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 5 minutes. Adelaide's math homework time A follows a Normal distribution with mean 50 minutes and standard deviation 10 minutes. Assume that E and A are independent random variables.

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.

b. Find the probability that Ed spent longer on his assignment than Adelaide did on hers.

Given :

E is the amount of time Ed spends on his arithmetic homework.

A is the amount of time Adelaide spends on her arithmetic homework.

E=25 minutes is the average time

E=5 minutes standard deviation

The average time is$A=50$minutes.

$A=10$minutes standard deviation

Both $\mathrm{E}$and A are unrelated to one another.

Concept used:

If two variables are independent,

$$\begin{gathered} Mean(\mathrm{X}-\mathrm{Y})=\mathrm{E}\left(\mathrm{X}\right)-\mathrm{E}\left(\mathrm{Y}\right) \\ Var(\mathrm{X}-\mathrm{Y})=Var\left(\mathrm{X}\right)+Var\left(\mathrm{Y}\right) \end{gathered}$$

Consider,

Mean of D

Variance of D

Standard deviation $\mathrm{D}=\sqrt{\mathrm{V}\left(\mathrm{D}\right)}=\sqrt{125}=11.180$ minutes

So,

Given :

E is the amount of time Ed spends on his arithmetic homework.

A is the amount of time Adelaide spends on her arithmetic homework.

$E=25$minutes is the average time

$E=5$minutes standard deviation

The average time is $A=50$minutes.

$A=10$ minutes standard deviation

Both E and A are unrelated to one another.

Concept used:

If two variables are independent,

$$\begin{gathered} Mean(\mathrm{X}-\mathrm{Y})=\mathrm{E}\left(\mathrm{X}\right)-\mathrm{E}\left(\mathrm{Y}\right) \\ Var(\mathrm{X}-\mathrm{Y})=Var\left(\mathrm{X}\right)+Var\left(\mathrm{Y}\right) \end{gathered}$$

Consider,

From z tables,

Therefore,