A survey of eating habits showed that approximately 4% of people in a certain ci

Question

A survey of eating habits showed that approximately 4% of people in a certain city are vegans Vegans do not eat meat, poultry, fish, seafood, eggs, or milk. A restaurant in the city expects 300 people on opening night, and the chef is deciding on the menu. Treat the patrons as a simple random sample from the city and the surrounding area, which has a population of about 600,000. If 15 vegan meals are available, what is the approximate probability that there will not be enough vegan meals-that is, the probability that 16 or more vegans will come to the restaurant? Assume the vegans are independent and there are no families of vegans. The probability is (Round to three decimal places as needed.)

Explanation / Answer

Mean = np = 300 * 0.04 = 12

Standard deviation = sqrt (npq) = sqrt(300*0.04*(1-0.04)) = 3.3941

P ( x>= 16 )

We use correction factor we get,

P (x > 15.5)

z = ( x – Mean) / Standard deviation

= (15.5 – 12)/3.3941

= 1.03

P ( z > 1.03 )

= 1 – P (z < 1.03 )

By using Normal Distribution Table we get,

= 1 – 0.8485

= 0.152

Answer: 0.152

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