Chapter 1.3, Problem 98E

(a)

To determine

To find: Standard deviation

Answer to Problem 98E

Here, 1.1547 is the Standard deviation

Explanation of Solution

Given:

The data values 7,7,9,9

Calculation:

The sample size is the number of values:

n = 4

The mean is the sum of all values divided by the sample size:

  $\overline{X}=\frac{7+7+9+9}{4}=\frac{32}{4}=8$

If $X_i$ are the data values, then $X_i-\overline{X}$ are the deviations, ${(X_i-\overline{X})}^{2^{}}$ the squared deviations:

$X_i$	$(X_i-X_{mean})$	${(X_i-X_{mean})}^2$
7	-1	1
7	-1	1
9	1	1
9	1	1

The variance is the sum of the squared deviations divided by n-1:

  $s_x^2=\frac{1+1+1+1}{4-1}=\frac{4}{3}$

The standard deviation is the square root of the variance:

  $s_x=\sqrt{\frac{4}{3}}\approx 1.1547$

Conclusion:

Hence, 1.1547 is the standard deviation.

(b)

To determine

To interpret: the value of $s_x$

Answer to Problem 98E

Here, 1.1547 is the expected distance.

Explanation of Solution

Calculation:

This means that the expected distance between a typical amount of sleep and the mean is 1.1547 hours.

Conclusion:

Hence, 1.1547 is the expected distance.

(c)

To determine

To find: Whether the mean amount of sleep for all 30n students in class is close to 8hours or not

Answer to Problem 98E

No, the results of a convenience samplecannot be generalized.

Explanation of Solution

Calculation:

No, because the sample is a convenience sample, since the students were NOT randomly chosen. The results of a convenience sample cannot be generalized to entire population.

For example, it is possible that the students, who arrive earliest to class, are the ones that need the least sleep.

Conclusion:

Hence, the results of a convenience sample cannot be generalized.