Question

A distribution of values is normal with a mean of 80 and a standard deviation of 18. From this distribution, you are drawing samples of size 12.

Find the interval containing the middle-most 88% of sample means:

Enter your answer using interval notation. In this context, either inclusive or exclusive intervals would be acceptable. Your numbers should be accurate to 1 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Solution:
Given in the question
A distribution of values is normal with
Mean () = 80
Standard deviation() = 18
Number of sample (n) = 12
We need to find interval containing the middle-most 88% of sample means with sampling distribution
Mean of sampling distribution of sample mean xbar = 80
Standard deviation of sampling distribution of sample mean xbar = /sqrt(n) = 18/sqrt(12) = 5.2
Here alpha = 1-0.88 = 0.12
so alpha/2 = 0.12/6 = 0.06
So we need to calculate interval with p-value for Lower bound = 0.06, p-value for upper bound = 0.94
So from Z table we found Z-score form Lower bound = -1.555, and Z-score from Upper bound = 1.555
P(-1.555<=Z<=1.555) = 0.88
So Numbers can be calculated as
Lower bound X = xbar + Z-score * xbar = 80 -1.555*5.2 = 80 - 8.1 = 71.9
Upper bound X = xbar + Z-score * xbar = 80 +1.555*5.2 = 80 + 8.1 = 88.1
So we can write as P(71.9<=xbar <= 88.1) = 0.88