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 meals—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.

Answer 1

Answer:

Given That:

• A restaurant in the city expects 400 people on opening​ night, and the chef is deciding on the menu.
• population of about​ 600,000.
• If 24 vegan meals are​ available, what is the approximate probability that there will not be enough vegan meals
• the probability that 25 or more vegans will come to the​ restaurant
• NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
  the PDF of normal distribution is = 1/σ √2π e ^ -(x-u)^2/ 2σ^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