Independent solution

How to solve this Central Limit Theorem question

Setup

Setup

Compute the mean and variance of one summand from its two equally likely values.

E[X]=0(0.5)+2(0.5)=1E[X]=0(0.5)+2(0.5)=1
E[X2]=02(0.5)+22(0.5)=2E[X^2]=0^2(0.5)+2^2(0.5)=2
Var⁡(X)=2−12=1\operatorname{Var}(X)=2-1^2=1

Model

Model

Independence makes the aggregate mean and variance additive, and the central limit theorem supplies a normal approximation.

E[S]=100E[X]=100E[S]=100E[X]=100
Var⁡(S)=100Var⁡(X)=100,SD⁡(S)=10\operatorname{Var}(S)=100\operatorname{Var}(X)=100,\qquad \operatorname{SD}(S)=10

Compute

Compute

Standardize the boundary and evaluate the upper standard-normal tail.

z=115−10010=1.5z=\frac{115-100}{10}=1.5
Pr⁡(S>115)≈1−Φ(1.5)=0.0668072\Pr(S>115)\approx 1-\Phi(1.5)=0.0668072

Answer

Answer

The requested approximation rounds to 0.067.

0.067(B)\boxed{0.067\quad\text{(B)}}