Suppose the time it takes for a purchasing agent to complete an online ordering process is normally distributed with a mean of 8 minutes and a standard deviation of 2
Minutes. Suppose a random sample of 25 ordering processes is selected. The standard
Deviation of the sampling distribution of mean times is
A) 0.4 minutes
B) 2 minutes
C) 0.08 minutes
D) 1.6 minutes
E) 0.12 minutes

Question

Quizplus · Accepted Answer

The standard deviation of the sampling distribution of the mean (often called the standard error) is calculated by dividing the population standard deviation by the square root of the sample size: $ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} $, where $ \sigma $ is the population standard deviation and $ n $ is the sample size. Here, $ \sigma = 2 $ minutes and $ n = 25 $, so $ \sigma_{\bar{x}} = \frac{2}{\sqrt{25}} = \frac{2}{5} = 0.4 $ minutes.

Suppose the Time It Takes for a Purchasing Agent to Complete