Question

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.

Short answer

1. Mean and standard deviation of D are 25 minutes and $11.180$minutes respectively.
2. Probability that Ed spent more time than Adelaide is $0.987322.$
   

Step-by-step solution

Part (a) Step 1: Given information

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,

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

Part (a) Step 2: Calculation

Consider,

$\mathrm{D}=\mathrm{A}-\mathrm{E}$

Mean of D

$\begin{array}{rcl}\mathrm{E}\left(\mathrm{D}\right)& =& \mathrm{E}(\mathrm{A}-\mathrm{E})\\ & =& \mathrm{E}\left(\mathrm{A}\right)-\mathrm{E}\left(\mathrm{E}\right)\\ & =& 50-25\\ & =& 25\text{minutes}\end{array}$

Variance of D

$\begin{array}{rcl}V(A-E)& =& V\left(A\right)+V\left(E\right)\\ & =& {10}^2+5^2\\ & =& 125\end{array}$

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

So,

$\mathrm{D}~\mathrm{N}(25,125)$

Part (b) Step 1: Given information

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,

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

Part (b) Step 2: Calculation

Consider,

$\begin{array}{rcl}\mathrm{P}(\mathrm{E}& >& \mathrm{A})=\mathrm{P}(\mathrm{E}-\mathrm{A}>0\left)\mathrm{P}\right(\mathrm{D}>0)\\ & =& \mathrm{P}\left(\frac{D-{\mu }_D}{{\sigma }_D}>\frac{0-25}{11.180}\right)\mathrm{P}(\mathrm{z}>-2.236)\\ & =& \mathrm{P}(\mathrm{z}<2.236)\end{array}$

From z tables,

$$\begin{gathered} P(z<2.23)=0.98713 \\ P(z<2.24)=0.98745 \\ P(z<2.236)=0.98713+(0.98745-0.98713)\times 0.6 \end{gathered}$$

Therefore,

$\mathrm{P}(\mathrm{z}<2.236)=0.987322$