A manufacturer who is considering the implementation of a one-week training program for all new employees decides to test the program with the next 100 employees hired, and then compare their productivity rate to the productivity rate of new employees based on past records–a rate that is normally distributed with a mean of 60 and a standard deviation of 8. The new program needs to produce a minimum improvement of 4 to be considered worthwhile. What is the comparison distribution's standard deviation?

0.8

Step-by-step explanation:

From the given information:

The population mean = 60

The standard deviation = 8

The sample size = 100

The sample distribution which follows a normal distribution is the set of values taken by this statistic on each sample from the same population.

The standard deviation of the comparison distribution's =

The standard deviation of the comparison distribution's =

The standard deviation of the comparison distribution's = 8/10

The standard deviation of the comparison distribution's = 0.8

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743