2 Marks Questions

Question 12 Marks

The runs scored by players in $10$ matches are $38, 70, 48, 34, 42, 55, 63, 46, 54, 44.$ Find the variance and standard deviation.

Answer

Arithmetic mean
$\begin{aligned}\bar{x} & =\frac{38+70+48+34+42+55+63+46+54+44}{10} \\& =\frac{494}{10}=49.4\end{aligned}$

$x_i$	$x_i-\bar{x}$	$\left(x_i-\bar{x}\right)^2$
38	-11.4	129.96
70	+20.6	424.36
48	-1.4	1.96
34	-15.4	237.16
42	-7.4	54.76
55	+5.6	31.36
63	+13.6	184.96
46	-3.4	11.56
54	+4.6	21.16
44	-5.4	29.16
		$\Sigma\left\|x_i-\bar{x}\right\|^2=1126.4$

$\therefore \quad$ Variance $\sigma^2=\frac{1}{n} \Sigma\left(x_i-\bar{x}\right)^2$
$\quad\quad\sigma^2=\frac{1126.4}{10}=112.64$
$\begin{array}{rlr}\text {and Standard derivation}\quad \sigma & =+\sqrt{\sigma^2}=\sqrt{112.64} \\ & =10.61 \end{array}$

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Question 22 Marks

The means of four students are as follows : $25,35,$ $45$ and $55,$ find the mean deviation from the mean.

Answer

Here, Arithmetic mean $\bar{x}=\frac{\Sigma x_i}{n}$
$=\frac{25+35+45+55}{4}$
$=\frac{160}{4}=40$

S.NO.	$x_i$	$\left\|x_i-40\right\|$
1.	25	15
2.	35	5
3.	45	5
4.	55	15
	Total	$\Sigma\left\|x_i-40\right\|=40$

$\therefore$ Standard deviation from mean
$
=\frac{1}{n} \Sigma\left|x_i-40\right|=\frac{40}{4}=10
$

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Question 32 Marks

In a series of $2 n$ observations, half of them equal to $a$ and remaining half equal $-a$. If the standard deviation of the observations is 2, then find the value of $|a|.$

Answer

Total number of observations $=2 n$
Half are equal to $a$ and half are equal to $-a$.
Therefore, mean $\bar{x}=0$
Standard deviation $=\sqrt{\frac{n(-a-0)^2+n(a-0)^2}{2 n}}$
$\Rightarrow \quad 2=\sqrt{\frac{n a^2+n a^2}{2 n}}$
$\Rightarrow \quad 2=\sqrt{a^2}$
$\Rightarrow \quad a=2$
$\Rightarrow \quad|a|=2.$

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Question 42 Marks

The mean of 150 items is 90 and standard deviation is 6. Find the sum of data items and the sum of squares of the data items.

Answer

$\begin{aligned}\text { Sum of items } & =\text { Number of items } \times \text { their mean } \\& =150 \times 90=13500\end{aligned}$
Here, $\bar{x}=90, n=150$ and standard deviation $(\sigma)=6$
$\begin{aligned}\because\quad \sigma & =\sqrt{\frac{\sum_{i=1}^n\left(x_i-\bar{x}\right)^2}{n}} \\ \sigma^2 n & =\sum_{i=1}^n\left(x_i-\bar{x}\right)^2\end{aligned}$
$\begin{aligned} 36 \times 150 & =\sum_{i=1}^{150} x_i^2-2 \bar{x} \sum_{i=1}^{150} x_i+(\bar{x})^2 \times 150 \\ 5400 & =\sum_{i=1}^{150} x_i^2-2 \times 90 \times 13500+150 \times 8100 \\ \sum_{i=1}^{150} x_i^2 & =5400+2430000-1215000 \\ \sum_{i=1}^{150} x_i^2 & =1220400\end{aligned}$

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Question 52 Marks

The weight of $9$ students of a class is given below and its median is $50$. Calculate the mean deviation from the median.

Answer

Here, Median $(M)=50$

S.No.	$x_i$	$\left\|x_i-50\right\|$
1	47	3
2	50	0
3	58	8
4	45	5
5	53	3
6	59	9
7	47	3
8	60	10
9	49	1
	Total	$\Sigma\left\|x_i-50\right\|=42$

$\begin{aligned} \therefore \text { Standard deviation from median } & =\frac{1}{n} \sum\left|x_i-50\right| \\ & =\frac{42}{9}=4.67\end{aligned}$

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