The average salary in a certain large Chicago suburb is 5000

The average salary in a certain large Chicago suburb is $50,000, with a standard deviation of $10,000. A sample of wage earners is taken. If the probability that the mean-sample salary is under $52,500 equals 84.13%, how large is the sample?

Solution

et xbar be the sample mean.
z = (xbar - ?) / (? / ? n )

P( xbar < 52,500) = P( z < (52,500-50,000) / (10000 / ? n ))

From normal probability table, P( z < 1 ) = 0.8413

Therefore, (52,500-50,000) / (10000 / ? n ) = 1
2,500? n / 10,000 = 1
0.25 ? n = 1
? n = 1/0.25 = 4
n=16