Independent solution

How to solve this Expectation and Variance question

Answer in brief

This is a variance-of-an-independent-sum calculation for a three-point count distribution. One individual count has variance 0.1456, so the 64-count total has variance 9.3184 and matches choice D.

Setup

Setup

Let X be one individual's count. Compute its first and second moments directly from its three possible values.

E⁡[X]=0(0.90)+1(0.08)+2(0.02)=0.12\operatorname{E}[X]=0(0.90)+1(0.08)+2(0.02)=0.12
E⁡[X2]=02(0.90)+12(0.08)+22(0.02)=0.16\operatorname{E}[X^2]=0^2(0.90)+1^2(0.08)+2^2(0.02)=0.16

Model

Model

Convert the two moments to an individual variance. Independence then makes the variance of the group total the sum of 64 equal variances.

Var⁡(X)=E⁡[X2]−E⁡[X]2\operatorname{Var}(X)=\operatorname{E}[X^2]-\operatorname{E}[X]^2
Var⁡ ⁣(∑i=164Xi)=∑i=164Var⁡(Xi)\operatorname{Var}\!\left(\sum_{i=1}^{64}X_i\right)=\sum_{i=1}^{64}\operatorname{Var}(X_i)

Compute

Compute

Evaluate the individual variance and multiply by the group size.

Var⁡(X)=0.16−(0.12)2=0.1456\operatorname{Var}(X)=0.16-(0.12)^2=0.1456
Var⁡ ⁣(∑i=164Xi)=64(0.1456)=9.3184\operatorname{Var}\!\left(\sum_{i=1}^{64}X_i\right)=64(0.1456)=9.3184

Answer

Answer

The total variance rounds to 9.32.

9.3184≈9.32(D)\boxed{9.3184\approx 9.32\quad\text{(D)}}