A certain flight arrives on time 81 % of the time. Suppose 148 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that….

Question

A certain flight arrives on time 81 % of the time. Suppose 148 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that 

​(a) exactly 108 flights are on time.

​(b) at least 108 flights are on time.

​(c) fewer than 127 flights are on time.

​(d) between 127 and 129, inclusive are on time.

Answer 1

Let:

p = probability that a flight is on time = .81

q = 1 - p = probability that a flight is not on time = .19

n = number of flights that are randomly selected = 148

The formula for binomial probability distribution is P(x) = (_nC_x)p^xq^n-x

where x is a number taken from n,

_nC_x = n!/(x!•(n-x)!) = is a binomial coefficient.

(a) x= 108

P(108) = (₁₄₈C₁₀₈)(.81)¹⁰⁸ (.19)⁴⁰

P(108) = 0.0044 = 0.44%

(b) P(108 ≤ x ≤ 148) => The least is 108 and the maximum is 148. So what you need to do is get the following probability: P(108), P(109), P(110), . . ., P(148) and then add them all up. Of course that would be very tedious.

But if you have a calculator with the sigma notation and the _nC_r that would be easy to calculate.

₁₄₈

∑ ₁₄₈C_k (0.81)^k(0.19)^148-k ≈ .9937 = 99.37%

_k=108

(c)Fewer than 127 means 0 ≤ x < 127. Make sure you don't include 127 for the x. In a calculator or any computing app:

₁₂₆

∑ ₁₄₈C_k (0.81)^k(0.19)^148-k ≈ .9208 = 92.08%

_k=0

(d) 127 ≤ x ≤ 129.

₁₂₉

∑ ₁₄₈C_k (0.81)^k(0.19)^148-k ≈ .0612 = 6.12%

_k=127