(A) For a discrete random variable,the sum of probabilities must be $1$. $\sum_{x=0}^{6} f(x) = 1$ $k(0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = 1$ $k

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(A) For a discrete random variable,the sum of probabilities must be $1$. $\sum_{x=0}^{6} f(x) = 1$ $k(0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = 1$ $k(0 + 1 + 4 + 9 + 16 + 25 + 36) = 1$ $91k = 1 \implies k = \frac{1}{91}$ The cumulative distribution function $F(4)$ is defined as $P(X \leq 4)$. $F(4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)$ $F(4) = k(0^2 + 1^2 + 2^2 + 3^2 + 4^2)$ $F(4) = k(0 + 1 + 4 + 9 + 16) = 30k$ Substituting $k = \frac{1}{91}$,we get: $F(4) = 30 \times \frac{1}{91} = \frac{30}{91}$

Class 12 Mathematics Probability Probability distribution

Medium

The p.d.f. of a discrete random variable $X$ is defined as $f(x) = \begin{cases} kx^2, & x \in \{0, 1, 2, 3, 4, 5, 6\} \\ 0, & \text{otherwise} \end{cases}$. Then the value of $F(4)$ (c.d.f.) is:

MHT CET 2023 All MHT CET

A
$\frac{30}{91}$
B
$\frac{30}{97}$
C
$\frac{15}{47}$
D
$\frac{15}{97}$

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The p.d.f. of a discrete random variable $X$ is defined as $f(x) = \begin{cases} kx^2, & x \in \{0, 1, 2, 3, 4, 5, 6\} \\ 0, & \text{otherwise} \end{cases}$. Then the value of $F(4)$ (c.d.f.) is:

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