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Let $X _{1}, X _{2}, \ldots, X _{18}$ be eighteen observations such that $\sum_{ i =1}^{18}\left( X _{ i }-\alpha\right)=36 \quad$ and $\sum_{i=1}^{18}\left(X_{i}-\beta\right)^{2}=90,$ where $\alpha$ and $\beta$ are distinct real numbers. If the standard deviation of these observations is $1,$ then the value of $|\alpha-\beta|$ is ...... .
$4$
$2$
$3$
$5$
Solution
$\sum_{i=1}^{18}\left(x_{i}-\alpha\right)=36, \sum_{i=1}^{18}\left(x_{i}-\beta\right)^{2}=90$
$\Rightarrow \sum_{i=1}^{18} x_{i}=18(\alpha+2), \sum_{i=1}^{18} x_{i}^{2}-2 \beta \sum_{i=1}^{18} x_{i}+18 \beta^{2}=90$
Hence $\sum x _{ i }^{2}=90-18 \beta^{2}+36 \beta(\alpha+2)$
Given $\frac{\sum x _{ i }^{2}}{18}-\left(\frac{\sum x _{ i }}{18}\right)^{2}=1$
$\Rightarrow 90-18 \beta^{2}+36 \beta(\alpha+2)-18(\alpha+2)^{2}=18$
$\Rightarrow 5-\beta^{2}+2 \alpha \beta+4 \beta-\alpha^{2}-4 \alpha-4=1$
$\Rightarrow(\alpha-\beta)^{2}+4(\alpha-\beta)=0 \Rightarrow|\alpha-\beta|=0$ or $4$
As $\alpha$ and $\beta$ are distinct $|\alpha-\beta|=4$
Similar Questions
Let $\mathrm{X}$ be a random variable with distribution.
| $\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
| $\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :
