Question

An article in Parenting magazine reported that 35% of Americans needed a vacation after visiting their families for the holidays. Suppose this is the true proportion of Americans who feel this way. A random sample of 200 Americans is taken. Let X be the number of people who feel that way in this random sample.

a) What is the mean of X?

b) Using the normal approximation, what is the probability that more than 80 people in the sample feel that they need a vacation after visiting their families for the holidays? (Round to 4 decimal places.)

Answer 1

Solution:

Given that,

P = 0.35

1 - P = 0.65

n = 200

Here, BIN ( n , P ) that is , BIN (200 , 0.35)

According to normal approximation binomial,

a)

X Normal

Mean = = n*P = 70

Standard deviation = =n*p*(1-p) = 45.5

We using continuity correction factor

P(x > a ) = P( X > a + 0.5)

P(x > 80.5) = 1 - P(x < 80.5)

= 1 - P((x - ) / < (80.5 - 70) / 45.5)

= 1 - P(z < 1.56)

= 1 - 0.9406

= 0.0594

Probability = 0.0594