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 $a, b \in \mathbb{R}$.

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:

Question

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 $a, b \in \mathbb{R}$.

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:

Accepted Answer

D. 1. Since $X$ takes values $0,1,2,3$ with the given probabilities, compute the **mean**:
$$
\mu=0\cdot\frac{2a+1}{30}+1\cdot\frac{8a-1}{30}+2\cdot\frac{4a+1}{30}+3b=\frac{16a+90b+1}{30}
$$

2. Note that $\sigma^2+\mu^2=E(X^2)$, since $\sigma^2=E(X^2)-\mu^2$. So the given condition $\sigma^2+\mu^2=2$ directly means:
$$
E(X^2)=2
$$

3. Computing $E(X^2)=\sum X_i^2P(X_i)$:
$$
0+1\cdot\frac{8a-1}{30}+4\cdot\frac{4a+1}{30}+9b=2
$$

4. Simplifying:
$$
8a+90b-19=0 \quad(i)
$$

5. Since the total probability must sum to 1:
$$
\frac{2a+1}{30}+\frac{8a-1}{30}+\frac{4a+1}{30}+b=1 \;\Rightarrow\; 14a+1+30b=30
$$

6. So:
$$
14a+30b-29=0 \quad(ii)
$$

7. Solving equations (i) and (ii) simultaneously:
$$
a=2,\qquad b=\frac{1}{30}
$$

8. Therefore:
$$
\frac{a}{b}=\frac{2}{1/30}=60
$$

Hence, the answer is **Option D**.