Independent solution

How to solve this Normal Distribution question

Setup

Setup

Aggregate the independent additions and removals separately.

A∼N(20(1.5), 20(0.252))=N(30,1.25)A\sim N\left(20(1.5),\,20(0.25^2)\right)=N(30,1.25)
R∼N(4(7.25), 4(0.502))=N(29,1)R\sim N\left(4(7.25),\,4(0.50^2)\right)=N(29,1)

Model

Model

The ending level exceeds the beginning level exactly when the net change D=A-R is positive.

D=A−R∼N(30−29, 1.25+1)=N(1,2.25)D=A-R\sim N(30-29,\,1.25+1)=N(1,2.25)

Compute

Compute

Standardize the zero threshold using standard deviation 1.5.

Pr⁡(D>0)=Pr⁡(Z>0−11.5)\Pr(D>0)=\Pr\left(Z>\frac{0-1}{1.5}\right)
Pr⁡(D>0)=Φ(23)=0.7475075\Pr(D>0)=\Phi\left(\frac23\right)=0.7475075

Answer

Answer

The probability rounds to 0.75.

0.75(D)\boxed{0.75\quad\text{(D)}}