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Naily [24]
3 years ago
10

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.
Mathematics
1 answer:
FrozenT [24]3 years ago
3 0

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.

<em>Let </em>\hat p<em> = sample proportion of airborne viruses</em>

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

<em>Now, in the z table the P(Z </em>\leq<em> 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.</em>

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

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