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

b

step-by-step explanation:

answer:   expression= p+p*d%

price reduction + sale price = $625

sale price: $p+$p*d%

where d% is the percentage of discount

step-by-step explanation:

original price = p

price went down by $150 price reduction $p*d%

sale price= $ 475

therefore the price before sale is: $625

and the expression that shows the price before sale is: p+p*d%