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 $2\left(\mathrm{x}_{1}-1\right)+4 \beta, 2\left(\mathrm{x}_{2}-1\right)+ 4 \beta, \ldots . ., 2\left(\mathrm{x}_{10}-1\right)+4 \beta$, then $\frac{\beta \mu}{\sigma^{2}}$ is equal to :

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Statistics

Let

x_{1}, x_{2}, \dots\dots 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

\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 :

Answer: (A)

From

\sum_{i = 1}^{10} (x_{i} - 2) = 30

\sum x_{i} = 50

, so the mean is

\overset{x}{ˉ} = 5

.

Since the variance is

4/5

\frac{\sum x _{i}^{2}}{10} - \overset{x}{ˉ}^{2} = \frac{4}{5} ⟹ \sum x_{i}^{2} = 10 (25 + 0.8) = 258.

Expanding

\sum (x_{i} - β)^{2} = 98

258 - 100 β + 10 β^{2} = 98 ⟹ β^{2} - 10 β + 16 = 0 ⟹ β = 2 or 8.

Since

β > 2

β = 8

.

The transformed observations are

2 (x_{i} - 1) + 4 β = 2 x_{i} + 30

, so their mean is

μ = 2 (5) + 30 = 40,

and their variance is

σ^{2} = 2^{2} \cdot \frac{4}{5} = \frac{16}{5} .

Therefore

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

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option