American Airlines flights from Dallas to Chicago are on time

American Airlines flights from Dallas to Chicago are on time 80% of the time, suppose 15 flights are randomly selected and the number of on-time flights is recorded. a.) Explain why this ?s a binomial experiment. b.) Find and interpret the mean. c.) Find and interpret the probability that exactly 10 flights are on time. d.) Find and interpret the probability that at least 10 flights are on time. e.) Find and interpret the probability that fewer than 10 flights are on time. f.) Find and interpret the probability between 8 and 10, inclusive, flights are on time.

Solution

a. This is a binomial experiments, because there are a fixed number of trials (15), each trial is independent, and the probability of being on time is the same for each trial (.8).

b. The mean is n*p, which is 15*.8 = 12 (answer)

Interpretation: We expect that in a random sample of 15 flights, 12 will be on time.

c. The formula is:

P(X = x) = nCx*p^x*(1-p)^(n-x)

P(X = 10) = 15C10*.8^10*(1-.8)^(15-10)

P(X = 10) = 0.1032 (answer)

Interpretation: There is a 0.1032 probability that in a random sample of 15 flights that 10 will be on time.

d. Probability at least 10

P(at least 10) = 1 - P(9 or less)

The probability of 9 or less is found by using a calculator:

binomcdf(15, .8, 9) = .0611

So probability of at least 10 = 1 - .0611 = 0.9389 (answer)

Interpretation: There is a .9389 probability that in a random sample of 15 flights, at least 10 will be on time.

e. Fewer than 10 flights.

This was calculated in part d:

binomcdf(15, .8, 9) = .0611 (answer)

Interpretation: There is a .0611 probability that in a random sample of 15 flights that fewer than 10 are on time.

f. Probability between 8 and 10.

This can be done using the formula,

P(X = x) = nCx*p^x*(1-p)^(n-x),

three times, each for 8, 9, and 10

P(X = 8) = 15C8*.8^8*(1-.8)^(15-8) = 0.0138

P(X = 9) = 15C9*.8^9*(1-.8)^(15-9) = 0.0439

P(X = 10) = 15C10*.8^10*(1-.8)^(15-10) = 0.1032

Then add them up: 0.0138 + 0.0439 + 0.1032 = 0.1600 (answer)

Interpretation: There is a 0.1600 probability that in a random sample of 15 flights, between 8 and 10 inclusive are on time.