Question

The United States Department of Agriculture (USDA) found that the proportion of young adults ages 20–39 who regularly skip eating breakfast is 0.2380. Suppose that Lance, a nutritionist, surveys the dietary habits of a random sample of size ?=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. You may find table of critical values helpful.

Express the result as a decimal precise to three places.

?(?>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

Answer)

N = 500

P = 0.2380

We need to find

P(x>122)

By continuity correction

P(x>122.5)

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 119

N*(1-p) = 381

Both the conditions are met so we can use standard normal z table to estimate the probability

Z = (x - mean)/s.d

Mean = n*p = 119

S.d = √{n*p*(1-p)} = 9.52249967182

Z = (122.5 - 119)/9.52249967182

Z = 0.37

From z table, P(z>0.37) = 0.3557

B)

P(x<103)

By continuity correction

P(x<102.5)

Z = (102.5 - 119)/9.52249967182 = -1.73

From z table, P(z<-1.68) = 0.0418