Question

In a certain city of several million people, 7.7% of the adults are unemployed. If a random sample of 300 adults in this city is selected, approximate the probability that at least 26 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 place

Answer 1

Answer: 0.2643

Explanation:

Given : The proportion of adults are unemployed : p=0.077

The sample size = 300

By suing normal approximation to the binomial , we have

`\mu=np=300*0.077=23.1`

`\sigma=√(np(1-p))=√(300*0.077(1-0.077))\\\\=4.61749932323\approx4.62`

Now, using formula
`z=(x-\mu)/(\sigma)`, the z-value corresponding to 26 will be :-

`z=(26-23.1)/(4.62)\approx0.63`

Using standard distribution table for z , we have

P-value=
`P(z\geq0.63)=1-P(z<0.63)`

`=1-0.7356527=0.2643473\approx0.2643`

Hence, the probability that at least 26 in the sample are unemployed =0.2643