Question

Your flight has been delayed: At Denver International Airport, 85% of recent flights have arrived on time. A sample of flights is studied. Round the probabilities to four decimal places. Part 1 of 4 (a) Find the probability that all of the flights were on time. The probability that all of the flights were on time is . Part 2 of 4 (b) Find the probability that exactly of the flights were on time. The probability that exactly of the flights were on time is . Part 3 of 4 (c) Find the probability that or more of the flights were on time. The probability that or more of the flights were on time is . Part 4 of 4 (d) Would it be unusual for or more of the flights to be on time

Answer 1

a)
`P(X = n) = (0.85)^(n)`

b)
`P(X = x) = C_(n,x).(0.85)^(x).(0.15)^(n-x)`

c)

`P(X \geq x) = P(X = x) + P(X = x+1) + ... + P(X = n)`

For each value, we use

`P(X = x) = C_(n,x).(0.85)^(x).(0.15)^(n-x)`

d) If
`x > E(X) + √(V(X))`, yes, otherwise no

Explanation:

For each flight, there are only two possible outcomes. Either it arrived on time, or it did not. Flights are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

A measure is said to unusual if it is more than 2.5 standard deviations from the mean.

85% of recent flights have arrived on time.

This means that
`p = 0.85`

(a) Find the probability that all of the flights were on time.

This is
`P(X = n)`, in which n is the size of the sample studied.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = n) = C_(n,n).(0.85)^(n).(0.15)^(n-n)`

`P(X = n) = (0.85)^(n)`

(b) Find the probability that exactly x of the flights were on time.

Exactly x of the flights: So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = x) = C_(n,x).(0.85)^(x).(0.15)^(n-x)`

(c) Find the probability that x or more of the flights were on time.

This is

`P(X \geq x) = P(X = x) + P(X = x+1) + ... + P(X = n)`

For each value, we use

`P(X = x) = C_(n,x).(0.85)^(x).(0.15)^(n-x)`

(d) Would it be unusual for x or more of the flights to be on time?

If
`x > E(X) + √(V(X))`, yes, otherwise no