Question

A scientist claims that 9% of viruses are airborne.

Answer 1

Answer:

The probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% is 0.006.

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 information provided is:

n = 596

p = 0.09

As the sample size is quite large, i.e. n = 596 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by a Normal distribution.

The mean and standard deviation are:

$\mu_{\hat p}=p=0.09\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.09(1-0.09)}{596}}=0.012$

Compute the probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% as follows:

$P(\hat p-p>0.03)=P(\hat p>0.12)$

                         $=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}>\frac{0.12-0.09}{0.012})\\\\=P(Z>2.5)\\\\=1-P(Z$

Thus, the probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% is 0.006.