Chapter 8.1, Problem 5E

(a)

To determine

To calculate:The shape, center, and the spread of the sampling distribution of the selected people appeared for the National Assessment of Educational Progress test.

Answer to Problem 5E

Shape is approximately normal, center is 280 and the spread is 2.0702.

Explanation of Solution

Given information:

Sample mean of the score = 280

Standard deviation = 60

Sample size of the people from the large population = 840

Calculation:

Sample mean $\overline{x}$ =280

Standard deviation $\sigma$ =60

Sample size = 840

For finding the shape of the sampling distribution, use the Central Limit.

Thus, the shape of the sampling distribution of the sample mean is approximately normal.

For center, from the given information the mean of the sampling distribution of the sample mean $\mu =280$ .

Thus, center equal to the mean $\mu =280$

And for the spread, use the formula

  $\text{Population}\text{standard}\text{deviation}=\frac{\text{Standard}\text{deviation}}{\text{Square}\text{root}\text{of}\text{sample}\text{size}}$

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{60}{\sqrt{840}}\\ \approx 2.0702\end{array}$

Thus, shape is approximately normal with mean 280 as center and the spread is the standard deviation which is equal 2.0702.

(b)

To determine

Todraw:The sampling distribution of the mean andmark three standard deviation value on each side of the mean.

Explanation of Solution

Given information:

Sample mean of the score = 280

Standard deviation = 60

Sample size of the people from the large population = 840

Sample mean $\overline{x}$ = 280

Standard deviation $\sigma$ = 60

Sample size = 840

  

Below figure shows the sampling distribution of $\overline{x}$ with mean $\mu$ as center and three standard deviation values on each sides.

  

(c)

To determine

To calculate: The distance m of the mean of the sampling distribution.

Answer to Problem 5E

Distance m of the mean of the sampling distribution is 4.140.

Explanation of Solution

Given information:

Sample mean of the score = 280

Standard deviation = 60

Sample size of the people from the large population = 840

Calculation:

Sample mean $\overline{x}$ = 280

Standard deviation $\sigma$ = 60

Sample size = 840

Population standard deviation ${\sigma }_{\overline{x}}=2.0702$

The mean of the sampling distribution of the sample mean is $\overline{x}$ = $\mu =280$ .

Find the standard deviation.

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{60}{\sqrt{840}}\\ \approx 2.0702\end{array}$

By 68-95-00.7% rule, about 95% of all values of $\overline{x}$ lie within a distance m of the mean i.e. twice the population standard deviations of the mean. Thus

  $\begin{array}{c}m=2\sigma \\ =2\times 2.070\\ =4.140\end{array}$

  

Hence, distance m of the mean of the sampling distribution is 4.140.

(d)

To determine

To calculate: The percent of all possible sample of $\mu$ is captured by interval.

Answer to Problem 5E

Sample intervals capture 95% of $\mu$ .

Explanation of Solution

Given information:

Sample mean of the score = 280

Standard deviation = 60

Sample size of the people from the large population = 840

Distance m of the mean of the sampling distribution.

95% of all values of $\overline{x}$ lie within a distance m of the mean of the sampling distribution.

Hence,95% of all possible sample capture $\mu$ .