Question

A normally distributed population has a mean of 65 and a standard deviation of 24. Sample averages from samples of size 19 are collected. What would be the lower end of the centered interval that contains 90% of all possible sample averages?

I know how to do a most of this, but I am confused on how I find the Z variable. Thanks!

Answer 1

Given that,

mean = = 65

standard deviation = = 24

n = 19

= 65

= / n = 24/19=5.5

Using standard normal table,

P(Z < z) = 90%

= P(Z <1.28 ) = 0.90

see the area 0.90 in table

z = 1.28 Using standard normal table,

Using z-score formula  

= z * +   

= 1.28*5.5+65

= 72.04