Question

Q4

1. A sales manager would like to compare the recent performance of two salespersons under his supervision. From their sales record, their daily sales were normally distributed with the same mean and standard deviations were $200 and $250. Amount of sales of 10 days in the sales records will be selected randomly for each salesperson, calculate the probability that their mean daily sales will have a difference of at least $150.

2. In evaluating the customers satisfaction level, a company would like to be 95% confident that the error in estimating the population mean does not exceed 0.05. If the population standard deviation is 0.8 what is the sample size required?

Answer 1

a.

Let X be the mean difference of sales between two salespersons.

E(X) = 0 (As the means of daily sales are equal for two salespersons)

Standard error of mean difference, SE(X) = = 101.2423

Probability that their mean daily sales will have a difference of at least $150

= P(|X| > 150)

= P(X < -150) + P(X > 150)

= P[Z < (-150 - 0)/101.2423] + P[Z > (150 - 0)/101.2423]

= P[Z < -1.48] + P[Z > 1.48]

= 0.0694 + 0.0694

= 0.1388

b.

Margin of Error , E = 0.05

= 0.8

Z score for 95% confidence level is 1.96

Required sample size, n =

=  983.4496 984 (Rounding to next integer)