In: Math
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.)
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.