- A$20$
- B$30$
- C$40$
- ✓$42$
| $x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $R -\{0,1,2,3,4\}$ |
| $P ( x )$ | $C$ | $m + c$ | $2 m + c$ | $3 m + c$ | $4 m+c$ | $0$ |
$\sum_{ x =0}^4( mx + c )=1 \Rightarrow 10 m +5 c =1 \Rightarrow 2 m + c =\frac{1}{5}$ $. . . (1)$
$\text { mean }=\sum x _{ i } P _{ i }=\sum_{ i =0}^4\left( mx _{ i }+ c \right) x _{ i }=30 m +10 c =\frac{5}{2}$
$\therefore 3 m + c =\frac{1}{4} \ldots(2)$
$\text { from (1) and (2) m= } \frac{1}{20}, c =\frac{1}{10}$
$\sum P _{ i } x _{ i }^2=\sum_{ i =0}^4\left( mx _{ i }+ c \right) x _1^2$
$=\sum_{ i =0}^4\left( mx _{ i }^3+ cx _{ i }^2\right) \Rightarrow 100 m +30 c (\text { Now putting } m \text { and } c )$
$\Rightarrow \Sigma P _{ i }^2=5+3=8$
$\text { Variance }=\Sigma P _{ i } x _{ i }^2-\left(\Sigma P _{ i } x _{ i }\right)^2=8-\left(\frac{5}{2}\right)^2=\frac{7}{4}$
$\therefore 24 \alpha=42$
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$I_n=\int_0^1\left(1-x^k\right)^n d x, n \in \mathbb{N} \text {, satisfies } 147 I_{20}=148 I_{21}$ is :