Let the mean of the data | x | 1 | 3 | 5 | 7 | 9 | |---|---|---|---|---|---| | Frequency(f) | 4 | 24 | 28 | $\alpha$ | 8 | be 5. 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 _____.

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Statistics

Let the mean of the data

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

be 5. If

m

and

σ^{2}

are respectively the mean deviation about the mean and the variance of the data, then

\frac{3 α}{m + σ ^{2}}

is equal to _____.

Answer: 8

Given mean

\overset{x}{ˉ} = 5

5 = \frac{\sum x _{i} f _{i}}{\sum f _{i}} = \frac{4 ( 1 ) + 24 ( 3 ) + 28 ( 5 ) + α ( 7 ) + 8 ( 9 )}{64 + α}

5 (64 + α) = 4 + 72 + 140 + 7 α + 72

320 + 5 α = 288 + 7 α

2 α = 32 \Rightarrow α = 16

Total frequency:

\sum f_{i} = 80

.

Mean deviation:

m = \frac{\sum f _{i} ∣ x _{i} - x ˉ ∣}{\sum f _{i}} = \frac{4 ( 4 ) + 24 ( 2 ) + 28 ( 0 ) + 16 ( 2 ) + 8 ( 4 )}{80} = \frac{16 + 48 + 0 + 32 + 32}{80} = \frac{128}{80} = \frac{8}{5}

Variance:

σ^{2} = \frac{\sum f _{i} ( x _{i} - x ˉ ) ^{2}}{\sum f _{i}} = \frac{4 ( 16 ) + 24 ( 4 ) + 28 ( 0 ) + 16 ( 4 ) + 8 ( 16 )}{80} = \frac{64 + 96 + 0 + 64 + 128}{80} = \frac{352}{80} = \frac{22}{5}

Therefore:

\frac{3 α}{m + σ ^{2}} = \frac{3 ( 16 )}{\frac{8}{5} + \frac{22}{5}} = \frac{48}{\frac{30}{5}} = \frac{48}{6} = 8

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer