3 Marks Question

Question 13 Marks

Find the mean and standard deviation of the age distribution given in the following table of 542 persons :

Age (In years)	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80	80 - 90
Number of Persons	3	61	132	153	140	51	2

Answer

Here, Assumed mean $=55$

Age Class Interval	Mid-value $\left(x_i\right)$	Frequency $\left(f_i\right)$	$\begin{array}{c}u_i=\frac{x_i-a}{h} (a=55,~ h=10)\end{array}$	$u_i^2$	$f_i u_i$	$f_i u_i^2$
20-30	25	3	-3	9	-9	27
30-40	35	61	-2	4	-122	244
40-50	45	132	-1	1	-132	132
50-60	55	153	0	0	0	0
60-70	65	140	1	1	140	140
70-80	75	51	2	4	102	204
80-90	85	2	3	9	6	18
Total		N = 542			-15	765

$\begin{aligned} \operatorname{Mean}(\bar{x}) & =a+h \frac{\sum f_i u_i}{N} \\ & =55+10 \times\left(\frac{-15}{542}\right) \\ & =55-\frac{150}{542} \\ & =55-0.28 \\ & =54.72\end{aligned}$
and Standard deviation $(\sigma)$
$\begin{array}{l}=h \sqrt{\left\{\frac{1}{N} \sum f_i u_i^2-\left(\frac{\sum f_i u_i}{N}\right)^2\right\}} \\ =10 \sqrt{\frac{765}{542}-\left(\frac{-15}{542}\right)^2} \\ =10 \sqrt{1.411-0.000}=10 \times \sqrt{1.411} \\ =10 \times 1.187=11.87\end{array}$

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Question 23 Marks

The frequency distribution of income of $100$ families is as follows :

Income (in ₹)	0 - 1000	1000 - 2000	2000 - 3000
Number of families	18	26	30
Income (in ₹)	3000 - 4000	4000 - 5000	5000 - 6000
Number of families	12	10	4

Find the standard deviation.

Answer

Let assumed mean $a=3500$ Class height $h=1000$

Class interval	$x_i$ mid value	Number of families	$\begin{array}{c}v_i= \frac{x_1-3500}{1000}\end{array}$	$v_i^2$	$f_i v_i$	$f_i v_i^2$
0 - 1000	500	18	-3	9	-54	162
1000 - 2000	1500	26	-2	4	-52	104
2000 - 3000	2500	30	-1	1	-30	30
3000 - 4000	3500	12	0	0	0	0
4000 - 5000	4500	10	1	1	10	10
5000 - 6000	5500	4	2	4	8	16
		$\begin{array}{c}100 =\Sigma f_i\end{array}$			$\begin{array}{l}\Sigma f_i v_i= -188\end{array}$	$\begin{array}{l}\Sigma f_i v_i^2 =322\end{array}$

$\begin{aligned}\therefore \text{Arithmetic mean}\quad \bar{x} & =a+\frac{h \times \Sigma f_i v_i}{\Sigma f} \\ & =3500+\frac{1000 \times(-118)}{100} \\ & =3500-1180=2320\end{aligned}$
$\begin{aligned} \therefore \text { Standard deviation } \sigma & =h \times \sqrt{\frac{1}{N} \sum f_i v_i^2-\left(\frac{\sum f_i v_i}{N}\right)^2} \\ \sigma & =1000 \times \sqrt{\frac{322}{100}-\frac{118 \times 118}{100 \times 100}} \\ \sigma & =1000 \times \sqrt{\frac{32200-13924}{100 \times 100}} \\ \sigma & =10 \sqrt{18276}=10 \times 135.188 \\ \sigma & =1351.88\end{aligned}$
$\therefore \text {Standard deviation} \quad \sigma=1351.88$

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Question 33 Marks

Find the mean deviation and standard deviation of the series $a, a+d, a+2 d, \ldots, a+2 n d$ from mean and prove that the variable is greater than the first one.

Answer

Given series $a, a+d, a+2 d, \ldots a+2 n d$ is an arithmetic progression.
$\begin{aligned}\text {So, its mean}\quad\bar{x} & =\frac{\text { First term }+ \text { last term }}{2} \\& =\frac{a+a+2 n d}{2}=\frac{2(a+n d)}{2} \\\bar{x} & =(a+n d)\end{aligned}$
We know that :
$\begin{array}{l}\text { Mean deviation M.D. }=\frac{\sum|x-\bar{x}|}{n} \\\text { Here number of terms }=(2 n+1)\end{array}$
$\therefore \quad\quad\quad\quad\quad\quad\ \text { M.D. }=\frac{\sum\left|x_i-\bar{x}\right|}{2 n+1}$
$\begin{array}{r}\{a-(a+n d)\}+\{a+d-(a+n d)\}+ \{a+2 d-(a+n d)\}+\ldots \ldots\end{array}$
$\text { M.D. }=\frac{\{(a+2 n d)-(a+n d)\}}{2 n+1}$
$=\frac{\begin{array}{r}(-n d)+d(1-n)+d(2-n) \ldots \ldots \\d(n-n)+d(n+1-n)+ \\d(n+2-n)+\ldots \ldots d(n+n-n)\end{array}}{2 n+1}$
$=\frac{2 d(1+2+3+4+\ldots \ldots n)}{(2 n+1)}=\frac{2 d \frac{n(n+1)}{2}}{(2 n+1)}$
Since sum of $n$ terms of natural numbers $=\frac{n(n+1)}{2}$
$\Rightarrow \quad \text { M.D. }=\frac{n(n+1) d}{(2 n+1)}\ldots\ldots (1)$
Standard deviation
$\sigma=\sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{n}}$
Here, $n=(2 n+1)$
$\therefore \sigma=\sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{(2 n+1)}}$
$=\sqrt{\begin{array}{l}(-n d)^2+\{(n-1) d\}^2+\{(n-2) d\}^2+\ldots \ldots . .+ \\ \frac{\{(n-n) d\}^2+\{n-(n+1) d\}^2+\ldots \ldots\left\{\{n-(n+n) d\}^2 \right.}{(2 n+1)}\end{array}}$
The above expression can be written as :
$\sigma=\sqrt{\frac{2 d^2\left\{1^2+2^2+3^2+4^2+\ldots \ldots+n^2\right\}}{(2 n+1)}}$
As we know that sum of arithmetic progression
$\left(1^2+2^2+3^2+4^2+\ldots+n^2\right) \text { is } \frac{n(n+1)(2 n+1)}{6}$
Putting values,
$\begin{array}{l}\sigma=\sqrt{\frac{2 d^2(n)(n+1)(2 n+1)}{6 \times(2 n+1)}} \\ \sigma=d \sqrt{\frac{n(n+1)}{3}}\ldots\ldots (2)\end{array}$
It is clear from equation (1) and (2), the variable is greater than the first one.
Hence proved.

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Question 43 Marks

The following data are based on the measurement of collar of a representative group of students of a college :

Mid value (in-inches)	12.0	12.5	13.0	13.5	14.0
Number of students	2	16	36	60	76
Mid value (in-inches)	14.5	15.0	15.5	16.0
Number of students	37	18	3	2

Find the mean and standard deviation of the above data.

Answer

Here, Let assumed mean $=14$

Mid-value (in inches) $\left(x_i\right)$	Frequency $\left(f_i\right)$	$\begin{array}{c}d_i=x_i-a (a=14)\end{array}$	$d_i{ }^2$	$f_i d_i$	$f_i d_i{ }^2$
12.0	2	-2	4	-4	8
12.5	16	-1.5	2.25	-24	36
13.0	36	-1	1	-36	36
13.5	60	-0.5	0.25	-30	15
14.0	76	0	0	0	0
14.5	37	0.5	0.25	18.5	9.25
15.0	18	1	1	18	18
15.5	3	1.5	2.25	4.5	6.75
16.0	2	2	4	4	8
Total	N=250			-49	137

$\operatorname{Mean}(\bar{x})=a+\frac{\sum f_i d_i}{N}=14+\frac{(-49)}{250}$
$\quad\quad\quad\quad\begin{array}{l}=14+(-0.2) \\ =14-0.2=13.8\end{array}$
and standard deviation
$=\sqrt{\left\{\frac{1}{N} \sum f_i d_i^2-\left(\frac{\sum f_i d_i}{N}\right)^2\right\}}=\sqrt{\frac{137}{250}-\left(\frac{-49}{250}\right)^2}$
$=\sqrt{0.548-0.038}=\sqrt{0.510}=0.713$

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