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

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

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}}=√(p(1-p)/n)`

`\sigma_{\hat{p}}=\sqrt{(0.04(1-0.04))/(662)}`

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

Now,

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

`=1-P(\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}<(0.06-0.04)/(0.0076))`

`=1-P(Z<2.63)`

`=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