Time lost due to employee absenteeism is an important problem for many companies

Question

Time lost due to employee absenteeism is an important problem for many companies. The human resources department of Western Electronics has studied the distribution of time lost due to absenteeism by individual employees. During a one-year period, the department found a mean of 21 days and a standard deviation of 10 days based on data for all the employees.

a) If you pick an employee at random, what is the probability that the number of absences for this one employee would exceed 25 days?

b) if a sample of 36 employees are taken and sample mean computed. What would be the mean, standard deviation and shape of the distribution of sample means for samples of size 36?

c) A group of 36 employees is selected at random to participate in a program that allows a flexible work schedule, which the human resources department hopes will decrease the employee absenteeism in the future. What is the probability that the mean for the sample of 36 employees randomly selected for the study would exceed 25 days?

Explanation / Answer

Let X be the number of absences for the employee.

X ~ N(mean = 21, sigma = 10)

a) If you pick an employee at random, what is the probability that the number of absences for this one employee would exceed 25 days?

That is we have to find P(X > 25).

First convert X = 25 into z-score.

z = (x - mean) / sigma

z = (25 - 21) / 10 = 0.4

That is now we have to find P(Z > 0.4).

P(Z > 0.4) = 1 - P(Z <= 0.4)

This probability we can find by using EXCEL.

syntax is,

=NORMSDIST(z)

where z is test statistic value.

P(Z > 0.4) = 1 - 0.6554

P(Z > 0.4) = 0.34

b) if a sample of 36 employees are taken and sample mean computed. What would be the mean, standard deviation and shape of the distribution of sample means for samples of size 36?

n = 36

We know that the distribution of sample mean is also normal with mean (mu) and standard deviation (sd) is sigma / sqrt(n).

mean (mu) = 21

sd = 10 / sqrt(36) = 1.67

Xbar ~ Normal(mean = 21, sd = 1.67).

c) A group of 36 employees is selected at random to participate in a program that allows a flexible work schedule, which the human resources department hopes will decrease the employee absenteeism in the future. What is the probability that the mean for the sample of 36 employees randomly selected for the study would exceed 25 days?

n = 36

Here we have to find P(Xbar > 25).

Convert Xbar = 25 into z-score.

Z = (25 - 21) / 1.67 = 2.4

That is now we have to find P(Z > 2.4).

P(Z > 2.4) = 1 - P(Z <=2.4)

= 1 - 0.99

= 0.01

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