Independent solution

How to solve this Exponential Distribution question

Answer in brief

This is a sum of five independent exponential waiting times. Each has variance 1000^2, so the total variance is 5,000,000 and its standard deviation is 1000 sqrt(5), approximately 2236, making choice D correct.

Setup

Setup

Let each successive lifetime be an independent exponential variable with mean 1000. The requested time is their five-term sum.

T=X1+X2+X3+X4+X5T=X_1+X_2+X_3+X_4+X_5
E[Xi]=1000,Var⁡(Xi)=10002E[X_i]=1000,\qquad \operatorname{Var}(X_i)=1000^2

Model

Model

Independence makes the variance of the sum equal to the sum of the five component variances.

Var⁡(T)=∑i=15Var⁡(Xi)\operatorname{Var}(T)=\sum_{i=1}^{5}\operatorname{Var}(X_i)

Compute

Compute

Add the variances and then take one square root because the question asks for standard deviation.

Var⁡(T)=5(10002)=5,000,000\operatorname{Var}(T)=5(1000^2)=5{,}000{,}000
SD⁡(T)=10005=2236.067977…\operatorname{SD}(T)=1000\sqrt{5}=2236.067977\ldots

Answer

Answer

The standard deviation rounds to 2236 hours.

2236(D)\boxed{2236\quad\text{(D)}}