Question

In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of employee absences. Based on a sample of 176 blue-collar workers, Martocchio estimated that the mean amount of paid time lost during a three-month period was 1.2 days per employee with a standard deviation of 1.1 days. Martocchio also estimated that the mean amount of unpaid time lost during a three-month period was 1.2 day per employee with a standard deviation of 1.9 days.

Suppose we randomly select a sample of 100 blue-collar workers. Based on Martocchio’s estimates:

(a) What is the probability that the average amount of paid time lost during a three-month period for the 100 blue-collar workers will exceed 1.5 days? (Use the rounded mean and standard error to compute the rounded Z-score used to find the probability. Round means to 1 decimal place, standard deviations to 2 decimal places, and probabilities to 4 decimal places. Round z-value to 2 decimal places.)

(b) What is the probability that the average amount of unpaid time lost during a three-month period for the 100 blue-collar workers will exceed 1.5 days? (Use the rounded mean and standard error to compute the rounded Z-score used to find the probability. Round standard deviations to 2 decimal places and probabilities to 4 decimal places. Round z-value to 2 decimal places.)

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	1.2
std deviation   =σ=	1.100
sample size       =n=	100
std error=σx̅=σ/√n=	0.11000

probability =P(X>1.5)=P(Z>(1.5-1.2)/0.11)=P(Z>2.73)=1-P(Z<2.73)=1-0.9968=0.0032

b)

sample size       =n=	100
std error=σx̅=σ/√n=	0.1900

probability =P(X>1.5)=P(Z>(1.5-1.2)/0.19)=P(Z>1.58)=1-P(Z<1.58)=1-0.9429=0.0571