Question

A large corporation employs 39955 individuals. The average income of all employees is $62530, with a standard deviation of $19296 and is skewed to the right. Consider this to be the population distribution.

You are given a data set consisting of the incomes of 180 randomly selected employees.

• The population mean is μ=
• The population standard deviation is σ=
• The sample size is n=
• Since the sample size is relatively large, the Central Limit Theorem tells us that the sample averages should have a sampling distribution that is (skewed to the rightapproximately normal).
• The sampling distribution of the sample means is centered at the (populationsample) mean.
• The sampling distribution has a standard deviation of SD=σ√n=

Answer 1

Solution :

Given that ,

The population mean = = 62530

The population standard deviation = = 19296

the sample size n = 180

The sampling distribution of the sample mean and standard deviation is

=   = 62530

= / n = 19296/ 180 =  1438.2389