36.6k views
3 votes
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.

1 Answer

4 votes

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

User Jessika
by
8.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories