Question

The mean diameter of a population of newly-settled periwinkles is 8.5 cm with a
standard deviation of 0.10 cm. What is the probability of selecting a random sample of
100 winkles that has a mean diameter greater than 8.52 cm?

Answer 1

$P(\bar X>8.52)=P(Z>\frac{8.52-8.5}{\frac{0.10}{\sqrt{100}}})=P(Z>2)=$

$=1-P(Z<2)=1-0.9772=0.0228.$