Suppose that amount people spend at the grocery store have a

Suppose that amount people spend at the grocery store have a mean of $55 and a standard deviation of $12. In a random sample of 36 people, find the probability that the average amount spent is more than $61.50.

Solution

Mean ( u ) =55
Standard Deviation ( sd )=12
Number ( n ) = 36
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
P(X > 61.5) = (61.5-55)/12/ Sqrt ( 36 )
= 6.5/2= 3.25
= P ( Z >3.25) From Standard Normal Table
= 0.0006