Question

The United States Department of Agriculture (USDA) found that the proportion of young adults ages 20–39 who regularly skip eating breakfast is 0.238 . Suppose that Lance, a nutritionist, surveys the dietary habits of a random sample of size n=500 of young adults ages 20–39 in the United States.

Apply the cnormal to find the probability that the number of individuals, X, in Lance's sample who regularly skip breakfast is greater than 122 .

Express the result as a decimal precise to three places

P(X<122)=

Apply the central limit theorem for the binomial distribution to find the probability that the number of individuals in Lance's sample who regularly skip breakfast is less than 103.

Express the result as a decimal precise to three places.

?(?<103)=

Answer 1

Here p=0.238 and n=500

Now using binomial distribution we need to find

Now here mean=np=500*0.238=119 and standard deviation is sd=sqrt{npq}=9.522

Now we need to find using normal approximation

As we know np>5 and nq>5

So we can convert x to z