ધારો કે માહિતીનો મધ્યક 5 છે. X 1 3 5 7 9 | vedclass

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(C) આપેલ મધ્યક $\bar{x} = 5$. આવૃત્તિઓનો સરવાળો $\sum f_i = 4 + 24 + 28 + \alpha + 8 = 64 + \alpha$.મધ્યક $\bar{x} = \frac{\sum x_i f_i}{\sum f_i} = \

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) આપેલ મધ્યક $\bar{x} = 5$. આવૃત્તિઓનો સરવાળો $\sum f_i = 4 + 24 + 28 + \alpha + 8 = 64 + \alpha$.મધ્યક $\bar{x} = \frac{\sum x_i f_i}{\sum f_i} = \frac{1(4) + 3(24) + 5(28) + 7(\alpha) + 9(8)}{64 + \alpha} = 5$.$\frac{4 + 72 + 140 + 7\alpha + 72}{64 + \alpha} = 5 \Rightarrow 288 + 7\alpha = 320 + 5\alpha \Rightarrow 2\alpha = 32 \Rightarrow \alpha = 16$.કુલ આવૃત્તિ $N = 64 + 16 = 80$.મધ્યક સાપેક્ષ સરેરાશ વિચલન $m = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{4|1-5| + 24|3-5| + 28|5-5| + 16|7-5| + 8|9-5|}{80} = \frac{128}{80} = 1.6$.વિચરણ $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N} = \frac{4(1-5)^2 + 24(3-5)^2 + 28(5-5)^2 + 16(7-5)^2 + 8(9-5)^2}{80} = \frac{352}{80} = 4.4$.તેથી,$\frac{3\alpha}{m + \sigma^2} = \frac{3(16)}{1.6 + 4.4} = \frac{48}{6} = 8$.

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

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