Question

Given a normal distribution with mean population of 51 and standard devitation of 8​, and given you select a sample of n=100​, complete parts​ (d).

d. There is a 30​% chance thatt X(w/bar above) is above what​ value? (Type an integer or decimal rounded to two decimal places as​ needed.)

Answer 1

Solution:
   We are given that: X follows a normal distribution with a mean population of 51 and a standard deviation of 8 .

That is:

n = sample size = 100

We have to find the sample mean value such that:

that is:

Then

Now find z value such that:
P( Z < z ) = 0.70

That is look for an area = 0.7000 or its closest area and find z value.

Area 0.6985 is closest to 0.7000 than 0.7019

and it corresponds to 0.5 and 0.02

Thus z = 0.52

Now use the following formula to find the sample mean value:

Thus there is a 30% chance that is above 51.42.