Question

A quality analyst wants to construct a sample mean chart for controlling a packaging process. He knows from past experience that whenever this process is in control, package weight is normally distributed with a mean of 20 ounces and a standard deviation of two ounces. Each day last week, he randomly selected four packages and weighed each: Day Weight (ounces) Monday 23 22 23 24 Tuesday 23 21 19 21 Wednesday 20 19 20 21 Thursday 18 19 20 19 Friday 18 20 22 20 What is the standard deviation of the sampling distribution of sample means for whenever this process is in control?

Answer 1

1 ounce

Explanation:

Given that :

mean = 20 ounces

standard deviation = 2 ounces

sample size = 4 ounces

The standard deviation of the sampling distribution =
`(Standard \ deviation)/(√(sample \ size))`

=
`(2)/(√(4))`

= 2/2

= 1 ounce