The standard deviation of sample means = 0.9/sqrt400
Hi,
I am new here, and I hope you can help.
Here is the question:
One year, the distribution of salaries for professional sports players had mean $1.5 million and standard deviation $0.9 million. Suppose a sample of 400 major leagu0e players was taken. Find the approximate probability that the average salary of the 400 players that year exceeded $1.1 million.
The correct answer is approximately 1, but I can't figure out how to work it out.
So far I have tried resolving it by following
z= 1,100,000-1,500,000/900,000 = 0.4 which according to the z table translates to .1554
Then I subtracted this number from .5 which resulted in .3446 which I take it to mean that approximately 34% of players earn more than 1.1million.
How do I use the 400 baseball players in this equation?
Thanks
0.9 / sqrt(400) = 0.045
but...I still don't know how to solve the original question. How do I get to the answer that approximately 1 player's salary exceeds 1.1 million? Please be as explicit as possible.
thanks.