Gujarati
13.Statistics
normal

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. . . . .

A

$20$

B

$30$

C

$40$

D

$42$

(IIT-2024)

Solution

Let equation of line is $y=m x+c$

$x$ $0$ $1$ $2$ $3$ $4$ $R -\{0,1,2,3,4\}$
$P ( x )$ $C$ $m + c$ $2 m + c$ $3 m + c$ $4 m+c$ $0$

$\sum_{ x =0}^4( mx + c )=1 \Rightarrow 10 m +5 c =1 \Rightarrow 2 m + c =\frac{1}{5}$   $. . . (1)$

$\text { mean }=\sum x _{ i } P _{ i }=\sum_{ i =0}^4\left( mx _{ i }+ c \right) x _{ i }=30 m +10 c =\frac{5}{2}$

$\therefore 3 m + c =\frac{1}{4} \ldots(2)$

$\text { from (1) and (2) m= } \frac{1}{20}, c =\frac{1}{10}$

$\sum P _{ i } x _{ i }^2=\sum_{ i =0}^4\left( mx _{ i }+ c \right) x _1^2$

$=\sum_{ i =0}^4\left( mx _{ i }^3+ cx _{ i }^2\right) \Rightarrow 100 m +30 c (\text { Now putting } m \text { and } c )$

$\Rightarrow \Sigma P _{ i }^2=5+3=8$

$\text { Variance }=\Sigma P _{ i } x _{ i }^2-\left(\Sigma P _{ i } x _{ i }\right)^2=8-\left(\frac{5}{2}\right)^2=\frac{7}{4}$

$\therefore 24 \alpha=42$

Standard 11
Mathematics

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