JEE Main 2024 (Online) 1st February Evening ShiftQuestion 1
Consider $10$ observations $x_1,x_2,\ldots,x_{10}$ 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. Then $\frac{\beta}{\alpha}$ is equal to :
Statistics
Solution
Mean $=\frac65\Rightarrow\sum x_i=12$. $\sum(x_i-\alpha)=12-10\alpha=2\Rightarrow\alpha=1$. Variance $\frac{84}{25}\Rightarrow\sum x_i^2=10\left(\frac{84}{25}+\frac{36}{25}\right)=48$. $\sum(x_i-\beta)^2=48-24\beta+10\beta^2=40\Rightarrow5\beta^2-12\beta+4=0\Rightarrow\beta=2$ (integer). $\beta/\alpha=2$. Option B.