8. Find the standard deviation, s, of sample data summarized in the frequency distribution table given below by using the formula below, where x represents the class midpoint, f represents the class frequency, and n represents the total number of sample values. Also, compare the computed standard deviation to the standard deviation obtained from the original list of data values, 9.0.

s=nfx2(fx)2n(n1)

Interval	30-39	40-49	50-59	60-69	70-79	80-89
Frequency	2	21	42	24	8	3

8a. Standard deviation=enter your response here (Round to one decimal place as needed.)

8b.Consider a difference of 20% between two values of a standard deviation to be significant. How does this computed value compare with the given standard deviation,9.0?

A.The computed value is not significantly different from the given value.

B.The computed value is significantly less than the given value.

C.The computed value is significantly greater than the given value.

15. Refer to the sample data for pre-employment drug screening shown below. If one of the subjects is randomly selected, what is the probability that the test result is a false positive? Who would suffer from a false positive result? Why?

	Positive test result	Negative test result
	Drug Use Is Indicated	Drug Use Is Not Indicated
Subject Uses Drugs	44	13
Subject Is Not a Drug User	1	33

15A. The probability of a false positive test result is =

(Round to three decimal places as needed.)

15B. Who would suffer from a false positive result? Why?

A.The employer would suffer because the person tested would not be suspected of using drugs when in reality he or she does use drugs.

B.The person tested would suffer because he or she would not be suspected of using drugs when in reality he or she does use drugs.

C.The employer would suffer because the person tested would be suspected of using drugs when in reality he or she does not use drugs.

D.The person tested would suffer because he or she would be suspected of using drugs when in reality he or she does not use drugs.