Let the mean of the data be 5. X 1 3 5 7 9 | vedclass

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(C) Given mean $\bar{x} = 5$. The sum of frequencies is $\sum f_i = 4 + 24 + 28 + \alpha + 8 = 64 + \alpha$.The mean is given by $\bar{x} = \frac{\sum

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$8$

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(C) Given mean $\bar{x} = 5$. The sum of frequencies is $\sum f_i = 4 + 24 + 28 + \alpha + 8 = 64 + \alpha$.The mean is given by $\bar{x} = \frac{\sum x_i f_i}{\sum f_i} = \frac{1(4) + 3(24) + 5(28) + 7(\alpha) + 9(8)}{64 + \alpha} = 5$.$\frac{4 + 72 + 140 + 7\alpha + 72}{64 + \alpha} = 5 \Rightarrow 288 + 7\alpha = 320 + 5\alpha \Rightarrow 2\alpha = 32 \Rightarrow \alpha = 16$.Total frequency $N = 64 + 16 = 80$.Mean deviation about mean $m = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{4|1-5| + 24|3-5| + 28|5-5| + 16|7-5| + 8|9-5|}{80} = \frac{4(4) + 24(2) + 0 + 16(2) + 8(4)}{80} = \frac{16 + 48 + 32 + 32}{80} = \frac{128}{80} = 1.6$.Variance $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N} = \frac{4(1-5)^2 + 24(3-5)^2 + 28(5-5)^2 + 16(7-5)^2 + 8(9-5)^2}{80} = \frac{4(16) + 24(4) + 0 + 16(4) + 8(16)}{80} = \frac{64 + 96 + 64 + 128}{80} = \frac{352}{80} = 4.4$.Thus,$\frac{3\alpha}{m + \sigma^2} = \frac{3(16)}{1.6 + 4.4} = \frac{48}{6} = 8$.

$X$	$1$	$3$	$5$	$7$	$9$
$f$	$4$	$24$	$28$	$\alpha$	$8$

Expenditure in $Rs. (x)$	$0-50$	$50-100$	$100-150$	$150-200$	$200-250$
No. of families $(f)$	$24$	$33$	$37$	$b$	$25$

Value	$4$	$5$	$8$	$9$	$6$	$12$	$11$
Frequency	$5$	$f_1$	$f_2$	$2$	$1$	$1$	$3$

List-$I$	List-$II$
$(P) \ 7f_1+9f_2$ is equal to	$(1) \ 146$
$(Q) \ 19\alpha$ is equal to	$(2) \ 47$
$(R) \ 19\beta$ is equal to	$(3) \ 48$
$(S) \ 19\sigma^2$ is equal to	$(4) \ 145$
	$(5) \ 55$

Let the mean of the data be $5$.

$X$ $1$ $3$ $5$ $7$ $9$

$f$ $4$ $24$ $28$ $\alpha$ $8$

If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data,then $\frac{3 \alpha}{m+\sigma^2}$ is equal to $..........$.

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Let the mean of the data be $5$. $X$ $1$ $3$ $5$ $7$ $9$ $f$ $4$ $24$ $28$ $\alpha$ $8$ If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data,then $\frac{3 \alpha}{m+\sigma^2}$ is equal to $..........$.