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
[tex]\mu_{\hat{p}}=p=4%=0.04[/tex]
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
[tex]\sigma_{\hat{p}}=\sqrt{p(1-p)/n}[/tex]
[tex]\sigma_{\hat{p}}=\sqrt{\frac{0.04(1-0.04)}{662}}[/tex]
[tex]\sigma_{\hat{p}}=0.0076[/tex]
Now,
[tex]P(\hat{p}>0.06)=1-P(\hat{p}<0.06)[/tex]
[tex]=1-P(\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}<\frac{0.06-0.04}{0.0076})[/tex]
[tex]=1-P(Z<2.63)[/tex]
[tex]=1-0.99573[/tex]
[tex]P(\hat{p}>0.06)=0.00427[/tex]
Hence, the probability that the proportion of airborne viruses in a sample of 662 viruses would be greater than 6%=0.00427
