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Let $X$ be a random variable, and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x)), x=0,1,2,3,4$, lie on a fixed straight line in the $x y$-plane, and $P(X=x)=0$ for all $x \in R$ $\{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24 \alpha$ is. . . . .
$20$
$30$
$40$
$42$
Solution
Let equation of line is $y=m x+c$
| $x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $R -\{0,1,2,3,4\}$ |
| $P ( x )$ | $C$ | $m + c$ | $2 m + c$ | $3 m + c$ | $4 m+c$ | $0$ |
$\sum_{ x =0}^4( mx + c )=1 \Rightarrow 10 m +5 c =1 \Rightarrow 2 m + c =\frac{1}{5}$ $. . . (1)$
$\text { mean }=\sum x _{ i } P _{ i }=\sum_{ i =0}^4\left( mx _{ i }+ c \right) x _{ i }=30 m +10 c =\frac{5}{2}$
$\therefore 3 m + c =\frac{1}{4} \ldots(2)$
$\text { from (1) and (2) m= } \frac{1}{20}, c =\frac{1}{10}$
$\sum P _{ i } x _{ i }^2=\sum_{ i =0}^4\left( mx _{ i }+ c \right) x _1^2$
$=\sum_{ i =0}^4\left( mx _{ i }^3+ cx _{ i }^2\right) \Rightarrow 100 m +30 c (\text { Now putting } m \text { and } c )$
$\Rightarrow \Sigma P _{ i }^2=5+3=8$
$\text { Variance }=\Sigma P _{ i } x _{ i }^2-\left(\Sigma P _{ i } x _{ i }\right)^2=8-\left(\frac{5}{2}\right)^2=\frac{7}{4}$
$\therefore 24 \alpha=42$
Similar Questions
The variance $\sigma^2$ of the data is $ . . . . . .$
| $x_i$ | $0$ | $1$ | $5$ | $6$ | $10$ | $12$ | $17$ |
| $f_i$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
