Question ID: #688
A random variable $X$ takes values 0, 1, 2, 3 with probabilities $\frac{2a+1}{30}, \frac{8a-1}{30}, \frac{4a+1}{30}, b$ respectively, where $a, b\in R$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^{2}+\mu^{2}=2$. Then $\frac{a}{b}$ is equal to:
- (1) 30
- (2) 3
- (3) 60
- (4) 12
Solution:
Sum of all probabilities in a probability distribution is equal to 1.
$$\sum P(X) = 1$$
$$\frac{2a+1}{30} + \frac{8a-1}{30} + \frac{4a+1}{30} + b = 1$$
$$\frac{(2a+1) + (8a-1) + (4a+1)}{30} + b = 1$$
$$\frac{14a+1}{30} + b = 1$$
$$14a + 1 + 30b = 30$$
$$14a + 30b = 29 \quad \dots (1)$$
The variance $\sigma^2$ is given by the formula $\sigma^2 = \sum X^2 P(X) – \mu^2$.
$$\sigma^2 + \mu^2 = \sum X^2 P(X)$$
Given that $\sigma^2 + \mu^2 = 2$:
$$\sum X^2 P(X) = 2$$
$$0^2\left(\frac{2a+1}{30}\right) + 1^2\left(\frac{8a-1}{30}\right) + 2^2\left(\frac{4a+1}{30}\right) + 3^2(b) = 2$$
$$0 + \frac{8a-1}{30} + 4\left(\frac{4a+1}{30}\right) + 9b = 2$$
$$\frac{8a-1 + 16a+4}{30} + 9b = 2$$
$$\frac{24a+3}{30} + 9b = 2$$
$$24a + 3 + 270b = 60$$
$$24a + 270b = 57$$
Divide the equation by 3:
$$8a + 90b = 19 \quad \dots (2)$$
Multiply equation (1) by 3:
$$42a + 90b = 87 \quad \dots (3)$$
Subtract equation (2) from equation (3):
$$(42a – 8a) + (90b – 90b) = 87 – 19$$
$$34a = 68$$
$$a = 2$$
Substitute $a = 2$ into equation (1):
$$14(2) + 30b = 29$$
$$28 + 30b = 29$$
$$30b = 1 \Rightarrow b = \frac{1}{30}$$
Calculate the value of $\frac{a}{b}$:
$$\frac{a}{b} = \frac{2}{1/30}$$
$$\frac{a}{b} = 60$$
Ans. (3)
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