Suppose that we have a population proportion P020 and a rand

Suppose that we have a population proportion P=0.20 and a random sample of size n=100 draw from the population, What is the propability that the sample portion is greater than 0.24

Round to 4 decimal places

Solution

The standard deviation of the proportion is

s(p^) = sqrt(p(1-p)/n) = sqrt(0.20*(1-0.20)/100) = 0.04

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.24      
u = mean =    0.2      
          
s = standard deviation =    0.04      
          
Thus,          
          
z = (x - u) / s =    1      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1   ) =    0.158655254 [ANSWER]