91影视

The given Mean and standard deviation,

For Fred:

${渭}_{X_1}=24,{蟽}_{X_1}=2$

For Leona:

${渭}_{X_2}=24,{蟽}_{X_2}=2$

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.

${渭}_{X_1-X_2}={渭}_{X_1}-{渭}_{X_2}=24-24=0$

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

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

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

${渭}_{X_1-X_2}={渭}_{X_1}-{渭}_{X_2}=24-24=0$

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

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

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

${蟽}_{X_1-X_2}=\sqrt{{蟽}_{X_1-X_2}^2}=\sqrt{8}鈮�2.8284$

Since,

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

For the Probability:

Find the z - score,

$z=\frac{-5-0}{2.8284}鈮�-1.77$

$\mathrm{Or}$

$z=\frac{5-0}{2.8284}鈮�1.77$

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.