Question

The mean diameter of a population of newly settled periwinkle is 8.5 cm with a standard deviation of 1.10 cm. What is the probability of selecting a random sample of 100 wrinkles that has a mean diameter greater than 8.52.

Answer 1

$P(\bar X>8.52)=P(Z>\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}})=P(Z>\frac{8.52-8.5}{\frac{1.10}{\sqrt{100}}})=P(Z>0.18)=0.4286.$