Let x 1 , x 2 , ldots ldots . x 10 be ten observations such that sum i=1 10 (x i -2 )=30, sum i=1 10 (x i -β ) 2 =98, β 2 and their variance is 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, ldots . ., 2 (x 10 -1 )+4 β , then frac β μ σ 2 is equal to

Mathematics · Probability

JEE Main 2025 — 29 January, Morning Shift — Question 62

Let $\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots \ldots . \mathrm{x}_{10}$ be ten observations such that

$\sum_{i=1}^{10}\left(x_{i}-2\right)=30, \sum_{i=1}^{10}\left(x_{i}-\beta\right)^{2}=98, \beta>2$ and

their variance is $\frac{4}{5}$ . If $\mu$ and $\sigma^{2}$ are respectively the mean and the variance of

$2\left(x_{1}-1\right)+4 \beta, 2\left(x_{2}-1\right)+$ $4 \beta, \ldots . ., 2\left(x_{10}-1\right)+4 \beta$ ,

then $\frac{\beta \mu}{\sigma^{2}}$ is equal to

$\frac{4}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{10}-\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{10}\right)^{2}$

$\frac{4}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{10}-25$

$\Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^{2}=258$

Now $\sum_{i=1}^{10}\left(x_{i}-\beta\right)^{2}=98$

$\sum_{i=1}^{10}\left(x_{i}^{2}-2 \beta . x_{i}+\beta^{2}\right)=98$

$258-2 \beta(50)+10 \beta^{2}=98 (\beta-8)(\beta-2)=0$

$\beta=8$ or $\beta=2 \quad($ as $\beta>2)$

$\therefore \beta=8$

Now,

2(x_1 - 1) + 4\beta,\; 2(x_2 - 1) + 4\beta,\; \ldots,\; 2(x_{10} - 1) + 4\beta

= 2x_1 + 30,\; 2x_2 + 30,\; \ldots,\; 2x_{10} + 30

\mu = 2(5) + 30 = 40

\sigma^2 = 2^2 \left(\frac{4}{5}\right) = \frac{16}{5}

\therefore \ \frac{B\mu}{\sigma^2} = \frac{8 \times 40}{16/5} = 100

Answer key and solution verified before publishing.

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