Question

In: Statistics and Probability

In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of...

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.)

Solutions

Expert Solution

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=σ=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=σ=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

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