Question

Let the moment generating function (m.g.f) of a random variable is of the form \(M_X(t)=(0.4e^t +0.6)^8\). Find the mean and variance of a random variable Y=3X+2.

• A. E(Y) =9.4 and V(Y) = 17.28
• B. E(Y) =9.6 and V(Y) = 17.26
• C. E(Y) =9.4 and V(Y) = 17.26
• D. E(Y) =9.6 and V(Y) = 17.28

Hint:

When applying linear transformations to variance, always square the multiplier:
\[ V(aX + b) = a^2 V(X) \]
Since \(a = 3\), the variance must be multiplied by \(9\).
\[ 9 \times 1.92 = 17.28 \]
This immediately eliminates options (B) and (C).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The moment generating function (MGF) of a random variable uniquely determines its distribution.
The given MGF has the characteristic form of a Binomial distribution:
\[ M_X(t) = (p e^t + q)^n \]
Where \(p\) is the probability of success, \(q = 1-p\) is the probability of failure, and \(n\) is the number of trials.
Key Formula or Approach:
For a Binomial distribution \(X \sim \text{Binomial}(n, p)\):
- Mean, \(E(X) = n p\)
- Variance, \(V(X) = n p q\)
For a linear transformation \(Y = aX + b\):
- Mean, \(E(Y) = a E(X) + b\)
- Variance, \(V(Y) = a^2 V(X)\)

Step 2: Detailed Explanation:
From the given MGF, \(M_X(t) = (0.4e^t + 0.6)^8\), we identify:
- \(n = 8\)
- \(p = 0.4\)
- \(q = 0.6\)
Now, calculate the mean and variance of \(X\):
\[ E(X) = n p = 8 \times 0.4 = 3.2 \]
\[ V(X) = n p q = 8 \times 0.4 \times 0.6 = 1.92 \]
We are given \(Y = 3X + 2\).
Calculate the variance of \(Y\):
\[ V(Y) = 3^2 V(X) = 9 \times 1.92 = 17.28 \]
Calculate the mean of \(Y\) strictly following the linear transformation:
\[ E(Y) = 3E(X) + 2 = 3(3.2) + 2 = 9.6 + 2 = 11.6 \]
Note: The official options write the mean as \(E(Y) = 9.6\), which corresponds to the unshifted transformation \(Y = 3X\) (excluding the additive constant \(+2\) in the mean calculation by typographical oversight).
Since the variance \(V(Y) = 17.28\) is exactly matching, Option (D) represents the intended key.

Step 3: Final Answer:
The correct option is (D).