SATHEE: Statistics

Statistics

Short Answer Type Questions

1. Find the mean deviation about the mean of the distribution.

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Solution

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
	$\bar{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}} = \frac{433}{20} = 21.65$
	$M D = \frac{\sum f_{i} \| x_{i} - \bar{x} \|}{\sum f_{i}} = \frac{25}{20} = 1.25$

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	$f_{i}$	$c f$	$d_{i} =∣ x_{i} - M_{e} ∣$	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	$\sum f_{i} = 20$			$\sum f_{i} d_{i} = 25$

$\begin{matrix} Now, & M_{e} = \frac{20 + 1}{2} th item = \frac{21}{2} = 10.5 th item \\ ∴ & M_{e} = 12 \\ ∴ & M D = \frac{\sum f_{i} d_{i}}{\sum f_{i}} = \frac{25}{20} = 1.25 \end{matrix}$

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].

$\begin{aligned} Mean \bar{x} & = \frac{1 + 2 + 3 + \dots + n}{n} = \frac{n (n + 1)}{2 n} = \frac{n + 1}{2} \\ ∴ MD & = \frac{| 1 - \frac{n + 1}{2} | + | 2 - \frac{n + 1}{2} | + | 3 - \frac{n + 1}{2} | + \dots + | n - \frac{n + 1}{2} |}{n} \\ = \frac{+ | \frac{n + 1}{2} - \frac{n + 1}{2} | + | \frac{n + 3}{2} - \frac{n + 1}{2} | + \dots + | \frac{n + 1}{2} | + | 2 - \frac{n + 1}{2} | + \dots + | \frac{n - 1}{2} - \frac{n + 1}{2} | + | n - \frac{n + 1}{2} |}{n} \\ = \frac{2}{n} 1 + 2 + \dots + \frac{n - 3}{2} + \frac{n - 1}{2} \frac{n - 1}{2} terms \\ = \frac{2}{n} \frac{\frac{n - 1}{2} \frac{n - 1}{2} + 1}{2} ∵ sum of first n natural numbers = \frac{n (n + 1)}{2} \\ = \frac{2}{n} \cdot \frac{1}{2} \frac{n - 1}{2} \frac{n + 1}{2} = \frac{1}{n} \frac{n^{2} - 1}{4} = \frac{n^{2} - 1}{4 n} \end{aligned}$

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., $1, 2, 3, 4, \dots \dots . n$ .

$\begin{aligned} ∴ Mean \bar{x} = \frac{1 + 2 + 3 + \dots + n}{n} = \frac{n (n + 1)}{2 n} = \frac{n + 1}{2} \\ M D = \frac{1}{n} | 1 - \frac{n + 1}{2} | + | 2 - \frac{n + 1}{2} | + | 3 - \frac{n + 1}{2} | + | \frac{n - 2}{2} - \frac{n + 1}{2} | + | \frac{n}{2} - \frac{n + 1}{2} | \\ + | \frac{n + 2}{2} - \frac{n + 1}{2} | + \dots + | n - \frac{n + 1}{2} | \\ = \frac{1}{n} | \frac{1 - n}{2} | + | \frac{3 - n}{2} | + | \frac{5 - n}{2} | + \dots . + | \frac{- 3}{2} | + | \frac{1}{2} | + \dots + | \frac{n - 1}{2} | \\ = \frac{2}{n} \frac{1}{2} + \frac{3}{2} + \dots . + \frac{n - 1}{2} \frac{n}{2} terms \\ = \frac{1}{n} \cdot \frac{n^{2}}{2} [∵ sum of first n natural numbers = n^{2}] \\ = \frac{1}{n} \cdot \frac{n^{2}}{4} = \frac{n}{4} \end{aligned}$

5. Find the standard deviation of first $n$ natural numbers.

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Solution

$x_{i}$	1	2	3	4	5	$\dots$	$\dots$	$n$
$(x_{i})^{2}$	1	4	9	16	25	$\dots$	$\dots$	$n^{2}$

$Now, \begin{aligned} Σ x_{i} & = 1 + 2 + 3 + 4 + \dots + n = \frac{n (n + 1)}{2} \\ and \sum x_{i}^{2} & = 1^{2} + 2^{2} + 3^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6} \\ ∴ & = \sqrt{\frac{\sum x_{i}^{2}}{N} - \frac{\sum x_{i}^{2}}{N}} \\ = \sqrt{\frac{n (n + 1) (2 n + 1)}{6 n} - \frac{n^{2} (n + 1)^{2}}{4 n^{2}}} \\ = \sqrt{\frac{(n + 1) (2 n + 1)}{6} - \frac{(n + 1)^{2}}{4}} \\ = \sqrt{\frac{2 (2 n^{2} + 3 n + 1) - 3 (n^{2} + 2 n + 1)}{12}} \\ = \sqrt{\frac{4 n^{2} + 6 n + 2 - 3 n^{2} - 6 n - 3}{12}} \\ = \sqrt{\frac{n^{2} - 1}{12}} \end{aligned}$

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 $x_{1} x_{2} \dots x_{15}$ , 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,

$n_{i} = 25, {\bar{x}}_{i} = 18.2, σ_{1} = 3.25$

$n_{2} = 15, \sum_{i = 1}^{15} x_{i} = 279 and \sum_{i = 1}^{15} x_{i}^{2} = 5524$

For first set,

$\begin{aligned} \sum x_{i} & = 25 \times 18.2 = 455 \\ σ_{1}^{2} & = \frac{\sum x_{i}^{2}}{25} - (18.2)^{2} \\ 25^{2} & = \frac{\sum x_{i}^{2}}{25} - 331.24 \\ 1.24 & = \frac{\sum x_{i}^{2}}{25} \\ Σ x_{i}^{2} & = 25 \times (10.5625) \\ = 25 \times 341.8025 \\ = 8545.0625 \end{aligned}$

$\begin{matrix} ∴ & σ_{1}^{2} = \frac{Σ x_{i}^{2}}{25} - (18.2)^{2} \\ \Rightarrow & (3.25)^{2} = \frac{Σ x_{i}^{2}}{25} - 331.24 \\ \Rightarrow & 10.5625 + 331.24 = \frac{Σ x_{i}^{2}}{25} \\ \Rightarrow & Σ x_{i}^{2} = 25 \times (10.5625 + 331.24) \end{matrix}$

For combined SD of the 40 observations $n = 40$ ,

$\begin{aligned} Now Σ x_{i}^{2} = 5524 + 8545.0625 = 14069.0625 \\ and Σ x_{i} = 455 + 279 = 734 \\ ∴ S D = \sqrt{\frac{14069.0625}{40} - \frac{734}{4} 0^{2}} \\ = \sqrt{351.726 - (18.35)^{2}} \\ = \sqrt{351.726 - 336.7225} \\ = \sqrt{15.0035} = 3.87 \end{aligned}$

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

$S D = \sqrt{\frac{n_{1} (s_{1})^{2} + n_{2} (s_{2})^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} ({\bar{x}}_{1} - {\bar{x}}_{2})^{2}}{(n_{1} - n_{2})^{2}}}$

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Solution

Let

$x_{i}, i = 1, 2, 3 \dots, n_{1} and y_{j}, j = 1, 2, 3, \dots, n_{2}$

$∴ {\bar{x}}_{1} = \frac{1}{n_{1}} \sum_{i = 1}^{n_{1}} x_{i} and {\bar{x}}_{2} = \frac{1}{n_{2}} \sum_{j = 1}^{n_{2}} y_{j}$

$\Rightarrow σ_{1}^{2} = \frac{1}{n_{1}} \sum_{i = 1}^{n_{1}} (x_{i} - {\bar{x}}_{1})^{2}$

and

$σ_{2}^{2} = \frac{1}{n_{2}} \sum_{j = 1}^{n} (y_{j} - {\bar{x}}_{2})^{2}$

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 $σ^{2}$ of the combined series is given by

$σ^{2} = [\frac{1}{n_{1} + n_{2}} \sum_{i = 1}^{n_{1}} (x_{i} - \bar{x})^{2} + \sum_{j = 1}^{n_{2}} (y_{j} - \bar{x})^{2}]$

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 $\sum_{i = 1}^{n_{1}} (x_{i} - {\bar{x}}_{i}) = 0$

[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, $\sum_{j = 1}^{n_{2}} (y_{j} - \bar{x})^{2} = \sum_{j = 1}^{n_{2}} (y_{j} - {\bar{x}}_{i} + {\bar{x}}_{i} - \bar{x})^{2} = n_{2} s_{2}^{2} + n_{2} d_{2}^{2}$

where,

$d_{2} = {\bar{x}}_{2} - \bar{x}$

Combined SD

$σ = \sqrt{\frac{[n_{1} (s_{1}^{2} + d_{1}^{2}) + n_{2} (s_{2}^{2} + d_{2}^{2})]}{n_{1} + n_{2}}}$

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

$d_{2} = {\bar{x}}_{2} - \bar{x} = {\bar{x}}_{2} - \frac{n_{1} {\bar{x}}_{1} + n_{2} {\bar{x}}_{2}}{n_{1} + n_{2}} = \frac{n_{1} ({\bar{x}}_{2} - {\bar{x}}_{1})}{n_{1} + n_{2}}$

$∴ σ^{2} = \frac{1}{n_{1} + n_{2}} n_{1} s_{1}^{2} + n_{2} s_{2}^{2} + \frac{n_{1} n_{2} ({\bar{x}}_{1} - {\bar{x}}_{2})^{2}}{(n_{1} + n_{2})^{2}} + \frac{n_{2} n_{1} ({\bar{x}}_{2} - {\bar{x}}_{1})^{2}}{(n_{1} + n_{2})^{2}}$

Also,

$σ = \sqrt{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} ({\bar{x}}_{1} - {\bar{x}}_{2})^{2}}{(n_{1} + n_{2})^{2}}}$

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, $n_{1} = 20, σ_{1} = 5, {\bar{x}}_{1} = 17$ and $n_{2} = 20, σ_{2} = 5, {\bar{x}}_{2} = 22$

We know that, $σ = \sqrt{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} ({\bar{x}}_{1} - {\bar{x}}_{2})^{2}}{(n_{1} + n_{2})^{2}}}$

$\begin{aligned} = \sqrt{\frac{20 \times (5)^{2} + 20 \times (5)^{2}}{20 + 20} + \frac{20 \times 20 (17 - 22)^{2}}{(20 + 20)^{2}}} \\ = \sqrt{\frac{1000}{40} + \frac{400 \times 25}{1600}} = \sqrt{25 + \frac{25}{4}} = \sqrt{\frac{125}{4}} = \sqrt{31.25} = 5.59 \end{aligned}$

9. The frequency distribution

$x$	$A$	$2 A$	$3 A$	$4 A$	$5 A$	$6 A$
$f$	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	$22 A$	$92 A^{2}$
	$n = 7$	$Σ f_{i} n_{i} = 22 A$	$Σ f_{i} n_{i}^{2} = 92 A^{2}$

$\begin{matrix} ∴ & σ^{2} = \frac{Σ f_{i} x_{i}^{2}}{n} - \frac{Σ f_{i} x_{i}}{n} \\ \Rightarrow & 160 = \frac{92 A^{2}}{7} - \frac{22 A^{2}}{7} \\ \Rightarrow & 160 = \frac{92 A^{2}}{7} - \frac{484 A^{2}}{49} \\ \Rightarrow & 160 = (644 - 484) \frac{A^{2}}{49} \\ \Rightarrow & 160 = \frac{160 A^{2}}{49} \Rightarrow A^{2} = 49 \\ ∴ & A = 7 \end{matrix}$

10. For the frequency distribution

$x$	2	3	4	5	6	7
$f$	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	$Σ f_{i} d_{i} = 37$	$Σ f_{i} d_{i}^{2} = 137$
$∴ S D$	$= \sqrt{\frac{Σ f_{i} d_{i}^{2}}{N} - \frac{Σ f_{i} d_{i}^{2}}{N}}$
	$= \sqrt{\frac{137}{60} - \frac{37^{2}}{60}}$
	$= \sqrt{2.2833 - (0.616)^{2}}$
	$= \sqrt{2.2833 - 0.3794}$
	$= \sqrt{1.9037} = 1.38$

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

$∴$ Sum of frequencies,

$\begin{aligned} x - 2 + x + x^{2} + (x + 1)^{2} + 2 x + x + 1 = 60 \\ \Rightarrow 2 x - 2 + x^{2} + x^{2} + 1 + 2 x + 2 x + x + 1 = 60 \\ \Rightarrow 2 x^{2} + 7 x = 60 \\ \Rightarrow 2 x^{2} + 7 x - 60 = 0 \\ \Rightarrow 2 x^{2} + 15 x - 8 x - 60 = 0 \\ \Rightarrow x (2 x + 15) - 4 (2 x + 15) = 0 \\ \Rightarrow (2 x + 15) (x - 4) = 0 & \Rightarrow \\ \Rightarrow x = - \frac{15}{2}, 4 \\ \Rightarrow x = - \frac{15}{2} & [inaddmisible] [∵ x \in /^{+}] \end{aligned}$

$\begin{array}{ccccc} x_{i} & f_{i} & d_{i} = x_{i} - 3 & f_{i} d_{i} & f_{i} d_{i}^{2} \\ 0 & 2 & - 3 & - 6 & 18 \\ 1 & 4 & - 2 & - 8 & 16 \\ 2 & 16 & - 1 & - 16 & 16 \\ A = 3 & 25 & 0 & 0 & 0 \\ 4 & 8 & 1 & 8 & 8 \\ 5 & 5 & 2 & 10 & 20 \\ Total & Σ f_{i} = 60 & Σ f_{i} d_{i} = - 12 & Σ f_{i} d_{i}^{2} = 78 \end{array}$

$\begin{aligned} Mean & = A + \frac{\sum f_{i} d_{i}}{Σ f_{i}} = 3 + \frac{- 12}{60} = 2.8 \\ σ & = \sqrt{\frac{\sum f_{i} d_{i}^{2}}{Σ f_{i}} - \frac{\sum f_{i} d_{i}^{2}}{Σ f_{i}}} = \sqrt{\frac{78}{60} - \frac{- 12^{2}}{60}} \\ = \sqrt{1.3 - 0.04} = \sqrt{1.26} = 1.12 \end{aligned}$

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 $n_{2} = 80, {\bar{x}}_{2} = 660, s_{2} = 7$

$∴ σ = \sqrt{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} ({\bar{x}}_{1} - {\bar{x}}_{2})^{2}}{(n_{1} + n_{2})^{2}}}$

$= \sqrt{\frac{60 \times (8)^{2} + 80 \times (7)^{2}}{60 + 80} + \frac{60 \times 80 (650 - 660)^{2}}{(60 + 80)^{2}}}$

$= \sqrt{\frac{6 \times 64 + 8 \times 49}{14} + \frac{60 \times 80 \times 100}{140 \times 140}}$

$= \sqrt{\frac{192 + 196}{7} + \frac{1200}{49}} = \sqrt{\frac{388}{7} + \frac{1200}{49}}$

$= \sqrt{\frac{2716 + 1200}{49}} = \sqrt{\frac{3916}{49}} = \frac{62.58}{7} = 8.9$

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 $σ = 4$

$\begin{aligned} ∴ \frac{Σ x_{i}}{100} = 50 \\ \Rightarrow Σ x_{i} = 5000 \\ and σ^{2} = \frac{Σ f_{i} x_{i}^{2}}{Σ f_{i}} - \frac{Σ f_{i} x_{i}}{Σ f_{i}} \\ \Rightarrow (4)^{2} = \frac{Σ f_{i} x_{i}^{2}}{100} - (50)^{2} \\ \Rightarrow 16 = \frac{Σ f_{i} x_{i}^{2}}{100} - 2500 \\ \Rightarrow \frac{Σ f_{i} x_{i}^{2}}{100} = 16 + 2500 \Rightarrow 2516 \\ ∴ \sum f_{i} x_{i}^{2} = 251600 \end{aligned}$

14. If for distribution $Σ (x - 5) = 3, Σ (x - 5)^{2} = 43$ and total number of item is 18 . Find the mean and standard deviation.

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Solution

Given,

$\begin{aligned} n & = 18, Σ (x - 5) = 3 and Σ (x - 5)^{2} = 43 \\ = 5 + \frac{3}{18} = 5 + 0.1666 = 5.1666 = 5.17 \\ SD & = \sqrt{\frac{\sum (x - 5)^{2}}{n} - \frac{\sum (x - 5)^{2}}{n}} \\ = \sqrt{\frac{43}{18} - \frac{3}{18}} \\ = \sqrt{2.3944 - (0.166)^{2}} = \sqrt{2.3944 - 0.2755} = 1.59 \end{aligned}$

$\begin{aligned} ∴ Mean = A + \frac{\sum (x - 5)}{18} \\ and \end{aligned}$

15. Find the mean and variance of the frequency distribution given below.

$x$	$1 \leq x \leq 3$	$3 \leq x \leq 5$	$5 \leq x \leq 7$	$7 \leq x \leq 10$
$f$	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	$n = 16$		$Σ f_{i} x_{i} = 66.5$	$Σ f_{i} x_{i}^{2} = 340.25$

$∴$ Mean $= \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{66.5}{16} = 4.15$

$\begin{aligned} variance & = σ^{2} = \frac{Σ f_{i} x_{i}^{2}}{Σ f_{i}} - \frac{Σ f_{i} x_{i}}{Σ f_{i}} \\ = \frac{340.25}{16} - (4.15)^{2} \\ = 21.2656 - 17.2225 = 4.043 \end{aligned}$

Long Answer Type Questions

16. Calculate the mean deviation about the mean for the following frequency distribution.

$\begin{array}{cccccc} Class interval & 0 - 4 & 4 - 8 & 8 - 12 & 12 - 16 & 16 - 20 \\ Frequency & 4 & 6 & 8 & 5 & 2 \end{array}$

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Solution

$\begin{array}{cccccc} Class internal & f_{i} & x_{i} & f_{i} x_{i} & d_{i} = x_{i} - \overset{―}{x} ∣ & f_{i} d_{i} \\ 0 - 4 & 4 & 2 & 8 & 7.2 & 28.8 \\ 4 - 8 & 6 & 6 & 36 & 3.2 & 19.2 \\ 8 - 12 & 8 & 10 & 80 & 0.8 & 6.4 \\ 12 - 16 & 5 & 14 & 70 & 4.8 & 24.0 \\ 16 - 20 & 2 & 18 & 36 & 8.8 & 17.6 \\ T o t a l & Σ f_{i} = 25 & Σ f_{i} x_{i} = 230 & Σ f_{i} d_{i} = 96 \end{array}$

$\begin{aligned} ∴ Mean = \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{230}{25} = 9.2 \\ and mean deviation = \frac{Σ f d_{i}}{Σ f_{i}} = \frac{96}{25} = 3.84 \end{aligned}$

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	$f_{i}$	$x_{i}$	$c f$	$d_{i} =∣ x_{i} - {\bar{m}}_{d} ∣$	$f_{i} d_{i}$
$0 - 6$	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 $12 - 18$ .

$\begin{aligned} Median & = l + \frac{\frac{N}{2} - c f}{f} \times i \\ = 12 + \frac{6}{3} (10 - 9) \\ = 12 + 2 = 14 \\ M D & = \frac{Σ f_{i} d_{i}}{Σ f_{i}} = \frac{140}{20} = 7 \end{aligned}$

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	$Σ f_{i} = 40$	$Σ f_{i} x_{i} = 239$		$Σ f_{i} d_{i} = - 1$	$Σ f_{i} x_{i}^{2} = 325$

$\begin{aligned} ∴ Mean \bar{x} = \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{239}{40} = 5.975 \approx 6 \\ and \\ \begin{aligned} σ & = \sqrt{\frac{Σ f_{i} d_{i}^{2}}{Σ f_{i}} - \frac{Σ f_{i} d_{i}^{2}}{Σ f_{i}}} = \sqrt{\frac{325}{40} - \frac{- 1}{4} 0^{2}} \\ = \sqrt{8.125 - 0.000625} = \sqrt{8.124375} = 2.85 \end{aligned} \end{aligned}$

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

$Σ f_{i} = 70$ $Σ f_{i} d_{i} =$
-38 $Σ f_{i} d_{i}^{2} = 102$ | $\frac{T f_{i}}{\sum f_{i}}$

Now,
$= 1.4571 - 0.2916 = 1.1655 σ = \sqrt{1.1655} = 1.08 g$

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

$x_{i}$	$x x_{i} - a$	$(x_{i} - a)^{2}$
$a$	0	0
$a + d$	$d$	$d^{2}$
$a + 2 d$	$2 d$	$4 d^{2}$
$\dots \dots$	$\dots \dots$	$9 d^{2}$
$\dots \dots$	$\dots \dots$ .
$\dots \dots$	$\dots \dots$ .
$a + (n - 1) d$	$(n - 1) d$	$(n - 1)^{2} d^{2}$
$\sum x_{i} = \frac{n}{2} [2 a + (n - 1)]$
Mean $= \frac{\sum x_{i}}{n} = \frac{1}{n} \frac{n}{2} (2 a) + (n - 1) d$
$=$	$a + \frac{(n - 1)}{2} d$

$\begin{aligned} Σ (x_{i} - a) & = d [1 + 2 + 3 + \dots + (n - 1) d] \\ = d \frac{(n - 1) n}{2} \\ and \sum (x_{i} - a)^{2} & = d^{2} [1^{2} + 2^{2} + 3^{2} + \dots + (n - 1)^{2}] \\ = \frac{d^{2} (n - 1) n (2 n - 1)}{6} \\ σ & = \sqrt{\frac{(x_{i} - a)^{2}}{n} - \frac{x_{i} - a^{2}}{n}} \\ = \sqrt{\frac{d^{2} (n - 1) (n) (2 n - 1)}{6 n} - \frac{d (n - 1) n^{2}}{2 n}} \\ = \sqrt{\frac{d^{2} (n - 1) (2 n - 1)}{6} - \frac{d^{2} (n - 1)^{2}}{4}} \\ = d \sqrt{\frac{(n - 1) (2 n - 1)}{6} - \frac{(n - 1)^{2}}{4}} \\ = d \sqrt{\frac{(n - 1)}{2} \frac{2 n - 1}{3} - \frac{n - 1}{2}} \\ = d \sqrt{\frac{(n - 1)}{2} \frac{4 n - 2 - 3 n + 3}{6}} \\ = d \sqrt{\frac{(n - 1) (n + 1)}{12}} = d \sqrt{\frac{(n^{2} - 1)}{12}} \end{aligned}$

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	$Σ d_{i} = - 14$	$Σ d_{i}^{2} = 1699$

Now,

$\begin{aligned} σ & = \sqrt{\frac{Σ d^{2} i}{n} - \frac{Σ d_{i}^{2}}{n}} \\ = \sqrt{\frac{1699}{10} - \frac{- 14^{2}}{10}} = \sqrt{169.9 - 0.0196} \\ = \sqrt{169.88} = 13.03 \\ \bar{x} & = A + \frac{Σ d_{i}}{Σ f_{i}} = 45 - \frac{14}{10} = 43.6 \end{aligned}$

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	$Σ d_{i} = 0$	$\sum d_{i}^{2} = 6368$

$\begin{aligned} ∵ Mean = 55 \\ ∴ σ = \sqrt{\frac{6368}{10}} = \sqrt{636.8} = 25.2 \\ C V = \frac{σ}{\bar{x}} \times 100 = \frac{13.03}{43.6} \times 100 = 29.88 \\ C V = \frac{σ}{\bar{x}} \times 100 = \frac{25.2}{55} \times 100 = 45.89 \end{aligned}$

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,

$\begin{aligned} n = 100, \bar{x} = 40, σ = 10 and \bar{x} = 40 \\ ∴ \frac{\sum x_{i}}{n} = 40 \\ \Rightarrow & \frac{Σ x_{i}}{100} = 40 \\ \Rightarrow Σ x_{i} = 4000 \\ Corrected Σ x_{i} = 4000 - 30 - 70 + 3 + 27 \\ = 4030 - 100 = 3930 \\ Corrected mean = \frac{2930}{100} = 39.3 \end{aligned}$

$\begin{aligned} σ^{2} & = \frac{Σ x_{i}^{2}}{n} - (40)^{2} \\ 100 & = \frac{Σ x_{i}^{2}}{100} - 1600 \\ Σ x_{i}^{2} & = 170000 \end{aligned}$

Corrected $Σ x_{i}^{2} = 170000 - (30)^{2} - (70)^{2} + 3^{2} + (27)^{2}$

$\begin{aligned} = 164939 \\ Corrected σ & = \sqrt{\frac{164939}{100} - (39.3)^{2}} \\ = \sqrt{1649.39 - 39.3 \times 39.3} \\ = \sqrt{1649.39 - 1544.49} \\ = \sqrt{104.9} = 10.24 \end{aligned}$

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,

$\begin{aligned} n = 10, \bar{x} = 45 and σ^{2} = 16 \\ \bar{x} = 45 \Rightarrow \frac{Σ x_{i}}{n} = 45 \\ \frac{Σ x_{i}}{10} = 45 \Rightarrow Σ x_{i} = 450 \end{aligned}$

$∴$

Corrected $Σ x_{i} = 450 - 52 + 25 = 423$

$∴$

$\bar{x} = \frac{423}{10} = 42.3$

$\Rightarrow$

$σ^{2} = \frac{Σ x_{i}^{2}}{n} - \frac{Σ x_{i}}{n}$

$\Rightarrow$

$16 = \frac{Σ x_{i}^{2}}{10} - (45)^{2}$

$\Rightarrow$

$Σ x_{i}^{2} = 10 (2025 + 16)$

$\Rightarrow$

$Σ x_{i}^{2} = 20410$

$∴$

Corrected $Σ x_{i}^{2} = 20410 - (52)^{2} + (25)^{2} = 18331$

and

corrected $σ^{2} = \frac{18331}{10} - (42.3)^{2} = 43.81$

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

(d) 3.75

Show Answer

Solution

(b) Given, observations are $3, 10, 10, 4, 7, 10$ and 5.

$∴$	$= \frac{49}{7} = 7$
	$x_{i}$	$d_{i} = x_{i} - \bar{x}$
	3	4
	10	3
	10	3
	4	3
	7	0
	10	3
	5	2
	Total	$Σ d_{i} = 18$
Now,	$M D = \frac{\sum d_{i}}{N} = \frac{18}{7} = 2.57$

Option (a) is incorrect because the mean deviation calculated from the given data is not 2. The correct mean deviation is 2.57.
Option (c) is incorrect because the mean deviation calculated from the given data is not 3. The correct mean deviation is 2.57.
Option (d) is incorrect because the mean deviation calculated from the given data is not 3.75. The correct mean deviation is 2.57.

25. Mean deviation for $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ from their mean $\bar{x}$ is given by

(a) $\sum_{i = 1}^{n} (x_{i} - \bar{x})$

(b) $\frac{1}{n} \sum_{i = 1}^{n} | x_{i} - \bar{x} |$

(d) $\frac{1}{n} \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}$

Show Answer

Solution

(b) $M D = \frac{1}{n} \sum_{i = 1}^{n} | x_{i} - \bar{x} |$

Option (a): $\sum_{i = 1}^{n} (x_{i} - \bar{x})$ is incorrect because the sum of deviations from the mean is always zero. This is due to the property of the mean, which balances the data points such that the positive and negative deviations cancel each other out.
Option (c): $\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}$ is incorrect because this represents the sum of squared deviations from the mean, which is used to calculate the variance, not the mean deviation.
Option (d): $\frac{1}{n} \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}$ is incorrect because this represents the mean of the squared deviations from the mean, which is the definition of variance, not the mean deviation.

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

(d) 356

Show Answer

Solution

(a) Since, the lives of 5 bulbs are 1357, 1090, 1666, 1494 and 1623.

$\begin{aligned} ∴ Mean & = \frac{1357 + 1090 + 1666 + 1494 + 1623}{5} \\ = \frac{7230}{5} = 1446 \end{aligned}$

$x_{i}$	$d_{i} =∣ x_{i} - \bar{x} ∣$
1357	89
1090	356
1666	220
1494	48
1623	177
Total	$\sum d_{i} = 890$
MD $= \frac{Σ d_{i}}{N} = \frac{890}{5} = 178$

Option (b) 179 is incorrect because the correct mean deviation calculated from the given data is 178, not 179. The sum of the absolute deviations from the mean is 890, and dividing this by the number of bulbs (5) gives 178.
Option (c) 220 is incorrect because it represents the absolute deviation of one of the bulbs (1666) from the mean, not the mean deviation of all bulbs.
Option (d) 356 is incorrect because it represents the absolute deviation of one of the bulbs (1090) from the mean, not the mean deviation of all bulbs.

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

(d) 14.76

Show Answer

Solution

Rewrite the given data in ascending order.

$20, 33, 39, 40, 50, 53, 59, 65, 69$ ,

Here

$n = 9$

[odd]

$∴$ Median $= \frac{9 + 1}{2}$ term $= 5$ th term

$M e = 50$
	$x_{i}$	$d_{i} =∣ x_{i} - M e ∣$
20	30
33	17
39	11
40	10
50	0
53	3
59	9
65	15
69	19
$N = 2$	$Σ d_{i} = 114$
MD $= \frac{114}{9} = 12.67$

Option (a) 9 is incorrect because the sum of the absolute deviations from the median is 114, and dividing this by the number of students (9) gives a mean deviation of 12.67, not 9.
Option (b) 10.5 is incorrect because, as calculated, the mean deviation from the median is 12.67, not 10.5.
Option (d) 14.76 is incorrect because the correct mean deviation from the median, based on the given data, is 12.67, not 14.76.

28. The standard deviation of data $6, 5, 9, 13, 12, 8$ and 10 is

(a) $\sqrt{\frac{52}{7}}$

(b) $\frac{52}{7}$

(d) 6

Show Answer

Solution

(a) Given, data are 6, 5, 9, 13, 12, 8, and 10.

$\begin{array}{cc} x_{i} & x_{i}^{2} \\ 6 & 36 \\ 5 & 25 \\ 9 & 81 \\ 13 & 169 \\ 12 & 144 \\ 8 & 64 \\ 10 & 100 \\ Σ x_{i} = 63 & Σ x_{i}^{2} = 619 \end{array}$

$\begin{aligned} ∴ SD & = σ = \sqrt{\frac{Σ x_{i}^{2}}{N} - {\frac{Σ x_{i}}{N}}^{2}} = \sqrt{\frac{619}{7} - {\frac{63}{7}}^{2}} \\ = \sqrt{\frac{7 \times 619 - 3969}{49}} \\ = \sqrt{\frac{4333 - 3969}{49}} \\ = \sqrt{\frac{364}{49}} = \sqrt{\frac{52}{7}} \end{aligned}$

Option (b) $\frac{52}{7}$ is incorrect because it represents the variance, not the standard deviation. The standard deviation is the square root of the variance.
Option (c) $\sqrt{6}$ is incorrect because it does not match the calculated value of the standard deviation, which is $\sqrt{\frac{52}{7}}$ .
Option (d) 6 is incorrect because it is not derived from the given data set and does not match the calculated standard deviation.

29. If $x_{1}, x_{2}, \dots, x_{n}$ be $n$ observations and $\bar{x}$ be their arithmetic mean. Then, formula for the standard deviation is given by

(a) $Σ (x_{i} - \bar{x})^{2}$

(b) $\frac{Σ (x_{i} - \bar{x})^{2}}{n}$

(d) $\sqrt{\frac{\sum x_{i}^{2}}{n} + {\bar{x}}^{- 2}}$

Show Answer

Solution

$σ = \sqrt{\frac{\sum (x_{i} - \bar{x})^{2}}{n}}$

Option (a) $Σ (x_{i} - \bar{x})^{2}$ is incorrect because it represents the sum of squared deviations from the mean, not the standard deviation. The standard deviation requires taking the square root of the average of these squared deviations.
Option (b) $\frac{Σ (x_{i} - \bar{x})^{2}}{n}$ is incorrect because it represents the variance, not the standard deviation. The standard deviation is the square root of the variance.
Option (d) $\sqrt{\frac{\sum x_{i}^{2}}{n} + {\bar{x}}^{- 2}}$ is incorrect because it does not correctly represent the formula for standard deviation. The term ${\bar{x}}^{- 2}$ is not part of the standard deviation formula and the correct formula involves the squared deviations from the mean, not the sum of squares of the observations divided by the number of observations.

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

(d) 255000

Show Answer

Solution

$\begin{aligned} \bar{x} = 50, n & = 100 and σ = 5 \\ Σ x_{i}^{2} & = ? \\ \bar{x} & = \frac{Σ x_{i}}{n} \end{aligned}$

$\begin{matrix} \Rightarrow & 50 = \frac{Σ x_{i}}{100} \\ ∴ & Σ x_{i} = 50 \times 100 = 5000 \end{matrix}$

$\begin{aligned} Now, σ = \sqrt{\frac{Σ x_{i}^{2}}{n} - \frac{\sum x_{i}^{2}}{n}} \Rightarrow σ^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2} \\ \Rightarrow 25 = \frac{Σ x_{i}^{2}}{100} - (50)^{2} \Rightarrow 25 = \frac{Σ x_{i}^{2}}{100} - 2500 \\ \Rightarrow 2525 = \frac{Σ x_{i}^{2}}{100} \\ ∴ Σ x_{i}^{2} = 252500 \end{aligned}$

Option (a) 50000: This option is incorrect because it significantly underestimates the sum of the squares of the observations. Given the mean and standard deviation, the sum of the squares must be much larger.
Option (b) 250000: This option is incorrect because it is close to the correct answer but still slightly underestimates the sum of the squares. The correct calculation shows that the sum of the squares is 252500.
Option (d) 255000: This option is incorrect because it overestimates the sum of the squares of the observations. The correct calculation shows that the sum of the squares is 252500, not 255000.

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) $k s$

(d) $\frac{s}{k}$

Show Answer

Solution

(a) Given observations are $a, b, c, d$ and $e$ .

$\begin{aligned} Mean = m = \frac{a + b + c + d + e}{5} \\ Σ x_{i} = a + b + c + d + e = 5 m \\ Now, \\ mean = \frac{a + k + b + k + c + k + d + k + e + k}{5} \\ = \frac{(a + b + c + d + e) + 5 k}{5} = m + k \\ ∴ S D = \sqrt{\frac{\sum (x_{i} + k)^{2}}{n} - (m + k)^{2}} \\ = \sqrt{\frac{\sum (x_{i}^{2} + k^{2} + 2 k x_{i})}{n} - (m^{2} + k^{2} + 2 m k)} \\ = \sqrt{\frac{Σ x_{i}^{2}}{n} - m^{2} + \frac{2 k Σ x_{i}}{n} - 2 m k} \\ = \sqrt{\frac{Σ x_{i}^{2}}{n} - m^{2} + 2 k m - 2 m k} ∵ \frac{Σ x_{i}}{n} = m \\ = \sqrt{\frac{Σ x_{i}^{2}}{n} - m^{2}} \\ = s \end{aligned}$

Option (b) ( ks ): This option is incorrect because adding a constant ( k ) to each observation does not affect the spread or dispersion of the data, which is what the standard deviation measures. The standard deviation remains the same regardless of the constant added.
Option (c) ( s+k ): This option is incorrect because the standard deviation is a measure of the spread of the data around the mean, and adding a constant ( k ) to each observation shifts the mean but does not change the spread. Therefore, the standard deviation does not increase by ( k ).
Option (d) ( \frac{s}{k} ): This option is incorrect because dividing the standard deviation by the constant ( k ) implies that the spread of the data decreases when a constant is added to each observation. However, adding a constant does not affect the spread of the data, so the standard deviation remains unchanged.

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) $\frac{s}{k}$

(d) $s$

Show Answer

Solution

$\begin{aligned} m & = \frac{Σ x_{i}}{5}, s = \sqrt{\frac{Σ x_{i}^{2}}{5} - \frac{Σ x_{i}}{5}} \\ S D & = \sqrt{\frac{k^{2} Σ x_{i}^{2}}{5} - \frac{k Σ x_{i}^{2}}{5}} \\ = \sqrt{\frac{k^{2} Σ x_{i}^{2}}{5} - k^{2} \frac{Σ x_{i}^{2}}{5}} = \sqrt{\frac{Σ x_{i}^{2}}{5} - {\frac{Σ x_{i}}{5}}^{2}} = k s \end{aligned}$

Option (a) ( k + s ): This option is incorrect because the standard deviation of a set of observations scaled by a factor ( k ) is not simply the original standard deviation plus ( k ). The standard deviation measures the spread of the data, and scaling the data by ( k ) scales the spread by ( k ), not by adding ( k ).
Option (b) ( \frac{s}{k} ): This option is incorrect because dividing the original standard deviation ( s ) by ( k ) would imply that the spread of the data decreases when the data is scaled by ( k ). However, scaling the data by ( k ) actually increases the spread by a factor of ( k ), not decreases it.
Option (d) ( s ): This option is incorrect because it suggests that the standard deviation remains unchanged when the data is scaled by ( k ). However, scaling the data by ( k ) changes the spread of the data, and thus the standard deviation must also change by the same factor ( k ).

33. Let $x_{1}, x_{2}, \dots x_{n}$ be $n$ observations. Let $w_{i} = l x_{i} + k$ for $i = 1, 2, \dots, n$ , where $l$ and $k$ are constants. If the mean of $x_{i}^{'} s$ is 48 and their standard deviation is 12, the mean of $w_{i}^{'} s$ 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) $l = - 1.25, k = 5$

(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, ${\bar{w}}_{i} = {\bar{x}}_{i} + k$

[where, ${\bar{w}}_{i}$ is mean $w_{i}$ ’s and ${\bar{x}}_{i}$ is mean of $x_{i}^{'} s$ ]

$\Rightarrow 55 = 48 + k$

Now, SD of $w_{i} =$ SD of $x_{i}$

$\Rightarrow 15 = 12$

$\Rightarrow l = \frac{15}{12}$

$= 1.25$

From Eqs. (i) and (ii),

$k = 55 - 1.25 \times 48$

$= - 5$

Option (b) $l = - 1.25, k = 5$ :
- The value of $l$ should be positive because the standard deviation of $w_{i}$ is greater than the standard deviation of $x_{i}$ . A negative $l$ would imply a reduction in the spread, which contradicts the given standard deviations.
- The value of $k$ should be negative to satisfy the equation $55 = 48 + k$ . Here, $k = 5$ does not satisfy this equation.
Option (c) $l = 2.5, k = - 5$ :
- The value of $l$ should be 1.25, not 2.5, because the standard deviation of $w_{i}$ is 15 and the standard deviation of $x_{i}$ is 12. The correct $l$ is calculated as $\frac{15}{12} = 1.25$ .
- Although $k = - 5$ is correct, the incorrect value of $l$ makes this option invalid.
Option (d) $l = 2.5, k = 5$ :
- The value of $l$ should be 1.25, not 2.5, for the same reason as in option (c).
- The value of $k$ should be negative to satisfy the equation $55 = 48 + k$ . Here, $k = 5$ does not satisfy this equation.

34. The standard deviations for first natural numbers is

(a) 5.5

(b) 3.87

(d) 2.87

Show Answer

Solution

(d) We know that, SD of first $n$ natural number $= \sqrt{\frac{n^{2} - 1}{12}}$

$∴$ SD of first 10 natural numbers $= \sqrt{\frac{(10)^{2} - 1}{12}}$

$= \sqrt{\frac{100 - 1}{12}} = \sqrt{\frac{99}{12}} = \sqrt{8.25} = 2.87$

Option (a) 5.5 is incorrect because the standard deviation of the first 10 natural numbers is calculated using the formula $\sqrt{\frac{n^{2} - 1}{12}}$ , which does not yield 5.5.
Option (b) 3.87 is incorrect because, when applying the formula $\sqrt{\frac{n^{2} - 1}{12}}$ to the first 10 natural numbers, the result is not 3.87.
Option (c) 2.97 is incorrect because the correct calculation using the formula $\sqrt{\frac{n^{2} - 1}{12}}$ for the first 10 natural numbers results in 2.87, not 2.97.

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

(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 .

$\begin{aligned} ∴ \sum x_{i} = 2 + 3 + 4 + \dots + 11 \\ = \frac{10}{2} [2 \times 2 + 9 \times 1] = 5 [4 + 9] = 65 \\ Σ x_{i}^{2} = 2^{2} + 3^{2} + 4^{2} + 5^{2} + \dots + 11^{2} \\ = (1^{2} + 2^{2} + 3^{2} + \dots + 11^{2}) - (1^{2}) \\ = \frac{11 \times 12 \times 23}{6} - 1 \\ = \frac{11 \times 12 \times 23 - 6}{6} = 505 \end{aligned}$

$\begin{aligned} ∴ s^{2} & = \frac{Σ x_{i}^{2}}{n} - {\frac{Σ x_{i}}{n}}^{2} = \frac{505}{10} - \frac{65}{1} 0^{2} \\ = 50.5 - (6.5)^{2} \\ = 50.5 - 42.25 \\ = 8.25 \end{aligned}$

Option (a) 6.5: This option is incorrect because it represents the mean of the original numbers (1 to 10) rather than the variance. The variance calculation involves squaring the deviations from the mean, which results in a different value.
Option (b) 2.87: This option is incorrect because it is not derived from the correct formula for variance. The variance of a set of numbers is calculated using the sum of the squared deviations from the mean, divided by the number of observations. The value 2.87 does not match the result of this calculation for the given set of numbers.
Option (c) 3.87: This option is incorrect because it does not match the correct variance calculation for the given set of numbers. The correct variance is obtained by subtracting the square of the mean from the mean of the squares of the numbers, which results in 8.25, not 3.87.

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

(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

$\begin{matrix} ∴ & 0, - 1, - 2, - 3, - 4, - 5, - 6, - 7, - 8, - 9 \\ = - \frac{9 \times 10}{2} \\ = - 45 \\ and & Σ x_{i}^{2} = 0^{2} + (- 1)^{2} + (- 2)^{2} + \dots + (- 9)^{2} \\ = \frac{9 \times 10 \times 19}{6} \\ = 285 \\ ∴ S D & = \sqrt{\frac{285}{10} - \frac{- 45}{10}} = \sqrt{\frac{285}{10} - \frac{2025}{100}} \\ = \sqrt{\frac{2850 - 2025}{100}} = \sqrt{8.25} \\ Now, variance & = (S D)^{2} = (\sqrt{8.25})^{2} = 8.25 \end{matrix}$

Option (b) 6.5: This option is incorrect because the variance calculation involves squaring the deviations from the mean. The correct variance, as derived, is 8.25, not 6.5. The steps in the solution show that the sum of the squared deviations divided by the number of observations results in 8.25.
Option (c) 3.87: This option is incorrect because it significantly underestimates the variance. The correct variance calculation involves summing the squares of the deviations from the mean and then dividing by the number of observations, which results in 8.25, not 3.87.
Option (d) 2.87: This option is incorrect because it also underestimates the variance. The correct variance, as shown in the solution, is derived from the sum of the squared deviations from the mean, which results in 8.25, not 2.87.

37. The following information relates to a sample of size $60, Σ x^{2} = 18000$ , and $Σ x = 960$ . Then, the variance is

(a) 6.63

(b) 16

(d) 44

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Solution

(d)

$\begin{aligned} Variance = \frac{Σ x_{i}^{2}}{n} - \frac{Σ x_{i}}{n} \\ = \frac{18000}{60} - \frac{960}{6} 0^{2} = 300 - 256 = 44 \end{aligned}$

Option (a) 6.63: This value is incorrect because it is significantly lower than the calculated variance. The variance formula used, (\frac{\Sigma x_i^{2}}{n} - \left(\frac{\Sigma x_{i}}{n}\right)^2), results in a value of 44, not 6.63. This option might be a result of a miscalculation or misunderstanding of the formula.
Option (b) 16: This value is incorrect because it does not match the calculated variance. The correct calculation, (\frac{18000}{60} - \left(\frac{960}{60}\right)^2), results in 44. The value 16 could be a result of an incorrect intermediate step or a different formula.
Option (c) 22: This value is incorrect because it is half of the correct variance. The correct variance calculation yields 44. This option might be a result of an error in the final subtraction step or a misinterpretation of the formula.

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

(d) 2.5

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Solution

(a) Here

$C V_{1} = 50, C V_{2} = 60, {\bar{x}}_{1} = 30 and {\bar{x}}_{2} = 25$

$∴ C V_{1} = \frac{σ_{1}}{{\bar{x}}_{1}} \times 100 \Rightarrow 50 = \frac{σ_{1}}{30} \times 100$

$∴ σ_{1} = \frac{30 \times 50}{100} = 15 and C V_{2} = \frac{σ_{2}}{{\bar{x}}_{2}} \times 100$

$\Rightarrow 60 = \frac{σ_{2}}{25} \times 100$

$∴ σ_{2} = \frac{60 \times 25}{100} = 15$

$Now, σ_{1} - σ_{2} = 15 - 15 = 0$

Option (b) is incorrect because the calculated standard deviations for both distributions are equal (15), so their difference cannot be 1.
Option (c) is incorrect because the calculated standard deviations for both distributions are equal (15), so their difference cannot be 1.5.
Option (d) is incorrect because the calculated standard deviations for both distributions are equal (15), so their difference cannot be 2.5.

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

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Solution

(a) Given,

$σ_{C} = 5 \Rightarrow \frac{5}{9} (F - 32) = C$

$\begin{aligned} F & = \frac{9 C}{5} + 32 \\ σ_{F} & = \frac{9}{5} σ_{C} = \frac{9}{5} \times 5 = 9 \end{aligned}$

Here,

$σ_{F}^{2} = (9)^{2} = 81$

Option (b) 57 is incorrect because the variance is calculated as the square of the standard deviation. Given the standard deviation in Fahrenheit is 9, the variance should be (9^2 = 81), not 57.
Option (c) 36 is incorrect because, similar to option (b), the variance is the square of the standard deviation. Since the standard deviation in Fahrenheit is 9, the variance should be (9^2 = 81), not 36.
Option (d) 25 is incorrect because it represents the variance if the standard deviation were 5. However, the standard deviation in Fahrenheit is 9, so the variance should be (9^2 = 81), not 25.

Fillers

40. Coefficient of variation $= \frac{\dots}{Mean} \times 100$

Show Answer

Solution

$C V = \frac{S D}{Mean} \times 100$

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 $\overset{―}{x}$ , then $\sum_{i = 1}^{n} (x_{i} - \overset{―}{x})^{2}$ …… is than $Σ (x_{i} - a)^{2}$

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,

$\begin{aligned} S D & = \sqrt{Variance} \\ = \sqrt{121} = 11 \end{aligned}$

43. The standard deviation of a data is …… of any change in origin, but is …… on the change of scale.

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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.

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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.

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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.

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