Question

$\mu(X)$ and $Var(X)$ represents the mean and variance of random variable $X$. If $Var(2X)=\frac{1}{8}$ and $\mu(x^2)$ is $\frac{1}{4}$, then $\mu(4x)$ is $\mu$. The value of $2\mu^2$ is equal to

• A. 7
• B. 14
• C. 49
• D. 98

Answer 1

Answer: (A)

7

Answer 2

Given $Var(2X) = \frac{1}{8}$ and $\mu(x^2) = E(X^2) = \frac{1}{4}$.

Using the property $Var(aX) = a^2 Var(X)$: $Var(2X) = 2^2 Var(X) = 4 Var(X)$. So, $4 Var(X) = \frac{1}{8}$, which implies $Var(X) = \frac{1}{32}$.

Using the property $Var(X) = E(X^2) - [E(X)]^2$: $\frac{1}{32} = \frac{1}{4} - [E(X)]^2$. Rearranging, we get $[E(X)]^2 = \frac{1}{4} - \frac{1}{32} = \frac{8}{32} - \frac{1}{32} = \frac{7}{32}$.

We are given $\mu(4x) = \mu$, which means $\mu = E(4X)$. Using the property $E(aX) = a E(X)$: $\mu = 4 E(X)$.

Now we need to find $2\mu^2$: $\mu^2 = (4 E(X))^2 = 16 [E(X)]^2$. Substitute the value of $[E(X)]^2$: $\mu^2 = 16 \times \frac{7}{32} = \frac{16}{32} \times 7 = \frac{1}{2} \times 7 = \frac{7}{2}$.

Finally, $2\mu^2 = 2 \times \frac{7}{2} = 7$.

The value of $2\mu^2$ is 7.