Question

The unemployment rate of persons with a disability is typically higher than for those with no disability. Recent statistics report that this rate is​ 13.4%. An advocacy group in a large city located in the southeastern region of the U.S. selected a random sample of 250 persons with a disability. What is the probability that no more than 25 persons in this sample are​ unemployed? (Round to four decimal places.)

Answer 1

Solution:

Given that,

P = 13.4% = 0.134

1 - P = 1 - 0.134 = 0.866

n = 250

Here, BIN ( n , P ) that is , BIN (250 , 0.134)

then,

n*p = 250*0.134 = 33.5 > 5

n(1- P) = 250*0.866 = 216.5 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 33.5

Standard deviation = =n*p*(1-p) = 250*0.134*0.866 = 29.011

We using countinuity correction factor

P( X a ) = P(X < a + 0.5)

P(x < 25.5) = P((x - ) / < (25.5 - 33.5) / 29.011)

= P(z < -1.485)

= 0.0688

Probability = 0.0688