Consider 10 observation $\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots, \mathrm{x}_{10}$. such that $\sum_{i=1}^{10}(x_{i}-\alpha)=2$ and $\sum_{i=1}^{10}(x_{i}-\beta)^{2}=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The $\frac{\beta}{\alpha}$ is equal to :

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Statistics

Consider 10 observation

x_{1}, x_{2}, \dots, x_{10}

. such that

\sum_{i = 1}^{10} (x_{i} - α) = 2

and

\sum_{i = 1}^{10} (x_{i} - β)^{2} = 40

, where

α, β

are positive integers. Let the mean and the variance of the observations be

\frac{6}{5}

and

\frac{84}{25}

respectively. The

\frac{β}{α}

is equal to :

Answer: (A)

Mean

= \frac{\sum x _{i}}{10} = \frac{6}{5} ⟹ \sum x_{i} = 12

\sum (x_{i} - α) = 2 ⟹ \sum x_{i} - 10 α = 2 ⟹ 12 - 10 α = 2 ⟹ α = 1

.

Variance

= \frac{1}{10} \sum x_{i}^{2} - (mean)^{2} = \frac{84}{25}

\frac{1}{10} \sum x_{i}^{2} = \frac{84}{25} + \frac{36}{25} = \frac{120}{25} = \frac{24}{5} ⟹ \sum x_{i}^{2} = 48.

Now use

\sum (x_{i} - β)^{2} = 40

\sum x_{i}^{2} - 2 β \sum x_{i} + 10 β^{2} = 40 ⟹ 48 - 24 β + 10 β^{2} = 40 ⟹ 10 β^{2} - 24 β + 8 = 0 ⟹ 5 β^{2} - 12 β + 4 = 0.

β = \frac{12 \pm 144 - 80}{10} = \frac{12 \pm 8}{10} ⟹ β = 2 or β = \frac{2}{5} .

Since

β

is a positive integer,

β = 2

\frac{β}{α} = \frac{2}{1} = 2.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option