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 553 viruses would be greater than 5%? Round your answer to four decimal places.

Answer 1

Answer:

The probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% is 0.1151.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

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

The sample size is, n = 553 > 30. Thus, the Central limit theorem is applicable in this case.

Compute the mean and standard deviation as follows:

$\mu_{\hat p}=p=0.04\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.04\times 0.96}{553}}=0.0083$

Compute the probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% as follows:

 $P(\hat p>0.05)=P(\frac{\hat p-\mu_{\hat p}}{\sigm_{\hat p}}>\frac{0.05-0.04}{0.0083})\\\\=P(Z>1.20)\\\\=1-P(Z$

Thus, the probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% is 0.1151.