If the mean of the following probability distribution of a random variable $X$: $X: 0, 2, 4, 6, 8$; $P(X): a, 2a, a+b, 2b, 3b$ is $\dfrac{46}{9}$, then the variance of the distribution is

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Probability

If the mean of the following probability distribution of a random variable

X

X : 0, 2, 4, 6, 8

;

P (X) : a, 2 a, a + b, 2 b, 3 b

\frac{46}{9}

, then the variance of the distribution is

Answer: (A)

\sum P (X) = 1

a + 2 a + (a + b) + 2 b + 3 b = 1 \Rightarrow 4 a + 6 b = 1

…(i)

E (X) = \frac{46}{9}

0 \cdot a + 2 (2 a) + 4 (a + b) + 6 (2 b) + 8 (3 b) = \frac{46}{9} \Rightarrow 4 a + 4 a + 4 b + 12 b + 24 b = \frac{46}{9} \Rightarrow 8 a + 40 b = \frac{46}{9}

…(ii)

From (i)

4 a = 1 - 6 b

, substitute into (ii):

2 (1 - 6 b) + 40 b = \frac{46}{9} \Rightarrow 2 + 28 b = \frac{46}{9} \Rightarrow 28 b = \frac{28}{9} \Rightarrow b = \frac{1}{9}, a = \frac{1}{12}

E (X^{2}) = 0 + 4 (2 a) + 16 (a + b) + 36 (2 b) + 64 (3 b) = 8 a + 16 a + 16 b + 72 b + 192 b = 24 a + 280 b = 24 (\frac{1}{12}) + 280 (\frac{1}{9}) = 2 + \frac{280}{9} = \frac{298}{9}

Var = E (X^{2}) - [E (X)]^{2} = \frac{298}{9} - (\frac{46}{9})^{2} = \frac{298 \times 9 - 4 6 ^{2}}{81} = \frac{2682 - 2116}{81} = \frac{566}{81}

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option