Mean103 and standard deviation10 and n4 There is a 66 X is a

Mean=103, and standard deviation=10, and n=4

There is a 66% X is above what value?

Solution

P(X>x)=0.66

--> P((X-mean)/s <(X-103)/10) =1-0.66 =0.34

--> P(Z<( X-103)/10) =0.34

--> ( X-103)/10 = -0.41 (from standard normal table)

So x= 103-0.41*10 =98.9