Question

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 95% sure that they don't run out of chicken dishes?

Answer 1

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 95% sure that they don't run out of chicken dishes

P(X < N) = 0.95

P(Z < (N - 120)/6.93) = 0.95

(N - 120)/6.93 = 1.64 (from standard normal distribution table)

N=120+6.93*1.64=131.3652

round the N to whole number we get

N = 131

Ans:131