Chapter 15, Problem 15.28E

a.

To determine

To find: The probability of the scores of the student is between 20 and 30.

Answer to Problem 15.28E

The probability of the student’s score in between 20 and 30 is 0.5588.

Explanation of Solution

Given info:

The mean and standard deviation for scores of an SRS of students which follows normal distribution is 25.0 and 6.5.

Calculation:

The formula for standard deviation of the sampling distribution of $\overline{x}$ is,

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Substitute 6.5 for $\sigma$ and 1 for n in above equation.

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{6.5}{\sqrt{1}}\\ =6.5\end{array}$

Therefore, the value of standard deviation for sampling space is 6.5.

The formula of z-score for sampling distribution is,

$z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$ (1)

For mean value 20:

Substitute 20 for $\overline{x}$, 25 for ${\mu }_{\overline{x}}$ and 6.5 for ${\sigma }_{\overline{x}}$ in equation (1)

$\begin{array}{c}{z}_{20}=\frac{20-25}{6.5}\\ =\frac{-5}{6.5}\\ =-0.77\end{array}$

For mean value 30:

Substitute 30 for $\overline{x}$, 25 for ${\mu }_{\overline{x}}$ and 6.5 for ${\sigma }_{\overline{x}}$ in equation (1)

$\begin{array}{c}{z}_{30}=\frac{30-25}{6.5}\\ =\frac{5}{6.5}\\ =0.77\end{array}$

The probability in between 20 and 30 is, $P\left(20<\overline{x}<30\right)=P\left(z_{20}<z<z_{30}\right)$

Substitute $-0.77$ for $z_{20}$ and 0.77 for $z_{30}$ in above equation.

$\begin{array}{c}P\left(20<\overline{x}<30\right)=P\left(-0.77<z<0.77\right)\\ =P\left(z<0.77\right)-P\left(z<-0.77\right)\end{array}$

From the Table A: Standard Normal Cumulative Proportions, the area to left of $-0.77$ is 0.2206 and the area to the left of 0.77 is 0.7794.

The probability of the student’s score in between 20 and 30 is,

$\begin{array}{c}P\left(20<\overline{x}<30\right)=P\left(z<0.77\right)-P\left(z<-0.77\right)\\ =0.7794-0.2206\\ =0.5588\end{array}$

Thus, the probability of the score of students score in between 20 and 30 is 0.5588.

b.

To determine

The sampling distribution of the average score $\overline{x}$.

Answer to Problem 15.28E

The value of standard deviation for sampling distribution is 1.3 for average score $\overline{x}$.

Explanation of Solution

Calculation:

The formula for standard deviation of the sampling distribution of $\overline{x}$ is,

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Substitute 6.5 for $\sigma$ and 25 for n in above equation.

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{6.5}{\sqrt{25}}\\ =\frac{6.5}{5}\\ =1.3\end{array}$

Hence, the value of standard deviation of the sampling distribution is 1.3.

c.

To determine

The probability of mean score of the sample is between 20 and 30.

Answer to Problem 15.28E

The probability of the student’s score between 20 and 30 is 0.9998.

Explanation of Solution

Calculation:

The formula of z score for sampling distribution is,

$z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$ (1)

For mean value 20:

Substitute 20 for $\overline{x}$, 25 for ${\mu }_{\overline{x}}$ and 1.3 for ${\sigma }_{\overline{x}}$ in equation (1), to obtain $z_{20}$.

$\begin{array}{c}{z}_{20}=\frac{20-25}{1.3}\\ =\frac{-5}{1.3}\\ =-3.85\end{array}$

Similarly, for mean value 30, Substitute 30 for $\overline{x}$, 25 for ${\mu }_{\overline{x}}$ and 1.3 for ${\sigma }_{\overline{x}}$ in equation (1)

$\begin{array}{c}{z}_{30}=\frac{30-25}{1.3}\\ =\frac{5}{1.3}\\ =3.85\end{array}$

The probability in between 20 and 30 is, $P\left(20<\overline{x}<30\right)=P\left(z_{20}<z<z_{30}\right)$

Substitute $-3.85$ for $z_{20}$ and 3.85 for $z_{30}$ in the equation.

$\begin{array}{c}P\left(20<\overline{x}<30\right)=P\left(-3.85<z<3.85\right)\\ =P\left(z<3.85\right)-P\left(z<-3.85\right)\end{array}$

From the Table A: Standard Normal Cumulative Proportions, the area to left of $-3.85$ is 0.0001 and the area to the left of 3.85 is 0.9999.

Thus, the probability of the student’s score is between 20 and 30 is,

$\begin{array}{c}P\left(20<\overline{x}<30\right)=P\left(z<0.77\right)-P\left(z<-0.77\right)\\ =0.9999-0.0001\\ =0.9998\end{array}$

Hence, the probability of the score of students score is between 20 and 30 is 0.9998.