
$

Therefore
$
\sum_{i=1}^{20}\left|x_i-\bar{x}\right|=124
$
and
$
\text { M.D. }(\bar{x})=\frac{124}{20}=6.2
$

Example 3

Find the mean deviation about the median for the following data:
$
3,9,5,3,12,10,18,4,7,19,21 .
$

Solution

Here the number of observations is 11 which is odd. Arranging the data into ascending order, we have $3,3,4,5,7,9,10,12,18,19,21$

Now
$
\text { Median }=\left(\frac{11+1}{2}\right)^{\text {th }} \text { or } 6^{\text {th }} \text { observation }=9
$

The absolute values of the respective deviations from the median, i.e., $\left|x_i-\mathrm{M}\right|$ are
$
6,6,5,4,2,0,1,3,9,10,12
$

Therefore
$
\sum_{i=1}^{11}\left|x_i-\mathrm{M}\right|=58
$
M.D. $(\mathrm{M})=\frac{1}{11} \sum_{i=1}^{11}\left|x_i-\mathrm{M}\right|=\frac{1}{11} \times 58=5.27$

Example 3

Find the mean deviation about the median for the following data:
$
3,9,5,3,12,10,18,4,7,19,21 \text {. }
$

Solution

Here the number of observations is 11 which is odd. Arranging the data into ascending order, we have $3,3,4,5,7,9,10,12,18,19,21$

Now
$
\text { Median }=\left(\frac{11+1}{2}\right)^{\text {th }} \text { or } 6^{\text {th }} \text { observation }=9
$

The absolute values of the respective deviations from the median, i.e., $\left|x_i-\mathrm{M}\right|$ are
$
6,6,5,4,2,0,1,3,9,10,12
$

Therefore
$
\sum_{i=1}^{11}\left|x_i-\mathrm{M}\right|=58
$
M.D. $(\mathrm{M})=\frac{1}{11} \sum_{i=1}^{11}\left|x_i-\mathrm{M}\right|=\frac{1}{11} \times 58=5.27$

Example 4

$\text {}\text { Find mean deviation about the mean for the following data : }$

Solution

Let us make a Table 13.1 of the given data and append other columns after calculations

$
\begin{aligned}
& \mathrm{N}=\sum_{i=1}^6 f_i=40, \quad \sum_{i=1}^6 f_i x_i=300, \quad \sum_{i=1}^6 f_i\left|x_i-\bar{x}\right|=92 \\
& \text { Therefore } \quad \bar{x}=\frac{1}{\mathrm{~N}} \sum_{i=1}^6 f_i x_i=\frac{1}{40} \times 300=7.5 \\
&
\end{aligned}
$
M. D. $(\bar{x})=\frac{1}{\mathrm{~N}} \sum_{i=1}^6 f_i\left|x_i-\bar{x}\right|=\frac{1}{40} \times 92=2.3$

Example 5

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

Solution

The given observations are already in ascending order. Adding a row corresponding to cumulative frequencies to the given data, we get (Table 13.2).

$\text { Now, } \mathrm{N}=30 \text { which is even. }$

Median is the mean of the $15^{\text {th }}$ and $16^{\text {th }}$ observations. Both of these observations lie in the cumulative frequency 18 , for which the corresponding observation is 13 .
Therefore, Median $\mathrm{M}=\frac{15^{\text {th }} \text { observation }+16^{\text {th }} \text { observation }}{2}=\frac{13+13}{2}=13$
Now, absolute values of the deviations from median, i.e., $\left|x_i-\mathrm{M}\right|$ are shown in Table 13.3.

We have
$
\begin{aligned}
& \sum_{i=1}^8 f_i=30 \text { and } \sum_{i=1}^8 f_i\left|x_i-\mathrm{M}\right|=149 \\
& \text { Therefore } \quad \text { M.D.(M) }=\frac{1}{\mathrm{~N}} \sum_{i=1}^8 f_i\left|x_i-\mathrm{M}\right| \\
& =\frac{1}{30} \times 149=4.97 \text {. } \\
&
\end{aligned}
$

Example 6

$\text {}\text { Find the mean deviation about the mean for the following data. }$

Solution

$\text {We make the following Table } 13.4 \text { from the given data : }$

Here
$
\mathrm{N}=\sum_{i=1}^7 f_i=40, \sum_{i=1}^7 f_i x_i=1800, \sum_{i=1}^7 f_i\left|x_i-\bar{x}\right|=400
$

Therefore
$
\bar{x}=\frac{1}{\mathrm{~N}} \sum_{i=1}^7 f_i x_i=\frac{1800}{40}=45
$
M.D. $(\bar{x})=\frac{1}{\mathrm{~N}} \sum_{i=1}^7 f_i\left|x_i-\bar{x}\right|=\frac{1}{40} \times 400=10$

Example 7

$\text {}\text { Calculate the mean deviation about median for the following data : }$

Solution

$\text {Form the following Table } 13.6 \text { from the given data : }$

The class interval containing $\frac{\mathrm{N}^{\text {th }}}{2}$ or $25^{\text {th }}$ item is 20-30. Therefore, $20-30$ is the median class. We know that
$
\text { Median }=l+\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{f} \times h
$

Here $l=20, \mathrm{C}=13, f=15, h=10$ and $\mathrm{N}=50$
Therefore,
$
\text { Median }=20+\frac{25-13}{15} \times 10=20+8=28
$

Thus, Mean deviation about median is given by
M.D. $(\mathrm{M})=\frac{1}{\mathrm{~N}} \sum_{i=1}^6 f_i\left|x_i-\mathrm{M}\right|=\frac{1}{50} \times 508=10.16$

Example 8

Find the variance of the following data:
$
6,8,10,12,14,16,18,20,22,24
$

Solution

From the given data we can form the following Table 13.7. The mean is calculated by step-deviation method taking 14 as assumed mean. The number of observations is $n=10$

Therefore
Mean $\bar{x}=$ assumed mean $+\frac{\sum_{i=1}^n d_i}{n} \times h=14+\frac{5}{10} \times 2=15$ and Variance $\left(\sigma^2\right)=\frac{1}{n} \sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2=\frac{1}{10} \times 330=33$

Thus Standard deviation $(\sigma)=\sqrt{33}=5.74$

Example 9

$\text {}\text { Find the variance and standard deviation for the following data: }$

Solution

Presenting the data in tabular form (Table 13.8), we get 

$
\begin{aligned}
& \mathrm{N}=30, \sum_{i=1}^7 f_i x_i=420, \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2=1374 \\
& \text { Therefore } \\
& \bar{x}=\frac{\sum_{i=1}^7 f_i x_i}{\mathrm{~N}}=\frac{1}{30} \times 420=14 \\
& \text { variance }\left(\sigma^2\right)=\frac{1}{\mathrm{~N}} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2 \\
& =\frac{1}{30} \times 1374=45.8 \\
&
\end{aligned}
$

Hence
Standard deviation $(\sigma)=\sqrt{45.8}=6.77$

Example 10

Calculate the mean, variance and standard deviation for the following distribution:

Solution

From the given data, we construct the following Table 13.9. 

Thus $\quad$ Mean $\bar{x}=\frac{1}{\mathrm{~N}} \sum_{i=1}^7 f_i x_i=\frac{3100}{50}=62$
Variance $\left(\sigma^2\right)=\frac{1}{\mathrm{~N}} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2$
$
=\frac{1}{50} \times 10050=201
$
and $\quad$ Standard deviation $(\sigma)=\sqrt{201}=14.18$
Example 11

Find the standard deviation for the following data :

Solution

Let us form the following Table 13.10: 

Now, by formula (3), we have
$
\begin{aligned}
\sigma & =\frac{1}{\mathrm{~N}} \sqrt{\mathrm{N} \sum f_i x_i^2-\left(\sum f_i x_i\right)^2} \\
& =\frac{1}{48} \sqrt{48 \times 9652-(614)^2} \\
& =\frac{1}{48} \sqrt{463296-376996}
\end{aligned}
$

$
=\frac{1}{48} \times 293.77=6.12
$

Therefore,
Standard deviation $(\sigma)=6.12$

Examples 12

Calculate mean, variance and standard deviation for the following distribution.

Solution

Let the assumed mean $\mathrm{A}=65$. Here $h=10$ We obtain the following Table 13.11 from the given data :

Therefore
$
\bar{x}=\mathrm{A}+\frac{\sum f_i y_i}{50} \times h=65-\frac{15}{50} \times 10=62
$

Variance
$
\begin{aligned}
\sigma^2 & =\frac{h^2}{\mathrm{~N}^2}\left[\mathrm{~N} \sum f_i y_i^2-\left(\sum f_i y_i\right)^2\right] \\
& =\frac{(10)^2}{(50)^2}\left[50 \times 105-(-15)^2\right] \\
& =\frac{1}{25}[5250-225]=201
\end{aligned}
$
and standard deviation $(\sigma)=\sqrt{201}=14.18$