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

Step-by-step explanation:

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

The sample size = 300

By suing normal approximation to the binomial , we have

$\mu=np=300\times0.077=23.1$

$\sigma=\sqrt{np(1-p)}=\sqrt{300\times0.077(1-0.077)}\\\\=4.61749932323\approx4.62$

Now, using formula $z=\dfrac{x-\mu}{\sigma}$, the z-value corresponding to 26 will be :-

$z=\dfrac{26-23.1}{4.62}\approx0.63$

Using standard distribution table for z , we have

P-value=$P(z\geq0.63)=1-P(z$

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

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