Let X be a random variable, and let \(\mathrm{P}(\mathrm{X}=\mathrm{x})\) denote the probability that X takes the value x. Suppose that the points ( \(\mathrm{x}, \mathrm{P}(\mathrm{X}=\mathrm{x})\) ), \(\mathrm{x}=0,1,2,3,4\), lie on a fixed straight line in the xy-plane, and \(\mathrm{P}(\mathrm{X}=\mathrm{x})=0\) for all \(\mathrm{x} \in \mathbb{R}-\{0,1,2,3,4\}\). If the mean of X is \(\frac{5}{2}\), and the variance of X is \(\alpha\), then the value of \(24 \alpha\) is .........

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Probability XII

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 R - {0, 1, 2, 3, 4}

. If the mean of X is

\frac{5}{2}

, and the variance of X is

α

, then the value of

24 α

is .........

Answer: 42

Let the line be

y = m x + c

. Then

P (X = x) = m x + c

for

x \in {0, 1, 2, 3, 4}

.

Sum of probabilities:

x = 0 \sum 4 (m x + c) = m (0 + 1 + 2 + 3 + 4) + 5 c = 10 m + 5 c = 1

2 m + c = \frac{1}{5} ...(1)

Mean:

E [X] = x = 0 \sum 4 x (m x + c) = m (0^{2} + 1^{2} + 2^{2} + 3^{2} + 4^{2}) + c (0 + 1 + 2 + 3 + 4) = 30 m + 10 c = \frac{5}{2}

3 m + c = \frac{1}{4} ...(2)

From (1) and (2):

m = \frac{1}{20}

c = \frac{1}{10}

.

Second moment:

E [X^{2}] = x = 0 \sum 4 x^{2} (m x + c) = m (0^{3} + 1^{3} + 2^{3} + 3^{3} + 4^{3}) + c (0^{2} + 1^{2} + 2^{2} + 3^{2} + 4^{2})

= m \cdot 100 + c \cdot 30 = \frac{100}{20} + \frac{30}{10} = 5 + 3 = 8

Variance:

α = E [X^{2}] - (E [X])^{2} = 8 - (\frac{5}{2})^{2} = 8 - \frac{25}{4} = \frac{7}{4}

Therefore:

24 α = 24 \cdot \frac{7}{4} = 42

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Integer answer