Answer:
Probability that the proportion of airborne viruses in a sample of 529 viruses would be greater than 8% is 0.9545.
Step-by-step explanation:
For the given population the distribution of air-borne viruses is given as 6%. Now for this the value of Z score is given as
Here
is the sample measure which is 8% or 0.08- p is the population measure which is 6% or 0.06
- n is the number of samples which is 529
So the Z score is given as
For this the score, the value of probability is calculated from the table as below:
Here value of p(Z<-1.6955) is given as 0.0455