Question

A scientist claims that 7%7% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 418418 viruses would be greater than 6%6%? Round your answer to four decimal places.

Answer 1

Answer:

Probability that the proportion of airborne viruses in a sample of 418 viruses would be greater than 6% is 0.8051.

Step-by-step explanation:

We are given that a scientist claims that 7% of viruses are airborne.

A sample of 418 viruses is taken.

Let $\hat p$ = sample proportion of airborne viruses

The z-score probability distribution for sample proportion is given by;

          Z = $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion

           p = population proportion of viruses that are airborne = 7%

           n = sample of viruses = 418

Probability that the proportion of airborne viruses in a sample of 418 viruses would be greater than 6% is given by = P( $\hat p$ > 0.06)

   P( $\hat p$ > 0.06) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ > $\frac{ 0.06-0.07}{\sqrt{\frac{0.06(1-0.06)}{418} } }$ ) = P(Z > -0.86) = P(Z < 0.86)

                                                                                          = 0.8051

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.86 in the z table which has an area of 0.8051.

Therefore, probability that the proportion of airborne viruses in a sample of 418 viruses would be greater than 6% is 0.8051.