Statistics
Short Answer Type Questions
1. Find the mean deviation about the mean of the distribution.
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Solution
| Size | 20 | $\frac{21}{4}$ | 22 | 24 | |
|---|---|---|---|---|---|
| Frequency | 6 | 5 | 4 | ||
| Size | Frequency | $f_{i} x_{i}$ | $d_{i}=|x_{i}-x|$ | $f_{i} d_{i}$ | |
| 20 | 6 | 120 | 1.65 | 9.90 | |
| 21 | 4 | 84 | 0.65 | 2.60 | |
| 22 | 5 | 110 | 0.35 | 1.75 | |
| 23 | 1 | 23 | 1.35 | 1.35 | |
| 24 | 4 | 96 | 2.35 | 9.40 | |
| Total | 20 | 433 | 25 | ||
| $\bar{x}=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{433}{20}=21.65$ | |||||
| $MD=\frac{\sum f_{i}|x_{i}-\bar{x}|}{\sum f_{i}}=\frac{25}{20}=1.25$ |
2. Find the mean deviation about the median of the following distribution.
| Marks obtained | 10 | 11 | 12 | 14 | 15 |
|---|---|---|---|---|---|
| Number of students | 2 | 3 | 8 | 3 | 4 |
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Solution
| Marks obtained | $f_i$ | $\boldsymbol{cf}$ | $d_i =\mid x_i-M_e\mid$ | f_id_i |
|---|---|---|---|---|
| 10 | 2 | 2 | 2 | 4 |
| 11 | 3 | 5 | 1 | 3 |
| 12 | 8 | 13 | 0 | 0 |
| 14 | 3 | 16 | 2 | 6 |
| 15 | 4 | 20 | 3 | 12 |
| Total | $\sum f_{i}=20$ | $\sum f_{i} d_{i}=25$ |
$ \begin{matrix} \text { Now, } & M_{e}=\frac{20+1}{2} \text { th item }=\frac{21}{2}=10.5 \text { th item } \\ \therefore & M_{e}=12 \\ \therefore & M D=\frac{\sum f_{i} d_{i}}{\sum f_{i}}=\frac{25}{20}=1.25 \end{matrix} $
3. Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.
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Solution
Consider first natural number when $n$ is an odd i.e., 1, 2, 3, 4, … $n$, [odd].
$ \begin{aligned} \text { Mean } \bar{x} & =\frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2} \\ \therefore \quad \text { MD } & =\frac{|1-\frac{n+1}{2}|+|2-\frac{n+1}{2}|+|3-\frac{n+1}{2}|+\ldots+|n-\frac{n+1}{2}|}{n} \\ & =\frac{+|\frac{n+1}{2}-\frac{n+1}{2}|+|\frac{n+3}{2}-\frac{n+1}{2}|+\ldots+|\frac{n+1}{2}|+|2-\frac{n+1}{2}|+\ldots+|\frac{n-1}{2}-\frac{n+1}{2}|+|n-\frac{n+1}{2}|}{n} \\ & =\frac{2}{n} 1+2+\ldots+\frac{n-3}{2}+\frac{n-1}{2} \quad \frac{n-1}{2} \text { terms } \\ & =\frac{2}{n} \frac{\frac{n-1}{2} \frac{n-1}{2}+1}{2} \quad \because \text { sum of first } n \text { natural numbers }=\frac{n(n+1)}{2} \\ & =\frac{2}{n} \cdot \frac{1}{2} \frac{n-1}{2} \frac{n+1}{2}=\frac{1}{n} \frac{n^{2}-1}{4}=\frac{n^{2}-1}{4 n} \end{aligned} $
4. Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an even number.
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Solution
Consider first $n$ natural number, when $n$ is even i.e., $1,2,3,4, \ldots \ldots . n$.
$ \begin{aligned} & \therefore \quad \text { Mean } \bar{x}=\frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2} \\ & MD=\frac{1}{n}|1-\frac{n+1}{2}|+|2-\frac{n+1}{2}|+|3-\frac{n+1}{2}|+|\frac{n-2}{2}-\frac{n+1}{2}|+|\frac{n}{2}-\frac{n+1}{2}| \\ & +|\frac{n+2}{2}-\frac{n+1}{2}|+\ldots+|n-\frac{n+1}{2}| \\ & =\frac{1}{n}|\frac{1-n}{2}|+|\frac{3-n}{2}|+|\frac{5-n}{2}|+\ldots .+|\frac{-3}{2}|+|\frac{1}{2}|+\ldots+|\frac{n-1}{2}| \\ & =\frac{2}{n} \frac{1}{2}+\frac{3}{2}+\ldots .+\frac{n-1}{2} \quad \frac{n}{2} \text { terms } \\ & =\frac{1}{n} \cdot \frac{n^{2}}{2} \quad[\because \text { sum of first } n \text { natural numbers }=n^{2}] \\ & =\frac{1}{n} \cdot \frac{n^{2}}{4}=\frac{n}{4} \end{aligned} $
5. Find the standard deviation of first $n$ natural numbers.
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Solution
| $x_i$ | 1 | 2 | 3 | 4 | 5 | $\ldots$ | $\ldots$ | $n$ |
|---|---|---|---|---|---|---|---|---|
| $(x_i)^2 $ | 1 | 4 | 9 | 16 | 25 | $\ldots$ | $\ldots$ | $n^{2}$ |
$ \text { Now, } \quad \begin{aligned} \Sigma x_{i} & =1+2+3+4+\ldots+n=\frac{n(n+1)}{2} \\ \text { and } \quad \sum x_i^{2} & =1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6} \\ \therefore \quad & =\sqrt{\frac{\sum x_i^{2}}{N}-\frac{\sum x_{i}{ }^{2}}{N}} \\ & =\sqrt{\frac{n(n+1)(2 n+1)}{6 n}-\frac{n^{2}(n+1)^{2}}{4 n^{2}}} \\ & =\sqrt{\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^{2}}{4}} \\ & =\sqrt{\frac{2(2 n^{2}+3 n+1)-3(n^{2}+2 n+1)}{12}} \\ & =\sqrt{\frac{4 n^{2}+6 n+2-3 n^{2}-6 n-3}{12}} \\ & =\sqrt{\frac{n^{2}-1}{12}} \end{aligned} $
6. The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results
Number of observation $=25$, mean $=18.2 s$, standard, deviation $=3.25 s$ Further, another set of 15 observations $x_1 x_2 \ldots x_{15}$, also in seconds, is now available and we have $\sum_{i=1}^{15} x_{i}=279$ and $\sum_{i=1}^{15} x_i^{2}=5524$. Calculate the standard derivation based on all 40 observations.
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Solution
Given,
$n_{i} =25, \bar x_i=18.2, \sigma _1=3.25 \\ $
$n_2 =15, \sum_{i=1}^{15} x_{i}=279 \text { and } \sum_{i=1}^{15} x_i^{2}=5524 $
For first set,
$ \begin{aligned} \sum x_{i} & =25 \times 18.2=455 \\ \sigma_1^{2} & =\frac{\sum x_i^{2}}{25}-(18.2)^{2} \\ 25^{2} & =\frac{\sum x_i^{2}}{25}-331.24 \\ 1.24 & =\frac{\sum x_i^{2}}{25} \\ \Sigma x_i^{2} & =25 \times(10.5625) \\ & =25 \times 341.8025 \\ & =8545.0625 \end{aligned} $
$ \begin{matrix} \therefore & \sigma_1^{2}=\frac{\Sigma x_i^{2}}{25}-(18.2)^{2} \\ \Rightarrow & (3.25)^{2}=\frac{\Sigma x_i^{2}}{25}-331.24 \\ \Rightarrow & 10.5625+331.24=\frac{\Sigma x_i^{2}}{25} \\ \Rightarrow & \Sigma x_i^{2}=25 \times(10.5625+331.24) \end{matrix} $
For combined SD of the 40 observations $n=40$,
$ \begin{aligned} & \text { Now } \quad \Sigma x_i^{2}=5524+8545.0625=14069.0625 \\ & \text { and } \quad \Sigma x_{i}=455+279=734 \\ & \therefore \quad SD=\sqrt{\frac{14069.0625}{40}-\frac{734}40^{2}} \\ & =\sqrt{351.726-(18.35)^{2}} \\ & =\sqrt{351.726-336.7225} \\ & =\sqrt{15.0035}=3.87 \end{aligned} $
7. The mean and standard deviation of a set of $n_1$ observations are $\bar x_1$ and $s_1$, respectively while the mean and standard deviation of another set of $n_2$ observations are $\bar x_2$ and $s_2$, respectively. Show that the standard deviation of the combined set of $(n_1+n_2)$ observations is given by
$ SD={\sqrt{\frac{n_1(s_1)^{2}+n_2(s_2)^{2}}{n_1+n_2}+\frac{n_1 n_2(\bar x_1- \bar x_2)^{2}}{(n_1-n_2)^{2}}}} $
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Solution
Let
$ x_{i}, i=1,2,3 \ldots, n_1 \text { and } y_{j}, j=1,2,3, \ldots, n_2 $
$ \therefore \bar x_1=\frac{1}{n_1} \sum_{i=1}^{n_1} x_{i} \text { and } \bar x_2=\frac{1}{n_2} \sum_{j=1}^{n_2} y_{j} \\ $
$\Rightarrow \sigma _1^2= \frac{1}{n_1} \sum _{i=1}^{n_1} (x_i - \bar x_1)^{2} $
and
$\sigma _2^{2}= \frac{1}{n_2} \sum _{j=1}^{n}(y_j - \bar x_2)^{2} $
Now, mean $\bar{x}$ of the given series is given by
$ \bar {x}= \frac{1}{n_1+n_2} \sum _{i=1}^{n_1} x_i + \sum _{j=1}^{n_2} y_j = \frac{n_1 \bar x_1+n_2 \bar x_2}{n_1+n_2} $
The variance $\sigma^{2}$ of the combined series is given by
$ \sigma^{2} =\Big[\frac{1}{n_1+n_2} \sum_{i=1}^{n_1}(x_{i}-\bar{x})^{2}+\sum_{j=1}^{n_2}(y_{j}-\bar{x})^{2}\Big] \\ $
Now,
$\sum_{i=1}^{n_1}(x_{i}-\bar{x})^{2} = \sum _{i=1}^{n_1} (x_i - \bar x_j - \bar x)^2 \\ $
But $\quad \sum _{i=1}^{n_1}(x_i - \bar x_i)=0 $
[algebraic sum of the deviation of values of first series from their mean is zero]
Also,
$ \sum _{i=1}^{n_1} (x_i-\bar{x})^2= n_1 s_1^2+n_1(\bar x_1-\bar{x})^2=n_1 s_1^2+n_1 d_1^2 $
Where,
$ d_1=(\bar x_1-\bar{x}) $
Similarly, $\quad \sum_{j=1}^{n_2}(y_{j}-\bar{x})^{2}=\sum _{j=1}^{n_2}(y_j-\bar x_i+\bar x_i-\bar{x})^2=n_2 s_2^2+n_2 d_2^2$
where,
$ d_2=\bar x_2-\bar x $
Combined SD
$ \sigma=\sqrt{\frac{[n_1(s_1^{2}+d_1^{2})+n_2(s_2^{2}+d_2^{2})]}{n_1+n_2}} $
where,
$d_1=\bar x_1-\bar x=\bar x_1-\frac{n_1 \bar x_1+n_2 \bar x_2}{n_1+n_2}=\frac{n_2(\bar x_1-\bar x_2)}{n_1+n_2} $
and
$d_2=\bar x_2-\bar{x}=\bar x_2-\frac{n_1 \bar x_1+n_2 \bar x_2}{n_1+n_2}=\frac{n_1(\bar x_2-\bar x_1)}{n_1+n_2} $
$\therefore \quad \sigma^{2}=\frac{1}{n_1+n_2} n_1 s_1^{2}+n_2 s_2^{2}+\frac{n_1 n_2(\bar x_1-\bar x_2)^{2}}{(n_1+n_2)^{2}}+\frac{n_2 n_1(\bar x_2-\bar x_1)^{2}}{(n_1+n_2)^{2}}$
Also,
$ \sigma=\sqrt{\frac{n_1 s_1^{2}+n_2 s_2^{2}}{n_1+n_2}+\frac{n_1 n_2(\bar x_1-\bar x_2)^{2}}{(n_1+n_2)^{2}}} $
8. Two sets each of 20 observations, have the same standard deviation 5. The first set has a mean 17 and the second mean 22 .
Determine the standard deviation of the $x$ sets obtained by combining the given two sets.
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Solution
Given, $n_1=20, \sigma _1=5, \bar x_1=17$ and $n_2=20, \sigma _2=5, \bar x_2=22$
We know that, $\sigma=\sqrt{\frac{n_1 s_1^{2}+n_2 s_2^{2}}{n_1+n_2}+\frac{n_1 n_2(\bar x_1-\bar x_2)^{2}}{(n_1+n_2)^{2}}}$
$ \begin{aligned} & =\sqrt{\frac{20 \times(5)^{2}+20 \times(5)^{2}}{20+20}+\frac{20 \times 20(17-22)^{2}}{(20+20)^{2}}} \\ & =\sqrt{\frac{1000}{40}+\frac{400 \times 25}{1600}}=\sqrt{25+\frac{25}{4}}=\sqrt{\frac{125}{4}}=\sqrt{31.25}=5.59 \end{aligned} $
9. The frequency distribution
| $\boldsymbol{x}$ | $A$ | $2 A$ | $3 A$ | $4 A$ | $5 A$ | $6 A$ |
|---|---|---|---|---|---|---|
| $\boldsymbol{f}$ | 2 | 1 | 1 | 1 | 1 | 1 |
where, $A$ is a positive integer, has a variance of 160 . Determine the value of $A$.
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Solution
| $x$ | $f_i$ | $f_i x_i$ | $f_i x_i^2$ |
|---|---|---|---|
| $A$ | 2 | $2 A$ | $2 A^{2}$ |
| $2 A$ | 1 | $2 A$ | $4 A^{2}$ |
| $3 A$ | 1 | $3 A$ | $9 A^{2}$ |
| $4 A$ | 1 | $4 A$ | $16 A^{2}$ |
| $5 A$ | 1 | $5 A$ | $25 A^{2}$ |
| $6 A$ | 1 | $6 A$ | $36 A^{2}$ |
| Total | 7 | $22 A$ | $92 A^{2}$ |
| $n=7$ | $\Sigma f_{i} n_{i}=22 A$ | $\Sigma f_{i} n_i^{2}=92 A^{2}$ |
$ \begin{matrix} \therefore & \sigma^{2}=\frac{\Sigma f_{i} x_i^{2}}{n}-\frac{\Sigma f_{i} x_{i}}{n} \\ \Rightarrow & 160=\frac{92 A^{2}}{7}-\frac{22 A^{2}}{7} \\ \Rightarrow & 160=\frac{92 A^{2}}{7}-\frac{484 A^{2}}{49} \\ \Rightarrow & 160=(644-484) \frac{A^{2}}{49} \\ \Rightarrow & 160=\frac{160 A^{2}}{49} \Rightarrow A^{2}=49 \\ \therefore & A=7 \end{matrix} $
10. For the frequency distribution
| $\boldsymbol{x}$ | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|
| $\boldsymbol{f}$ | 4 | 9 | 16 | 14 | 11 | 6 |
Find the standard distribution.
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Solution
| $x_i$ | $f_i$ | $d_i = x_i - 4$ | $f_i d_i$ | $f_i d_i^2$ |
|---|---|---|---|---|
| 2 | 4 | -2 | -8 | 16 |
| 3 | 9 | -1 | -9 | 9 |
| 4 | 16 | 0 | 0 | 0 |
| 5 | 14 | 1 | 14 | 14 |
| 6 | 11 | 2 | 22 | 44 |
| 7 | 6 | 3 | 18 | 54 |
| Total | 60 | $\Sigma f_{i} d_{i}=37$ | $\Sigma f_{i} d_i^{2}=137$ | |
| $\therefore \quad SD$ | $=\sqrt{\frac{\Sigma f_{i} d_i^{2}}{N}-\frac{\Sigma f_{i} d_i^{2}}{N}}$ | |||
| $=\sqrt{\frac{137}{60}-\frac{37^{2}}{60}}$ | ||||
| $=\sqrt{2.2833-(0.616)^{2}}$ | ||||
| $=\sqrt{2.2833-0.3794}$ | ||||
| $=\sqrt{1.9037}=1.38$ |
11. There are 60 students in a class. The following is the frquency distribution of the marks obtained by the students in a test.
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Frequency | $x-2$ | $x$ | $x^{2}$ | $(x+1)^{2}$ | $2 x$ | $x+1$ |
where, $x$ is positive integer. Determine the mean and standard deviation of the marks
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Solution
$\therefore$ Sum of frequencies,
$\begin{aligned} & x-2+x+x^2+(x+1)^2+2 x+x+1=60 \\ & \Rightarrow \quad 2 x-2+x^2+x^2+1+2 x+2 x+x+1=60 \\ & \Rightarrow \quad 2 x^2+7 x=60 \\ & \Rightarrow \quad 2 x^2+7 x-60=0 \\ & \Rightarrow \quad 2 x^2+15 x-8 x-60=0 \\ & \Rightarrow \quad x(2 x+15)-4(2 x+15)=0 \\ & \Rightarrow \quad(2 x+15)(x-4)=0 \ & \Rightarrow \\ & \Rightarrow x=-\frac{15}{2}, 4 \\ & \Rightarrow x=-\frac{15}{2} \ & \text { [inaddmisible] }\left[\because x \in /^{+}\right] \\ & \end{aligned}$
$\begin{array}{|c|c|c|c|c|} \hline x_{i} & f_{i} & d_{i}=x_{i}-3 & f_{i} d_{i} & f_{i} d_i^2 \\ \hline 0 & 2 & -3 & -6 & 18 \\ 1 & 4 & -2 & -8 & 16 \\ 2 & 16 & -1 & -16 & 16 \\ A=3 & 25 & 0 & 0 & 0 \\ 4 & 8 & 1 & 8 & 8 \\ 5 & 5 & 2 & 10 & 20 \\ \hline \text { Total } & \Sigma f_i=60 & & \Sigma f_i d_i=-12 & \Sigma f_i d_i^2=\mathbf{7 8} \\ \hline \end{array}$
$\begin{aligned} \text { Mean } & =A+\frac{\sum f_i d_i}{\Sigma f_i}=3+\frac{-12}{60}=2.8 \\ \sigma & =\sqrt{\frac{\sum f_i d_i^2}{\Sigma f_i}-\frac{\sum f_i d_i^2}{\Sigma f_i}}=\sqrt{\frac{78}{60}-\frac{-12^2}{60}} \\ & =\sqrt{1.3-0.04}=\sqrt{1.26}=1.12\end{aligned}$
12. The mean life of a sample of 60 bulbs was $650 h$ and the standard deviation was $8 h$. If a second sample of 80 bulbs has a mean life of 660 $h$ and standard deviation $7 h$, then find the over all standard deviation.
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Solution
Here, $n_1=60, \bar x_1=650, s_1=8$ and $n_2=80, \bar x_2=660, s_2=7$
$\therefore \quad \sigma =\sqrt{\frac{n_1 s_1{ }^{2}+n_2 s_2{ }^{2}}{n_1+n_2}+\frac{n_1 n_2(\bar x_1-\bar x_2)^{2}}{(n_1+n_2)^{2}}} \\ $
$=\sqrt{\frac{60 \times(8)^{2}+80 \times(7)^{2}}{60+80}+\frac{60 \times 80(650-660)^{2}}{(60+80)^{2}}} \\ $
$=\sqrt{\frac{6 \times 64+8 \times 49}{14}+\frac{60 \times 80 \times 100}{140 \times 140}} \\ $
$=\sqrt{\frac{192+196}{7}+\frac{1200}{49}}=\sqrt{\frac{388}{7}+\frac{1200}{49}} \\ $
$=\sqrt{\frac{2716+1200}{49}}=\sqrt{\frac{3916}{49}}=\frac{62.58}{7}=8.9 $
13. If mean and standard deviation of 100 items are 50 and 4 respectively, then find the sum of all the item and the sum of the squares of item.
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Solution
Here, $\bar{x}=50, n=100$ and $\sigma=4$
$ \begin{aligned} & \therefore \quad \frac{\Sigma x_{i}}{100}=50 \\ & \Rightarrow \quad \Sigma x_{i}=5000 \\ & \text { and } \quad \sigma^{2}=\frac{\Sigma f_{i} x_i^{2}}{\Sigma f_{i}}-\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}} \\ & \Rightarrow \quad(4)^{2}=\frac{\Sigma f_{i} x_i^{2}}{100}-(50)^{2} \\ & \Rightarrow \quad 16=\frac{\Sigma f_{i} x_i^{2}}{100}-2500 \\ & \Rightarrow \quad \frac{\Sigma f_{i} x_i^{2}}{100}=16+2500 \Rightarrow 2516 \\ & \therefore \quad \sum f_{i} x_i^{2}=251600 \end{aligned} $
14. If for distribution $\Sigma(x-5)=3, \Sigma(x-5)^{2}=43$ and total number of item is 18 . Find the mean and standard deviation.
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Solution
Given,
$ \begin{aligned} n & =18, \Sigma(x-5)=3 \text { and } \Sigma(x-5)^{2}=43 \\ & =5+\frac{3}{18}=5+0.1666=5.1666=5.17 \\ \text { SD } & =\sqrt{\frac{\sum(x-5)^{2}}{n}-\frac{\sum(x-5)^{2}}{n}} \\ & =\sqrt{\frac{43}{18}-\frac{3}{18}} \\ & =\sqrt{2.3944-(0.166)^{2}}=\sqrt{2.3944-0.2755}=1.59 \end{aligned} $
$ \begin{aligned} & \therefore \quad \text { Mean }=A+\frac{\sum(x-5)}{18} \\ & \text { and } \end{aligned} $
15. Find the mean and variance of the frequency distribution given below.
| $\boldsymbol{x}$ | $1 \leq x \leq 3$ | $3 \leq x \leq 5$ | $5 \leq x \leq 7$ | $7 \leq x \leq 10$ |
|---|---|---|---|---|
| $\boldsymbol{f}$ | 6 | 4 | 5 | 1 |
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Solution
| $x$ | $f_i$ | $x_i$ | $f_i x_i$ | $f_i x_i^2$ |
|---|---|---|---|---|
| $1-3$ | 6 | 2 | 12 | 24 |
| $3-5$ | 4 | 4 | 16 | 64 |
| $5-7$ | 5 | 6 | 30 | 180 |
| $7-10$ | 1 | 8.5 | 8.5 | 72.25 |
| Total | $n=16$ | $\Sigma f_{i} x_{i}=66.5$ | $\Sigma f_{i} x_i^{2}=340.25$ |
$\therefore \quad$ Mean $=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{66.5}{16}=4.15$
$ \begin{aligned} \text { variance } & =\sigma^{2}=\frac{\Sigma f_{i} x_i^{2}}{\Sigma f_{i}}-\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}} \\ & =\frac{340.25}{16}-(4.15)^{2} \\ & =21.2656-17.2225=4.043 \end{aligned} $
Long Answer Type Questions
16. Calculate the mean deviation about the mean for the following frequency distribution.
$\begin{array}{|c|c|c|c|c|c|} \hline \text { Class interval } & 0-4 & 4-8 & 8-12 & 12-16 & 16-20 \\ \hline \text { Frequency } & 4 & 6 & 8 & 5 & 2 \\ \hline \end{array}$
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Solution
$$ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class internal} & f _i & x _i & f_i x _i & d _i=x _i- \overline x \mid & f _i d _i \\ \hline 0-4 & 4 & 2 & 8 & 7.2 & 28.8 \\ 4-8 & 6 & 6 & 36 & 3.2 & 19.2 \\ 8-12 & 8 & 10 & 80 & 0.8 & 6.4 \\ 12-16 & 5 & 14 & 70 & 4.8 & 24.0 \\ 16-20 & 2 & 18 & 36 & 8.8 & 17.6 \\ \hline Total & \Sigma f _i=25 & & \Sigma f _i x _i=230 & & \Sigma f _i d _i=96 \\ \hline \end{array} $$
$\begin{aligned} & \therefore \quad \text { Mean }=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{230}{25}=9.2 \\ & \text { and } \quad \text { mean deviation }=\frac{\Sigma f d_i}{\Sigma f_i}=\frac{96}{25}=3.84 \\ & \end{aligned}$
17. Calculate the mean deviation from the median of the following data.
| Class interval | $0-6$ | $6-12$ | $12-18$ | $18-24$ | $24-30$ |
|---|---|---|---|---|---|
| Frequency | 4 | 5 | 3 | 6 | 2 |
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Solution
| Class interval | $f_i$ | $x_i$ | $cf$ | $d_i = \mid x_i - \bar m_d \mid$ | $f_i d_i$ | |
|---|---|---|---|---|---|---|
| $0-6$ | 4 | 3 | 4 | 11 | 44 | |
| 6-12 | 5 | 9 | 9 | 5 | 25 | |
| $12-18$ | 3 | 15 | 12 | 1 | 3 | |
| $18-24$ | 6 | 21 | 18 | 7 | 42 | |
| $24-30$ | 2 | 27 | 20 | 13 | 26 | |
| Total | $N=20$ | $\frac{N}{2}=\frac{20}{2}=10$ |
So, the median class is $12-18$.
$ \begin{aligned} \quad \text { Median } & =l+\frac{\frac{N}{2}-c f}{f} \times i \\ & =12+\frac{6}{3}(10-9) \\ & =12+2=14 \\ MD & =\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}=\frac{140}{20}=7 \end{aligned} $
18. Determine the mean and standard deviation for the following distribution.
| Marks | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Frequency | 1 | 6 | 6 | 8 | 8 | 2 | 2 | 3 | 0 | 2 | 1 | 0 | 0 | 0 | 1 |
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Solution
| Marks | $f_i$ | $f_i x_i$ | $d_i = x_i - \bar x$ | $f_i d_i$ | $f_i d_i^2$ |
|---|---|---|---|---|---|
| 2 | 1 | 2 | $2-6=-4$ | -4 | 16 |
| 3 | 6 | 18 | $3-6=-3$ | -18 | 54 |
| 4 | 6 | 24 | $4-6=-2$ | -12 | 24 |
| 5 | 8 | 40 | $5-6=-1$ | -8 | 8 |
| 6 | 8 | 48 | $6-6=0$ | 0 | 0 |
| 7 | 2 | 14 | $7-6=1$ | 2 | 2 |
| 8 | 2 | 16 | $8-6=2$ | 4 | 8 |
| 9 | 3 | 27 | $9-6=3$ | 9 | 27 |
| 10 | 0 | 0 | $10-6=4$ | 0 | 0 |
| 11 | 2 | 22 | $11-6=5$ | 10 | 50 |
| 12 | 1 | 12 | $12-6=6$ | 6 | 36 |
| 13 | 0 | 0 | $13-6=7$ | 0 | 0 |
| 14 | 0 | 0 | $14-6=8$ | 0 | 0 |
| 15 | 0 | 0 | $15-6=9$ | 0 | 0 |
| 16 | 1 | 16 | $16-6=10$ | 10 | 100 |
| Total | $\boldsymbol{\Sigma} f_{i}=40$ | $\Sigma f_{i} x_{i}=239$ | $\Sigma f_{i} d_{i}=-1$ | $\Sigma f_{i} x_i^{2}=325$ |
$ \begin{aligned} & \therefore \quad \text { Mean } \bar{x}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{239}{40}=5.975 \approx 6 \\ & \text { and } \\ & \begin{aligned} \sigma & =\sqrt{\frac{\Sigma f_{i} d_i^{2}}{\Sigma f_{i}}-\frac{\Sigma f_{i} d_i^{2}}{\Sigma f_{i}}}=\sqrt{\frac{325}{40}-\frac{-1}40^{2}} \\ & =\sqrt{8.125-0.000625}=\sqrt{8.124375}=2.85 \end{aligned} \end{aligned} $
19. The weights of coffee in 70 jars is shown in the following table
| Weight (in g) | Frequency |
|---|---|
| $200-201$ | 13 |
| $201-202$ | 27 |
| $202-203$ | 18 |
| $203-204$ | 10 |
| $204-205$ | 1 |
| $205-206$ | 1 |
Determine variance and standard deviation of the above distribution.
Show Answer
Solution
| Cl | $f_{i}$ | $x_{i}$ | $d_{i}=x_{i}-\bar{x}$ | $f_{i} d_{i}$ | $f_{i} d_i^{2}$ |
|---|---|---|---|---|---|
| 200-201 | 13 | 200.5 | -2 | -26 | 52 |
| $201-202$ | 27 | 201.5 | -1 | -27 | 27 |
| 202-203 | 18 | 202.5 | 0 | 0 | 0 |
| 203-204 | 10 | 203.5 | 1 | 10 | 10 |
| 204-205 | 1 | 204.5 | 2 | 2 | 4 |
| 205-206 | 1 | 205.5 | 3 | 3 | 9 |
$\Sigma f_{i}=70$ $\Sigma f_{i} d_{i}=$
-38 $\Sigma f_{i} d_i^{2}=102$
| $\frac{T f_{i}}{\sum f_{i}}$
Now,
$=1.4571-0.2916=1.1655 \\ \sigma =\sqrt{1.1655}=1.08 g$
20. Determine mean and standard deviation of first $n$ terms of an AP whose first term is a and common difference is $d$.
Show Answer
Solution
| $x_i$ | $xx_i - a$ | $(x_i - a)^2$ |
|---|---|---|
| $a$ | 0 | 0 |
| $a+d$ | $d$ | $d^{2}$ |
| $a+2 d$ | $2 d$ | $4 d^{2}$ |
| $\ldots \ldots$ | $\ldots \ldots$ | $9 d^{2}$ |
| $\ldots \ldots$ | $\ldots \ldots$. | |
| $\ldots \ldots$ | $\ldots \ldots$. | |
| $a+(n-1) d$ | $(n-1) d$ | $(n-1)^{2} d^{2}$ |
| $\sum x_{i}=\frac{n}{2}[2 a+(n-1)]$ | ||
| Mean $=\frac{\sum x_{i}}{n}=\frac{1}{n} \frac{n}{2}(2 a)+(n-1) d$ | ||
| $=$ | $a+\frac{(n-1)}{2} d$ |
$ \begin{aligned} \Sigma(x_{i}-a) & =d[1+2+3+\ldots+(n-1) d] \\ & =d \frac{(n-1) n}{2} \\ \text { and } \quad \sum(x_{i}-a)^{2} & =d^{2}[1^{2}+2^{2}+3^{2}+\ldots+(n-1)^{2}] \\ & =\frac{d^{2}(n-1) n(2 n-1)}{6} \\ \sigma & =\sqrt{\frac{(x_{i}-a)^{2}}{n}-\frac{x_{i}-a^{2}}{n}} \\ & =\sqrt{\frac{d^{2}(n-1)(n)(2 n-1)}{6 n}-\frac{d(n-1) n{ }^{2}}{2 n}} \\ & =\sqrt{\frac{d^{2}(n-1)(2 n-1)}{6}-\frac{d^{2}(n-1)^{2}}{4}} \\ & =d \sqrt{\frac{(n-1)(2 n-1)}{6}-\frac{(n-1)^{2}}{4}} \\ & =d \sqrt{\frac{(n-1)}{2} \frac{2 n-1}{3}-\frac{n-1}{2}} \\ & =d \sqrt{\frac{(n-1)}{2} \frac{4 n-2-3 n+3}{6}} \\ & =d \sqrt{\frac{(n-1)(n+1)}{12}}=d \sqrt{\frac{(n^{2}-1)}{12}} \end{aligned} $
21. Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests
| Ravi | 25 | 50 | 45 | 30 | 70 | 42 | 36 | 48 | 35 | 60 |
|---|---|---|---|---|---|---|---|---|---|---|
| Hashina | 10 | 70 | 50 | 20 | 95 | 55 | 42 | 60 | 48 | 80 |
Who is more intelligent and who is more consistent?
Show Answer
Solution
For Ravi,
| $x_i$ | $d_i = x_i - 45$ | $d_i^2$ |
|---|---|---|
| 25 | -20 | 400 |
| 50 | 5 | 25 |
| 45 | 0 | 0 |
| 30 | -15 | 225 |
| 70 | 25 | 625 |
| 42 | -3 | 9 |
| 36 | -9 | 81 |
| 48 | 3 | 9 |
| 35 | -10 | 100 |
| 60 | 15 | 225 |
| Total | $\boldsymbol{\Sigma} d_{i}=-14$ | $\Sigma d^{2} _{i}=1699$ |
Now,
$ \begin{aligned} \sigma & =\sqrt{\frac{\Sigma d^{2} i}{n}-\frac{\Sigma d_{i}{ }^{2}}{n}} \\ & =\sqrt{\frac{1699}{10}-\frac{-14^{2}}{10}}=\sqrt{169.9-0.0196} \\ & =\sqrt{169.88}=13.03 \\ \bar{x} & =A+\frac{\Sigma d_{i}}{\Sigma f_{i}}=45-\frac{14}{10}=43.6 \end{aligned} $
For Hashina,
| $x_i$ | $d_i = x_i - 55$ | $d_i^2$ |
|---|---|---|
| 10 | -45 | 2025 |
| 70 | 25 | 625 |
| 50 | -5 | 25 |
| 20 | -35 | 1225 |
| 95 | 40 | 1600 |
| 55 | 0 | 0 |
| 42 | -13 | 169 |
| 60 | 5 | 25 |
| 48 | -7 | 49 |
| 80 | 25 | 625 |
| Total | $\boldsymbol{\Sigma} d_{i}=0$ | $\sum d_i^{2}=6368$ |
$ \begin{aligned} & \because \quad \text { Mean }=55 \\ & \therefore \quad \sigma=\sqrt{\frac{6368}{10}}=\sqrt{636.8}=25.2 \\ & CV=\frac{\sigma}{\bar{x}} \times 100=\frac{13.03}{43.6} \times 100=29.88 \\ & CV=\frac{\sigma}{\bar{x}} \times 100=\frac{25.2}{55} \times 100=45.89 \end{aligned} $
Hence, Hashina is more consistent and intelligent.
22. Mean and standard deviation of 100 observations were found to be 40 and 10 , respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, then find the correct standard deviation.
Show Answer
Solution
Given,
$ \begin{aligned} & n=100, \bar{x}=40, \sigma=10 \text { and } \bar{x}=40 \\ & \therefore \frac{\sum x_{i}}{n}=40 \\ \Rightarrow & \frac{\Sigma x_{i}}{100}=40 \\ & \Rightarrow \Sigma x_{i}=4000 \\ & \text { Corrected } \Sigma x_{i}=4000-30-70+3+27 \\ & =4030-100=3930 \\ & \text { Corrected mean }=\frac{2930}{100}=39.3 \end{aligned} $
$ \begin{aligned} \sigma^{2} & =\frac{\Sigma x_i^{2}}{n}-(40)^{2} \\ 100 & =\frac{\Sigma x_i^{2}}{100}-1600 \\ \Sigma x_i^{2} & =170000 \end{aligned} $
Corrected $\Sigma x_i^{2}=170000-(30)^{2}-(70)^{2}+3^{2}+(27)^{2}$
$ \begin{aligned} & =164939 \\ \text { Corrected } \sigma & =\sqrt{\frac{164939}{100}-(39.3)^{2}} \\ & =\sqrt{1649.39-39.3 \times 39.3} \\ & =\sqrt{1649.39-1544.49} \\ & =\sqrt{104.9}=10.24 \end{aligned} $
23. While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25 . He obtained the mean and variance as 45 and 16, respectively. Find the correct mean and the variance.
Show Answer
Solution
Given,
$ \begin{aligned} & n=10, \bar{x}=45 \text { and } \sigma^{2}=16 \\ & \bar{x}=45 \Rightarrow \frac{\Sigma x_{i}}{n}=45 \\ & \frac{\Sigma x_{i}}{10}=45 \Rightarrow \Sigma x_{i}=450 \end{aligned} $
$ \therefore $
Corrected $\Sigma x_{i}=450-52+25=423$
$\therefore$
$\bar{x}=\frac{423}{10}=42.3$
$\Rightarrow$
$\sigma^{2}=\frac{\Sigma x_i^{2}}{n}-\frac{\Sigma x_{i}}{n}$
$\Rightarrow$
$16=\frac{\Sigma x_i^{2}}{10}-(45)^{2}$
$\Rightarrow$
$\Sigma x_i^{2}=10(2025+16)$
$\Rightarrow$
$\Sigma x_i^{2}=20410$
$\therefore$
Corrected $\Sigma x_i^{2}=20410-(52)^{2}+(25)^{2}=18331$
and
corrected $\sigma^{2}=\frac{18331}{10}-(42.3)^{2}=43.81$
Objective Type Questions
24. The mean deviation of the data $3,10,10,4,7,10,5$ from the mean is
(a) 2
(b) 2.57
(c) 3
(d) 3.75
Show Answer
Solution
(b) Given, observations are $3,10,10,4,7,10$ and 5.
| $\therefore$ | $=\frac{49}{7}=7$ | |
|---|---|---|
| $x_{i}$ | $d_{i}=x_{i}-\bar{x}$ | |
| 3 | 4 | |
| 10 | 3 | |
| 10 | 3 | |
| 4 | 3 | |
| 7 | 0 | |
| 10 | 3 | |
| 5 | 2 | |
| Total | $\Sigma d_{i}=18$ | |
| Now, | $MD=\frac{\sum d_{i}}{N}=\frac{18}{7}=2.57$ |
25. Mean deviation for $n$ observations $x_1, x_2, \ldots, x_{n}$ from their mean $\bar{x}$ is given by
(a) $\sum_{i=1}^{n}(x_{i}-\bar{x})$
(b) $\frac{1}{n} \sum_{i=1}^{n}|x_{i}-\bar{x}|$
(c) $\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}$
(d) $\frac{1}{n} \sum_{i=1}^{n}(x_{i}-\bar{x})^{2}$
Show Answer
Solution
(b) $MD=\frac{1}{n} \sum_{i=1}^{n}|x_{i}-\bar{x}|$
26. When tested, the lives (in hours) of 5 bulbs were noted as follows
$ 1357,1090,1666,1494,1623 $
The mean deviations (in hours) from their mean is
(a) 178
(b) 179
(c) 220
(d) 356
Show Answer
Solution
(a) Since, the lives of 5 bulbs are 1357, 1090, 1666, 1494 and 1623.
$ \begin{aligned} \therefore \quad \text { Mean } & =\frac{1357+1090+1666+1494+1623}{5} \\ & =\frac{7230}{5}=1446 \end{aligned} $
| $x_{\boldsymbol{i}}$ | $d_i = \mid x_i - \bar x \mid$ |
|---|---|
| 1357 | 89 |
| 1090 | 356 |
| 1666 | 220 |
| 1494 | 48 |
| 1623 | 177 |
| Total | $\sum d_{i}=890$ |
| MD $=\frac{\Sigma d_{i}}{N}=\frac{890}{5}=178$ |
27. Following are the marks obtained by 9 students in a mathematics test $50,69,20,33,53,39,40,65,59$
The mean deviation from the median is
(a) 9
(b) 10.5
(c) 12.67
(d) 14.76
Show Answer
Solution
(c) Since, marks obtained by 9 students in Mathematics are 50, 69, 20, 33, 53, 39, 40, 65 and 59.
Rewrite the given data in ascending order.
$20,33,39,40,50,53,59,65,69$,
Here
$ n=9 $
[odd]
$\therefore \quad$ Median $=\frac{9+1}{2}$ term $=5$ th term
| $M e=50$ | ||
|---|---|---|
| $x_i$ | $d_i = \mid x_i - Me \mid $ | |
| 20 | 30 | |
| 33 | 17 | |
| 39 | 11 | |
| 40 | 10 | |
| 50 | 0 | |
| 53 | 3 | |
| 59 | 9 | |
| 65 | 15 | |
| 69 | 19 | |
| $N=2$ | $\Sigma d_{i}=114$ | |
| MD $=\frac{114}{9}=12.67$ |
28. The standard deviation of data $6,5,9,13,12,8$ and 10 is
(a) $\sqrt{\frac{52}{7}}$
(b) $\frac{52}{7}$
(c) $\sqrt{6}$
(d) 6
Show Answer
Solution
(a) Given, data are 6, 5, 9, 13, 12, 8, and 10.
$\begin{array}{|c|c|}\hline x _i & x _i^2 \\ \hline 6 & 36 \\ 5 & 25 \\ 9 & 81 \\ 13 & 169 \\ 12 & 144 \\ 8 & 64 \\ 10 & 100 \\ \hline \Sigma x _i=6 3 & \Sigma x _ i^ 2=619 \\ \hline \end{array}$
$\begin{aligned} \therefore \quad \mathrm{SD} & =\sigma=\sqrt{\frac{\Sigma x_i^2}{N}-{\frac{\Sigma x_i}{N}}^2}=\sqrt{\frac{619}{7}-\frac{63}{7}^2} \\ & =\sqrt{\frac{7 \times 619-3969}{49}} \\ & =\sqrt{\frac{4333-3969}{49}} \\ & =\sqrt{\frac{364}{49}}=\sqrt{\frac{52}{7}}\end{aligned}$
29. If $x_1, x_2, \ldots, x_{n}$ be $n$ observations and $\bar{x}$ be their arithmetic mean. Then, formula for the standard deviation is given by
(a) $\Sigma(x_{i}-\bar{x})^{2}$
(c) $\sqrt{\frac{\sum(x_{i}-\bar{x})^{2}}{n}}$
(b) $\frac{\Sigma(x_{i}-\bar{x})^{2}}{n}$
(d) $\sqrt{\frac{\sum x_i^{2}}{n}+\bar x^{-2}}$
Show Answer
Solution
(c) SD is given by
$ \sigma=\sqrt{\frac{\sum(x_{i}-\bar{x})^{2}}{n}} $
30. If the mean of 100 observations is 50 and their standard deviation is 5 , than the sum of all squares of all the observations is
(a) 50000
(b) 250000
(c) 252500
(d) 255000
Show Answer
Solution
(c) Given,
$ \begin{aligned} \bar{x}=50, n & =100 \text { and } \sigma=5 \\ \Sigma x_i^{2} & =? \\ \bar{x} & =\frac{\Sigma x_{i}}{n} \end{aligned} $
$ \begin{matrix} \Rightarrow & 50=\frac{\Sigma x_{i}}{100} \\ \therefore & \Sigma x_{i}=50 \times 100=5000 \end{matrix} $
$ \begin{aligned} & \text { Now, } \quad \sigma=\sqrt{\frac{\Sigma x_i^{2}}{n}-\frac{\sum x_{i}{ }^{2}}{n}} \Rightarrow \sigma^{2}=\frac{\sum x_i^{2}}{n}-(\bar{x})^{2} \\ & \Rightarrow \quad 25=\frac{\Sigma x_i^{2}}{100}-(50)^{2} \Rightarrow 25=\frac{\Sigma x_i^{2}}{100}-2500 \\ & \Rightarrow \quad 2525=\frac{\Sigma x_i^{2}}{100} \\ & \therefore \quad \Sigma x_i^{2}=252500 \end{aligned} $
31. If $a, b, c, d$ and $e$ be the observations with mean $m$ and standard deviation $s$, then find the standard deviation of the observations $a+k$, $b+k, c+k, d+k$ and $e+k$ is
(a) $s$
(b) $k s$
(c) $s+k$
(d) $\frac{s}{k}$
Show Answer
Solution
(a) Given observations are $a, b, c, d$ and $e$.
$ \begin{aligned} & \text { Mean }=m=\frac{a+b+c+d+e}{5} \\ & \Sigma x_{i}=a+b+c+d+e=5 m \\ & \text { Now, } \\ & \text { mean }=\frac{a+k+b+k+c+k+d+k+e+k}{5} \\ & =\frac{(a+b+c+d+e)+5 k}{5}=m+k \\ & \therefore \quad SD=\sqrt{\frac{\sum(x_{i}+k)^{2}}{n}-(m+k)^{2}} \\ & =\sqrt{\frac{\sum(x_i^{2}+k^{2}+2 k x_{i})}{n}-(m^{2}+k^{2}+2 m k)} \\ & =\sqrt{\frac{\Sigma x_i^{2}}{n}-m^{2}+\frac{2 k \Sigma x_{i}}{n}-2 m k} \\ & =\sqrt{\frac{\Sigma x_i^{2}}{n}-m^{2}+2 k m-2 m k} \quad \because \frac{\Sigma x_{i}}{n}=m \\ & =\sqrt{\frac{\Sigma x_i^{2}}{n}-m^{2}} \\ & =s \end{aligned} $
32. If $x_1, x_2, x_3, x_4$ and $x_5$ be the observations with mean $m$ and standard deviation $s$ then, the standard deviation of the observations $k x_1, k x_2$, $k x_3, k x_4$ and $k x_5$ is
(a) $k+s$
(b) $\frac{s}{k}$
(c) $k s$
(d) $s$
Show Answer
Solution
(c) Here,
$ \begin{aligned} m & =\frac{\Sigma x_{i}}{5}, s=\sqrt{\frac{\Sigma x_i^{2}}{5}-\frac{\Sigma x_{i}}{5}} \\ SD & =\sqrt{\frac{k^{2} \Sigma x_i^{2}}{5}-\frac{k \Sigma x_{i}{ }^{2}}{5}} \\ & =\sqrt{\frac{k^{2} \Sigma x_i^{2}}{5}-k^{2} \frac{\Sigma x_i^{2}}{5}}=\sqrt{\frac{\Sigma x_i^{2}}{5}-{\frac{\Sigma x_{i}}{5}}^{2}}=k s \end{aligned} $
33. Let $x_1, x_2, \ldots x_{n}$ be $n$ observations. Let $w_{i}=l x_{i}+k$ for $i=1,2, \ldots, n$, where $l$ and $k$ are constants. If the mean of $x_{i}{ }^{\prime} s$ is 48 and their standard deviation is 12, the mean of $w_{i}{ }^{\prime} s$ is 55 and standard deviation of $w_{i}$ ’s is 15 , then the value of $l$ and $k$ should be
(a) $l=1.25, k=-5$
(b) $l=-1.25, k=5$
(c) $l=2.5, k=-5$
(d) $l=2.5, k=5$
Show Answer
Solution
(a) Given, $w_{i}=x_{i}+k, \bar x_i=48, s x_{i}=12, w_{i}=55$ and $s w_{i}=15$
Then, $\quad \bar w_i=\bar x_i+k$
[where, $\bar w_i$ is mean $w_{i}$ ’s and $\bar x_i$ is mean of $x_{i}{ }^{\prime} s$ ]
$\Rightarrow \quad 55=48+k$
Now, $\quad$ SD of $w_{i}=$ SD of $x_{i}$
$\Rightarrow \quad 15=12$
$\Rightarrow \quad l=\frac{15}{12}$
$ =1.25 $
From Eqs. (i) and (ii),
$ k=55-1.25 \times 48 $
$ =-5 $
34. The standard deviations for first natural numbers is
(a) 5.5
(b) 3.87
(c) 2.97
(d) 2.87
Show Answer
Solution
(d) We know that, SD of first $n$ natural number $=\sqrt{\frac{n^{2}-1}{12}}$
$\therefore \quad$ SD of first 10 natural numbers $=\sqrt{\frac{(10)^{2}-1}{12}}$
$ =\sqrt{\frac{100-1}{12}}=\sqrt{\frac{99}{12}}=\sqrt{8.25}=2.87 $
35. Consider the numbers $1,2,3,4,5,6,7,8,9$, and 10 . If 1 is added to each number the variance of the numbers, so obtained is
(a) 6.5
(b) 2.87
(c) 3.87
(d) 8.25
Show Answer
Solution
(d) Given numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
If 1 is added to each number, then observations will be 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11 .
$ \begin{aligned} & \therefore \quad \sum x_{i}=2+3+4+\ldots+11 \\ & =\frac{10}{2}[2 \times 2+9 \times 1]=5[4+9]=65 \\ & \Sigma x_i^{2}=2^{2}+3^{2}+4^{2}+5^{2}+\ldots+11^{2} \\ & =(1^{2}+2^{2}+3^{2}+\ldots+11^{2})-(1^{2}) \\ & =\frac{11 \times 12 \times 23}{6}-1 \\ & =\frac{11 \times 12 \times 23-6}{6}=505 \end{aligned} $
$ \begin{aligned} \therefore \quad s^{2} & =\frac{\Sigma x_i^{2}}{n}-{\frac{\Sigma x_{i}}{n}}^{2}=\frac{505}{10}-\frac{65}10^{2} \\ & =50.5-(6.5)^{2} \\ & =50.5-42.25 \\ & =8.25 \end{aligned} $
36. Consider the first 10 positive integers. If we multiply each number by -1 and, then add 1 to each number, the variance of the numbers, so obtained is
(a) 8.25
(b) 6.5
(c) 3.87
(d) 2.87
Show Answer
Solution
(a) Since, the first 10 positive integers are1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
On multiplying each number by -1 , we get
$ -1,-2,-3,-4,-5,-6,-7,-8,-9,-10 $
On adding 1 in each number, we get
$ \begin{matrix} \therefore \quad & 0,-1,-2,-3,-4,-5,-6,-7,-8,-9 \\ & =-\frac{9 \times 10}{2} \\ & =-45 \\ \text { and } \quad & \Sigma x_i^{2}=0^{2}+(-1)^{2}+(-2)^{2}+\ldots+(-9)^{2} \\ & =\frac{9 \times 10 \times 19}{6} \\ & =285 \\ \therefore \quad SD & =\sqrt{\frac{285}{10}-\frac{-45}{10}}=\sqrt{\frac{285}{10}-\frac{2025}{100}} \\ & =\sqrt{\frac{2850-2025}{100}}=\sqrt{8.25} \\ \text { Now, } \quad \text { variance } & =(SD)^{2}=(\sqrt{8.25})^{2}=8.25 \end{matrix} $
37. The following information relates to a sample of size $60, \Sigma x^{2}=18000$, and $\Sigma x=960$. Then, the variance is
(a) 6.63
(b) 16
(c) 22
(d) 44
Show Answer
Solution
(d)
$ \begin{aligned} & \text { Variance }=\frac{\Sigma x_i^{2}}{n}-\frac{\Sigma x_{i}}{n} \\ & =\frac{18000}{60}-\frac{960}60^{2}=300-256=44 \end{aligned} $
38. If the coefficient of variation of two distributions are 50,60 and their arithmetic means are 30 and 25 respectively, then the difference of their standard deviation is
(a) 0
(b) 1
(c) 1.5
(d) 2.5
Show Answer
Solution
(a) Here
$ CV_1=50, CV_2=60, \bar x_1=30 \text { and } \bar x_2=25 $
$ \therefore CV_1=\frac{\sigma _1}{\bar x_1} \times 100 \Rightarrow 50=\frac{\sigma _1}{30} \times 100 \\ $
$\therefore \sigma _1=\frac{30 \times 50}{100}=15 \text { and } CV_2=\frac{\sigma _2}{\bar x_2} \times 100 \\ $
$\Rightarrow 60=\frac{\sigma_2}{25} \times 100 \\ $
$\therefore \sigma _2=\frac{60 \times 25}{100}=15 \\ $
$\text { Now, } \sigma _1-\sigma _2=15-15=0 $
39. The standard deviation of some temperature data in ${ }^{\circ} C$ is 5 . If the data were converted into ${ }^{\circ} F$, then the variance would be (a) 81 (b) 57 (c) 36 (d) 25
Show Answer
Solution
(a) Given,
$ \sigma_{C}=5 \Rightarrow \frac{5}{9}(F-32)=C $
$ \begin{aligned} F & =\frac{9 C}{5}+32 \\ \sigma_{F} & =\frac{9}{5} \sigma_{C}=\frac{9}{5} \times 5=9 \end{aligned} $
Here,
$ \sigma_F^{2}=(9)^{2}=81 $
Fillers
40. Coefficient of variation $=\frac{\cdots}{\text { Mean }} \times 100$
Show Answer
Solution
$C V=\frac{S D}{\text { Mean }} \times 100$
41. If $\bar{x}$ is the mean of $n$ values of $x$, then $\sum_{i=1}^{n}(x_1-\bar{x})$ is always equal to ……
If $a$ has any value other than $\overline{\boldsymbol{x}}$, then $\sum_{i=1}^{n}(x_i-\overline{\boldsymbol{x}})^{2}$ …… is than $\Sigma(x_{i}-a)^{2}$
Show Answer
Solution
If $\bar{x}$ is the mean of $n$ values of $x$, then $\sum_{i=1}^{n}(x_{i}-\bar{x})=0$ and if $a$ has any value other than $\bar{x}$, then $\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}$ is less than $\sum(x_{i}-a)^{2}$
42. If the variance of a data is 121 , then the standard deviation of the data is ……
Show Answer
Solution
If the variance of a data is 121 .
Then,
$ \begin{aligned} S D & =\sqrt{\text { Variance }} \\ & =\sqrt{121}=11 \end{aligned} $
43. The standard deviation of a data is …… of any change in origin, but is …… on the change of scale.
Show Answer
Solution
The standard deviation of a data is independent of any change in origin but is dependent of change of scale.
44. The sum of squares of the deviations of the values of the variable is …… when taken about their arithmetic mean.
Show Answer
Solution
The sum of the squares of the deviations of the values of the variable is minimum when taken about their arithmetic mean.
45. The mean deviation of the data is …… when measured from the median.
Show Answer
Solution
The mean deviation of the data is least when measured from the median.
46. The standard deviation is …… to the mean deviation taken from the arithmetic mean.
Show Answer
Solution
The SD is greater than or equal to the mean deviation taken from the arithmetic mean.
