Let the mean of the data x 1 3 5 7 9… - JEE Main 2023 Mathematics PYQ

JEE Main 2023 — Mathematics Probability & Statistics

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Let the mean of the data

$x$	$1$	$3$	$5$	$7$	$9$
Frequency $(f)$	$4$	$24$	$28$	$\alpha$	$8$

5.

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

_______.

Official previous-year question

Held on 13 Apr 2023 · Verified 6 Jul 2026.

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Solution

Given,

The mean of the below data is $5$

$x$	$1$	$3$	$5$	$7$	$9$
Frequency $(f)$	$4$	$24$	$28$	$\alpha$	$8$

Now finding mean we get,

$\frac{4+72+28\times 5+7\alpha +72}{64+\alpha }=5$

$\Rightarrow \alpha =16$

So, total number of observation will be, $\displaystyle \sum {f}_{i}=80$

Now finding, mean deviation we get,

$M.D=\frac{\sum {f}_{i}|{x}_{i}-5|}{\sum {f}_{i}}$

$=\frac{4\times 4+24\times 2+0+16\times 2+8\times 4}{80}$ $=\frac{8}{5}$

So, $m=\frac{8}{5}$

Now finding, variance we get,

${\sigma }^{2}=\frac{\sum {f}_{i}{|{x}_{i}-5|}^{2}}{\sum {f}_{i}}$

$\Rightarrow {\sigma }^{2}=\frac{4\times 16+24\times 4+0+16\times 4+8\times 16}{80}$

$\Rightarrow {\sigma }^{2}=\frac{22}{5}$

Hence, the value of $\frac{3\alpha }{m+{\sigma }^{2}}=\frac{3\times 16}{\frac{8}{5}+\frac{22}{5}}=8$

Your note

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