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

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

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{(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{(p(1-p))/(n)}=\sqrt{(0.04* 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((\hat p-\mu_(\hat p))/(\sigm_(\hat p))>(0.05-0.04)/(0.0083))\\\\=P(Z>1.20)\\\\=1-P(Z<1.20)\\\\=1-0.88493\\\\=0.11507\\\\\approx 0.1151`

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