A sample survey interviews an SRS of 139 college women Suppo

A sample survey interviews an SRS of 139 college women. Suppose that 70% of all college women have been on a diet within the last 12 months. What is the probability that 78% or more of the women in the sample have been on a diet?

Use 4 decimal places.

Solution

Normal Distribution
Proportion ( P ) =0.7
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.7*0.3/139)
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X < 0.78) = (0.78-0.7)/0.0389
= 0.08/0.0389= 2.0566
= P ( Z <2.0566) From Standard Normal Table
= 0.9801                  
P(X>=0.78) = 1 - 0.9801 = 0.0199