An airline offers customers a choice between a chicken and a pasta dish on international flights. From experience, they estimate that 60% of the customers choose the chicken dish, the remaining 40% choose the pasta dish. On a flight with 200 passengers, how many chicken dishes should the airline stock to be 99% sure that they don't run out of chicken dishes?
a. 
104 

b. 
200 

c. 
136 

d. 
131 
Number of passengers, n = 200
P(chicken dish), p = 0.6
q = 1  p = 0.4
Using normal approximation for binomial distribution,
P(X < A) = P(Z < (A  mean)/standard deviation)
Mean = np
= 200 x 0.6
= 120
Standard deviation =
=
= 6.93
Let N be the number of chicken dishes the airline should stock to be 99% sure that they don't run out of chicken dishes
P(X < N) = 0.99
P(Z < (N  120)/6.93) = 0.99
(N  120)/6.93 = 2.33 (from standard normal distribution table)
N = 136
Ans: c. 136
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