Q99 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
If a random variable $X$ has a probability mass function $P(x) = \begin{cases} kx^2, & \text{for } x = 1, 2, 3, 4 \\ 0, & \text{otherwise} \end{cases}$, then the mean of $X$ is ...
- A$3.232$
- B$2.221$
- Correct answerC$3.333$
- D$2.222$
Solution
The sum of all probabilities must be equal to $1$.
$\sum_{x=1}^{4} P(x) = 1$
$k(1^2 + 2^2 + 3^2 + 4^2) = 1$
$k(1 + 4 + 9 + 16) = 1$
$30k = 1 \Rightarrow k = \dfrac{1}{30}$
The mean of the random variable $X$ is given by $E(X) = \sum_{x=1}^{4} x P(x)$.
$E(X) = \sum_{x=1}^{4} x(kx^2) = k \sum_{x=1}^{4} x^3$
$E(X) = \dfrac{1}{30} (1^3 + 2^3 + 3^3 + 4^3)$
$E(X) = \dfrac{1}{30} (1 + 8 + 27 + 64)$
$E(X) = \dfrac{100}{30} = \dfrac{10}{3} = 3.333$
Answer: $3.333$
$\sum_{x=1}^{4} P(x) = 1$
$k(1^2 + 2^2 + 3^2 + 4^2) = 1$
$k(1 + 4 + 9 + 16) = 1$
$30k = 1 \Rightarrow k = \dfrac{1}{30}$
The mean of the random variable $X$ is given by $E(X) = \sum_{x=1}^{4} x P(x)$.
$E(X) = \sum_{x=1}^{4} x(kx^2) = k \sum_{x=1}^{4} x^3$
$E(X) = \dfrac{1}{30} (1^3 + 2^3 + 3^3 + 4^3)$
$E(X) = \dfrac{1}{30} (1 + 8 + 27 + 64)$
$E(X) = \dfrac{100}{30} = \dfrac{10}{3} = 3.333$
Answer: $3.333$
Hence, the correct answer is option: 'C'.