Question

1. (10 points) According to an airline, flights on a certain route are on time 82% of the time. Suppose 12 flights are randomly selected and the number of on time flights is recorded. (a) Find the mean and standard deviation for the number of on time flights. (b) What is the probability that exactly 9 of the 12 flights will be on time? (c) What is the probability that at most 10 of the flights will be on time?

Answer 1

Given,

Flights on a certain route are on time 82% of the time

Probability that a flight on a certain route is on time = 82% = 82/100 = 0.82

12 flights are randomly selected

Let X be a random variable representing the number of flights that are on time in the sample of 12

Answer a:
Since, there are only to possible outcomes (the flight is on time or it is not on time)

X follows Binomial Distribution with parameters n = 12 and probability of success, p = 0.82

The Probability Mass Function, P.M.F. of X, f(x) = (nCx) x (p^x) x (1 – p)^(n – x)    x = 0, 1, 2, ..n

N- & P [xzx] - () to0-1) x=0,1 2. n. Expected value, 2. P[x= x] 720 n! F(x) = ê x. n- fr ( xo (n.2 x (n-1)! 2-1 - (x - 1) n. p &W i po (1-P) x=1 (n- (-))! (x-1)! mp

Elx) S ? P[**] (x-1+1) m! nox pa (-p) (2-x)/x! 19 Ex(x - 1)(n-1) (-2)! no(x-2) p²px-2 (1-P) (71-6 - 2))! (x-2)! *(x-1) 2-1 32. m)! n-(x-1) (1-P) + 2 & pl (n-(x-1))! 9.(2-1)! x-2 22 2- men-> p ( (2-2) + Co-p)*62-->) + mp ((:) *** (-p) ** (**») -1) p2 & rp Therefore, var. (x) = E(x2) - {E (x)}² ela-1) + 2 + up - nepe n/p² – np2 tap - n/p² np loop) Standard deviation of x = var. (*)

Mean of number of on time flights = np = 12 x 0.82 = 9.84 = 10 (approximately)

Standard Deviation of the number of on time flights = (np(1 - p))^0.5 = 1.3309 = 1 (approximately)

Answer b:

The required probability, Prob.[X = 9] = (12C9) x (0.82^9) x (1 – 0.82)^(12 – 9)

                                                                  = 0.2151

Therefore, Probability that exactly 9 of the 12 flights will be on time is 0.2151

Answer c:
The required probability,

Prob.[X ≤ 10] = 1 – Prob.[X > 10] = 1 – {P[X = 11] + P[X = 12]}

                  = 1 – {[(12C11) x (0.82^11) x (1 – 0.82)^(12 – 11)] + [(0.82^12) x 1]}

                   = 0.6641

Therefore, Probability that at most 10 of the 12 flights will be on time is 0.6641

(All probabilities are calculated up to 4 decimal places)