Question

If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 477 viruses would differ from the population proportion by more than 3%.

Answer 1

Answer:

0.01596

Step-by-step explanation:

A scientist claims that 8% of the viruses are airborne

Given that:

The population proportion p = 8%

The sample size = 477

We can calculate the standard deviation of the population proportion by using the formula:

$\sigma_p = \sqrt{\dfrac{p(1-p)}{n}}$

$\sigma_p = \sqrt{\dfrac{0.8(1-0.8)}{477}}$

$\sigma_p = \sqrt{\dfrac{0.0736}{477}}$

$\sigma_p = 0.02098$

The required probability can be calculated as:

$P(| \hat p - p| > 0.03) = P(\hat p - p< -0.03 \ or \ \hat p - p > 0.03)$

$= P \bigg ( \dfrac{\hat p -p }{\sqrt{\dfrac{p(1-p)}{n}}} < -\dfrac{0.03}{0.0124} \bigg ) + P \bigg ( \dfrac{\hat p -p }{\sqrt{\dfrac{p(1-p)}{n}}} >\dfrac{0.03}{0.0124} \bigg )$

= P(Z < -2.41) + P(Z > 2.41)

= P(Z < -2.41) + P(Z < -2.41)

= 2P( Z< - 2.41)

From the  Z-tables;

$P(| \hat p - p| > 0.03)$ = 2 ( 0.00798

$P(| \hat p - p| > 0.03)$ = 0.01596

Thus, the required probability = 0.01596