Question

In a division of a company with over 200 employees, 40% of whom are male, the company randomly selects 8 employees each month to have lunch with the CEO. What are the mean and standard deviation of the number of males selected each month?

a) Mean = 3.2, Standard Deviation = 1.48
b) Mean = 3.2, Standard Deviation = 1.28
c) Mean = 4.8, Standard Deviation = 1.28
d) Mean = 4.8, Standard Deviation = 1.48

Answer 1

The mean and standard deviation of the number of males selected each month in a division of a company can be found using the properties of the binomial distribution.

Step-by-step explanation:

To find the mean and standard deviation of the number of males selected each month, we can use the properties of the binomial distribution.

In this case, the probability of selecting a male is 0.4 and the number of trials is 8.

The mean of a binomial distribution is given by np, where n is the number of trials and p is the probability of success.

Therefore, the mean number of males selected each month is 8 * 0.4 = 3.2.

The standard deviation of a binomial distribution is given by sqrt(np(1-p)), so the standard deviation of the number of males selected each month is sqrt(8 * 0.4 * (1-0.4)) = 1.28.

Therefore, the correct answer is b) Mean = 3.2, Standard Deviation = 1.28.