A humane society reports that 19 of all pet dogs were adopte

A humane society reports that 19% of all pet dogs were adopted from an animal shelter. Assuming the truth of this assertion, find the probability that in a random sample of 80 pet dogs, between 15% and 20% were adopted from a shelter. You may assume that the normal distribution applies.

Solution

The distirbution of the population proportion has

mean = u(p) = 0.19
standard deviation = s(p) = sqrt(p(1-p)/n) = sqrt(0.19*(1-0.19)/80) = 0.043860575

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.15      
x2 = upper bound =    0.2      
u = mean =    0.19      
          
s = standard deviation =    0.043860575      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.911980748      
z2 = upper z score = (x2 - u) / s =    0.227995187      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.180889426      
P(z < z2) =    0.590175009      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.409285583   [ANSWER]