Given a population where the probability of success is p 40

Given a population where the probability of success is

p= .40, calculate the probabilities below if a sample of 300 is taken.

A. Calculate the probability the proportion of successes in the sample will be less than .42 (round 4 decimals)

B. What is the probability that the proportion of successes in the sample will be greater than .44 (round 4 decimals)

Solution

a)

The mean of the proportions is u = p = 0.40, and the standard deviation of it is

sp = sqrt(p (1-p)/n) = sqrt(0.40*(1-0.40)/300) = 0.028284271

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.42      
u = mean =    0.4      
          
sp = standard deviation =    0.028284271      
          
Thus,          
          
z = (x - u) / sp =    0.707106787      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   0.707106787   ) =    0.760249941 [answer]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.44      
u = mean =    0.4      
          
s = standard deviation =    0.028284271      
          
Thus,          
          
z = (x - u) / s =    1.414213575      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.414213575   ) =    0.078649602 [answer]