Loser buys the pizza Leona and Fred are friendly competitors in high school. Both are about to take the ACT college entrance examination. They agree that if one of them scores 5 or more points better than the other, the loser will buy the winner a pizza. Suppose that, in fact, Fred and Leona have equal ability so that each score on a randomly selected test varies Normally with mean $24$and standard deviation $2$. (The variation is due to luck in guessing and the accident of the specific questions being familiar to the student.) The two scores are independent. What is the probability that the scores differ by 5 or more points in either direction?

The given Mean and standard deviation,

For Fred:

${\mu }_{X_1}=24,{\sigma }_{X_1}=2$

For Leona:

${\mu }_{X_2}=24,{\sigma }_{X_2}=2$

Consider,

$X_1$: Fred's grade

$x_2$: Leona'sgrade

Then

$X_1-X_2$is the difference between Fred and Leona's scores

${\mathrm{X}}_1$and ${\mathrm{X}}_2$are independent.

Since $X_1$and $X_2$have a Normal distribution, their difference $\left(X_1-X_2\right)$also has a Normal distribution.

Now,

${\mathrm{X}}_1:$Fred's score

${\mathrm{X}}_2$: Leona's score

Then

${\mathrm{X}}_1-{\mathrm{X}}_2$is the difference in scores of Fred and Leona

${\mathrm{X}}_1$and ${\mathrm{X}}_2$are independent.

Since ${\mathrm{X}}_1$and ${\mathrm{X}}_2$have a Normal distribution, their difference $\left({\mathrm{X}}_1-{\mathrm{X}}_2\right)$also has a Normal distribution.

Now, if two variables are present, the difference between their means equals the difference between their means.

${\mu }_{X_1-X_2}={\mu }_{X_1}-{\mu }_{X_2}=24-24=0$

When the random variables are independent, the variance of the difference is equal to the total of their variances.

${\sigma }_{X_1-X_2}^2={\sigma }_{X_1}^2+{\sigma }_{X_2}^2=(2{)}^2+(2{)}^2=4+4=8$

The mean of the difference between any two variables is equal to the difference between their means.

${\mu }_{X_1-X_2}={\mu }_{X_1}-{\mu }_{X_2}=24-24=0$

When the random variables are independent, the variance of the difference is equal to the total of their variances.

${\sigma }_{X_1-X_2}^2={\sigma }_{X_1}^2+{\sigma }_{X_2}^2=(2{)}^2+(2{)}^2=4+4=8$

The standard deviation is equal to the square root of the variance.

${\sigma }_{X_1-X_2}=\sqrt{{\sigma }_{X_1-X_2}^2}=\sqrt{8}\approx 2.8284$

Since,

The variation$\left(X_2-X_1\right)$follows a Normal distribution.

For the Probability:

Find the z - score,

$z=\frac{-5-0}{2.8284}\approx -1.77$

$\mathrm{Or}$

$z=\frac{5-0}{2.8284}\approx 1.77$

Use the normal probability table in the appendix to get the corresponding probability.

In a standard normal probability table,$P(z<-1.77),$look for the row that begins with $-1.7$and the column that begins with.07.

$\begin{array}{rcl}P\left(X_1-X_2<-5\text{or}{X}_1-X_2>5\right)& =& P(z<-1.77\text{or}z>1.77)\\ & =& 2\left(P\right(z<-1.77\left)\right)\\ & =& 2(0.0384)\\ & =& 0.0768\end{array}$

As a result, the chances of the scores varying by more than 5 points in any direction are 0.0768.