Question

In a certain city of several million people, 6.4% of the adults are unemployed. If a random sample of 220 adults in this city is selected, approximate the probability that at most 11 in the sample are unemployed. Use the normal approximation to the binomial with a correction for continuity.Round your answer to at least three decimal places. Do not round any intermediate steps.

Answer 1

From the information given,

probability of success, p = 6.4% = 6.4/100 = 0.064

q = 1 - p = 1 - p = 1 - 0.064 = 0.936

sample size, n = 220

Mean, μ = np = 0.064 * 220 = 14.08

standard deviation, σ = √npq = √220 * 0.064 * 0.936 = 3.63

We want to find P(x ≤ 11). We would apply the continuity correction factor.

P(x ≤ 11) = P(x ≤ 11.5)

We would calculate the z score by applying the formula,

z = (x - μ)/σ

z = (11.5 - 14.08)/3.63 = - 0.71

The area to the left of z = - 0.71 on the normal distribution table is 0.239

the probability that at most 11 in the sample are unemployed is 0.239