A survey of eating habits showed that approximately 55% of people in a certain c

Question

A survey of eating habits showed that approximately 55% 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 400 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 24 vegan meals are available, what is the approximate probability that there will not be enough vegan mealslong dash—that is, the probability that 25 or more vegans will come to the restaurant? Assume the vegans are independent and there are no families of vegans.

Explanation / Answer

NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 400 * 0.05 = 20
standard deviation ( npq )= 400*0.05*0.95 = 4.3589
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
P(X < 25) = (25-20)/4.3589
= 5/4.3589= 1.1471
= P ( Z <1.1471) From Standard NOrmal Table
= 0.8743
P(X > = 25) = (1 - P(X < 25))
= 1 - 0.8743 = 0.1257

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