A random variable X takes values 0, 1, 2, 3 with probabilities 2a+130, 8a-130, 4a+130, b r…

Question

A random variable $X$ takes values $0, 1, 2, 3$ with probabilities $\dfrac{2a+1}{30}, \dfrac{8a-1}{30}, \dfrac{4a+1}{30}, b$ respectively, where $a, b \in \mathbb{R}$.

Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^2 + \mu^2 = 2$. Then $\dfrac{a}{b}$ is equal to:

Accepted Answer

Since probabilities sum to 1:
$$\frac{2a+1}{30} + \frac{8a-1}{30} + \frac{4a+1}{30} + b = 1 \implies 14a + 1 + 30b = 30 \implies 14a + 30b - 29 = 0 \quad \text{...(ii)}$$

Mean: $\mu = 0\cdot\frac{2a+1}{30} + 1\cdot\frac{8a-1}{30} + 2\cdot\frac{4a+1}{30} + 3b = \frac{16a+90b+1}{30}$

Using $\sigma^2 + \mu^2 = \sum x^2 p(x) = 2$:
$$\frac{8a-1}{30} + \frac{4(4a+1)}{30} + 9b = 2 \implies 8a + 90b - 19 = 0 \quad \text{...(i)}$$

Solving (i) and (ii): multiply (ii) by 3: $42a+90b=87$; subtract (i): $34a=68 \Rightarrow a=2$.
From (ii): $28+30b=29 \Rightarrow b = \frac{1}{30}$.

$$\frac{a}{b} = \frac{2}{1/30} = 60$$

Solutions · 1