Question 1161682: The average salary for graduates entering the actuarial field is $40,000. If the salaries are normally distributed with a standard deviation of $5000, find the probability that
b. A group of nine graduates will have a group average over $45,000.
Z = value - mean / st dv
Z = (45,000 - 40,000) / (5,0000/sqrt(9)) =
The correct answer is .0013
I am looking for the correct answer. The .0013 is the answer after looking at the Z table but I am having a hard time find the correct answer
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i'm not exactly sure what you're looking for.
from what i see:
population mean = 40,000
population standard deviation = 5000
sample size = 9
sample mean = 45,000
standard deviation of the distribution of sample means is 5000 / sqrt(9) = (5000 / 3).
this is called the standard error.
z-score = (sample mean minus population mean) / standard error = (45,000 - 40,000) / (5000 / 3) = 3.
look up in the z-score table and it tells you that a z-score of 3 has an area under the normal distribution curve of .99865 to the left of it.
the area to the right of it is 1 - .99865 = .00135.
that would round to .0014, if you worked only from the z-score table.
using a calculator, i found the area to the right of a z-score of 3 equal to .0013499673,
that rounds to .0013.
if that is supposed to be the correct answer, then you have it.
if not, then what answer are you looking for?
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