Question

A scientist claims that 9% of viruses are airborne.

If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 596 viruses would differ from the population proportion by greater than 3%?

Answer 1

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

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 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{(p(1-p))/(n)}=\sqrt{(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((\hat p-\mu_(\hat p))/(\sigma_(\hat p))>(0.12-0.09)/(0.012))\\\\=P(Z>2.5)\\\\=1-P(Z<2.5)\\\\=1-0.99379\\\\=0.00621\\\\\approx 0.006`

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