JEE Main 2025 Jan 29 Shift 1, Mathematics Q18: Measure of Dispersion

JEE Main2025Jan 29, Shift 1Mathematics

Let $x_{1}, x_{2}, \dots, x_{10}$ be ten observations such that $i = 1 \sum 10 (x_{i} - 2) = 30,$ $i = 1 \sum 10 (x_{i} - β)^{2} = 98, β > 2$ and their variance is $\frac{4}{5} .$ If $μ$ and $σ^{2}$ are respectively the mean and the variance of $2 (x_{1} - 1) + 4 β, 2 (x_{2} - 1) + 4 β, \dots, 2 (x_{10} - 1) + 4 β$ , then $\frac{β μ}{σ ^{2}}$ is equal to :

A
$100$
B
$110$
C
$120$
D
$90$

Solution

From $\sum (x_{i} - 2) = 30$ , $\sum x_{i} = 50$ , so $\overset{x}{ˉ} = 5.$

Variance: $\frac{\sum x _{i}^{2}}{10} - \overset{x}{ˉ}^{2} = \frac{4}{5} \Rightarrow \sum x_{i}^{2} = 10 (\frac{4}{5} + 25) = 258.$

Expand the second sum: $\sum (x_{i} - β)^{2} = \sum x_{i}^{2} - 2 β \sum x_{i} + 10 β^{2} = 258 - 100 β + 10 β^{2} = 98.$

So $β^{2} - 10 β + 16 = 0 \Rightarrow (β - 2) (β - 8) = 0.$ Since $β > 2$ , $β = 8.$

For $y_{i} = 2 (x_{i} - 1) + 4 β = 2 x_{i} + 30$ : $μ = 2 \overset{x}{ˉ} + 30 = 40$ and $σ^{2} = 4 \cdot \frac{4}{5} = \frac{16}{5} .$

$\frac{β μ}{σ ^{2}} = \frac{8 \cdot 40}{16/5} = \frac{320 \cdot 5}{16} = 100.$

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JEE Main 2025 Jan 29 Shift 1, Mathematics Q18: Measure of Dispersion

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