Question

A scientist claims that 4% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 662 viruses would be greater than 6%

Answer 1

Answer:

The probability that the proportion of airborne viruses in a sample of 662 viruses would be greater than 6%=0.00427

Step-by-step explanation:

We are given that

$\mu_{\hat{p}}=p=4%=0.04$

n=662

We have to find the probability that the proportion of airborne viruses in a sample of 662 viruses would be greater than 6%.

q=1-p=1-0.04=0.96

$\sigma_{\hat{p}}=\sqrt{p(1-p)/n}$

$\sigma_{\hat{p}}=\sqrt{\frac{0.04(1-0.04)}{662}}$

$\sigma_{\hat{p}}=0.0076$

Now,

$P(\hat{p}>0.06)=1-P(\hat{p}$

$=1-P(\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}$

$=1-P(Z$

$=1-0.99573$

$P(\hat{p}>0.06)=0.00427$

Hence, the probability that the proportion of airborne viruses in a sample of 662 viruses would be greater than 6%=0.00427