Question

About 90% of American adults had chickenpox before adulthood. We now consider a random sample of 150 American adults I only need help with B) and c) part of this question. a) How many people in this sample would you expect to have had chickenpox in their childhood? (Enter your answer as a whole number.) b) Would you be surprised if there were 132 people who have had chickenpox in their childhood? (Round your answer to two decimal places.) Since z = ?c) What is the probability that 132 or fewer people in this sample have had chickenpox in their childhood?

Answer 1

a) We expect to have 135 people to have had chickenpox in their childhood.

b) No, I would not be surprised, as the probability is not low.

z = -0.6812

c) P(X≤132)=0.248

Explanation:

a) This is a binomial random variable, with n=150 and p=0.9, so we can calculate the expected value as:

`E(X)=np=150\cdot 0.9=135`

b) To answer this we can calculate the probability that 132 people or fewer have had chickenpox in their childhood.

This binomial random variable can be approximated by a normal distribution, with the following parameters:

`\mu=np=150\cdot 0.9=135\\\\\sigma=√(np(1-p))=√(150\cdot 0.9\cdot 0.1)=√(13.5)=3.67`

To calculate the probability that 132 people or fewer have had chickenpox in their childhood we compute the z-score for X=132.5 (as we apply the continuity factor) and calculate the probability using the standard normal distribution:

`z=(X-\mu)/(\sigma)=(132.5-135)/(3.67)=(-2.5)/(3.67)=-0.6812\\\\\\P(X<132.5)=P(z<-0.6812)=0.248`