A toy company claims that 91% of its Good Guys dolls are defect-free. If a random sample of 180 Good Guys dolls is selected, what is the probability that more than 85% of the sample will be defect-free? Round your final answer to four decimal places.

Answer:
Given:
The population proportion $ (p)=0.91 $

The sample size $(n) = 180$

Solution:
The probability that more than 85% of the sample will be defect free:
$P(\hat{p} > 0.85) = P\left( \frac{\hat{p} – p}{\sqrt{\frac{p(1-p)}{n}}} > \frac{0.85 – 0.91}{\sqrt{\frac{0.91(1-0.91)}{180}}} \right) = P(z > -2.813) = 0.9975 $

Final answer:
The probability that more than 85% of the sample will be defect free = 0.9975