normal distrubution help me

Question

suppose annual salries of workers in X city are normally distributed with mean of $54000 and a standard deviation of $9000

a)what is probability of finding a group of 5 workers in santa clara making an average of more than $60000

B) what is the average salary coreesponds to the 90th percentile for a group of 10 workers in X city ?

Answer 1

Population mean=$54000 and pop SD=$9000.  

a) We can use the t test with an adjusted SD for the sample of 5 workers=9000/√5=$4024.92. t=(60000-54000)/4024.92=1.49. A sample of 5 has 4 degrees of freedom. We are looking for salaries greater than $60000 so it’s a one-tailed test. The probability associated with 1.49 is 0.1052 or about 10.5%.

b) The adjusted SD for 10 workers is 9000/√10=$2846.05. For 9 degrees of freedom, the t value for the 90th percentile is 1.383. So, if the salary is X:

(X-54000)/2846.05=1.383, X=54000+3936.09=$57936.09. Round this to the nearest $1000: $58000. Workers at this salary level or higher are in the 90th percentile.