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faltersainse [42]
3 years ago
13

A scientist claims that 6% of viruses are airborne. (a) If the scientist is accurate, what is the probability that the proportio

n of airborne viruses in a sample of 529 viruses would be greater than 8%? Round your answer to four decimal places.
Mathematics
1 answer:
soldi70 [24.7K]3 years ago
7 0

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

Z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}

Here

  • \hat{p} 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

Z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}\\Z=\dfrac{0.08-0.06}{\sqrt{\dfrac{0.08(1-0.08)}{529}}}\\Z=\dfrac{0.02}{\sqrt{\dfrac{0.08(0.92)}{529}}}\\Z=\dfrac{0.02}{\sqrt{\dfrac{0.0736}{529}}}\\Z=\dfrac{0.02}{0.01179}\\Z=1.6955

For this the score, the value of probability is calculated from the table as below:

P(\hat{p}>0.08)=p(Z>1.6955)\\=1-p(Z

Here value of p(Z<-1.6955) is given as 0.0455

P(\hat{p}>0.08)=p(Z>1.6955)\\P(\hat{p}>0.08)=1-p(Z0.08)=1-0.0455\\P(\hat{p}>0.08)=0.9545

 

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