JEE Advanced 2024 · Paper 1 · Mathematics Q13

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MathematicsInteger+4Question 13 of 51

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 $xy$ -plane, and $P(X = x) = 0$ for all $x \in \mathbb{R} - \{0, 1, 2, 3, 4\}$ . If the mean of $X$ is $\dfrac{5}{2}$ , and the variance of $X$ is $\alpha$ , then the value of $24\alpha$ is _____.

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Answer and worked solution

Official answer: 42

Let $P(X = x) = mx + k$ for $x = 0,\dots,4$ .

Total probability: $10m + 5k = 1$ .
Mean: $\sum x(mx+k) = 30m + 10k = \frac52$ .

Solving: $m = \frac{1}{20}$ , $k = \frac{1}{10}$ . The probabilities are $0.10, 0.15, 0.20, 0.25, 0.30$ .

$E[X^2] = 0.15 + 4(0.20) + 9(0.25) + 16(0.30) = 8$ , so $\alpha = 8 - \left(\frac52\right)^2 = \frac74$ .

$24\alpha = 42$ .

Topics:Random variables and distributions Variance and standard deviation

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JEE Advanced 2024 · Paper 1, Mathematics question 13