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><em> = sample proportion of airborne viruses</em>
The z-score probability distribution for sample proportion is given by;
Z = ~ N(0,1)
where, = 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( > 0.06)
P( > 0.06) = P(
>
) = P(Z > -0.86) = P(Z < 0.86)
= 0.8051
<em>Now, in the z table the P(Z </em><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.