Ed and Adelaide attend the same high school, but are in different math classes. The time E that it takes Ed to do his math homework follows a Normal distribution with mean $25$ minutes and standard deviation of $5$ minutes. Adelaide’s math homework time A follows a Normal distribution with mean of $50$ minutes and standard deviation $10$ minutes.

(a) Randomly select one math assignment of Ed’s and one math assignment of Adelaide’s. Let the random variable D be the difference in the amount of time each student spent on their assignments: $D=A-E$. Find the mean and the standard deviation of D. Show your work.

(b) Find the probability that Ed spent longer on his assignment than Adelaide did on hers. Show your work

${\mu }_A=50$

${\sigma }_A=10$

${\mu }_E=25$

${\sigma }_E=5$

Properties mean and variance:

${\mu }_{aX+bY}=a{\mu }_X+b{\mu }_Y$

${\sigma }_{aX+bY}^2=a^2{\sigma }_X^2+b^2{\sigma }_Y^2$

We can then determine the mean and variance of $D=A-E$:

localid="1650079367506" $$\begin{gathered} {\mu }_D={\mu }_{A-E} \\ ={\mu }_A-{\mu }_E \\ =50-25 \\ =25 \end{gathered}$$

localid="1650079383160" $$\begin{gathered} {\sigma }_D^2={\sigma }_{A-E}^2 \\ ={\sigma }_A^2+{\sigma }_E^2 \\ ={10}^2+5^2 \\ =125 \end{gathered}$$

The standard deviation is the square root of the variance:

localid="1650079394752" $$\begin{gathered} {\sigma }_D=\sqrt{{\sigma }_D^2} \\ =\sqrt{125} \\ \approx 11.1803 \end{gathered}$$

${\mu }_A=50$

${\sigma }_A=10$

${\mu }_E=25$

${\sigma }_E=5$

Properties mean and variance:

${\mu }_{aX+bY}=a{\mu }_X+b{\mu }_Y$

${\sigma }_{aX+bY}^2=a^2{\sigma }_X^2+b^2{\sigma }_Y^2$

We can then determine the mean and variance of $D=A-E$:

localid="1650079432625" $$\begin{gathered} {\mu }_D={\mu }_{A-E} \\ ={\mu }_A-{\mu }_E \\ =50-25 \\ =25 \end{gathered}$$

The standard deviation is the square root of the variance:

localid="1650079466152" $$\begin{gathered} {\sigma }_D=\sqrt{{\sigma }_D^2} \\ =\sqrt{125} \\ \approx 11.1803 \end{gathered}$$

The z-score is the value decreased by the mean, divided by the standard deviation:

$z=\frac{x-\mu }{\sigma }=\frac{0-25}{11.1803}\approx -2.24$

$z=\frac{x-\mu }{\sigma }=\frac{0-25}{11.1803}\approx -2.24$

Determine the corresponding probability using table A:

localid="1650079488846" $$\begin{gathered} P(E>A)=P(D<0) \\ =P(Z<-2.24) \\ =0.0125 \\ =1.25\% \end{gathered}$$