A chef is opening a new restaurant in Portland, Oregon, and wants to know how ma

Question

A chef is opening a new restaurant in Portland, Oregon, and wants to know how many vegan meals he should prepare. 300 people will attend opening night. He knows that about 4% of people in Portland are vegan, and the approximate population of Portland is 600,000.

1.     What is the probability that more than 5% of the people attending opening night will be vegans?

2.     What is the probability that less than 3% of the people attending opening night will be vegans?

3.     What is the probability that between 2.5% and 5.5% of the people attending opening night will be vegans?

Explanation / Answer

Ans:

Sampling distribution of sample proportion:

mean=p=0.04

standard deviation=sqrt(0.04*(1-0.04)/300)=0.0113

1)

z=(0.05-0.04)/0.0113

z=0.884

P(z>0.884)=0.1884

2)

z=(0.03-0.04)/0.0113

P(z<-0.884)=0.1884

3)

z(0.025)=(0.025-0.04)/0.0113=-1.326

z(0.055)=(0.055-0.04)/0.0113=1.326

P(-1.326<z<1.326)=P(z<1.326)-P(z<-1.326)=0.9076-0.0924=0.8152

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