Answer:
(a) Yes, the above experiment is a binomial distribution.
(b) Probability that exactly 18 flights are on time is 28.5%.
(c) Probability that at least 18 flights are on time is 67.7%.
(d) Probability that fewer than 18 flights are on time is 32.3%.
(e) Probability that between 17 and 19 flights, inclusive, are on time is 74.5%.
Step-by-step explanation:
We are given that according to an airline, flights on a certain route are on time 90% of the time.
Suppose 20 20 flights are randomly selected and the number of on time flights is recorded.
The above situation can be represented through binomial distribution;
where, n = number trials (samples) taken = 20 flights
r = number of success
p = probability of success which in our question is probability that
flights on a certain route are on time, i.e; p = 0.90
<em>Let X = Number of flights on a certain route that are on time</em>
So, X ~ Binom(n = 20, p = 0.90)
(a) Yes, the above experiment is a binomial distribution as the probability of success is the probability in a single trial.
And also, each flight is independent of another.
(b) Probability that exactly 18 flights are on time is given by = P(X = 18)
P(X = 18) =
=
= 0.285
<em>Therefore, probability that exactly 18 flights are on time is 28.5%.</em>
<em />
(c) Probability that at least 18 flights are on time is given by = P(X 18)
P(X 18) = P(X = 18) + P(X = 19) + P(X = 20)
=
=
= 0.677
<em>Therefore, probability that at least 18 flights are on time is 67.7%.</em>
<em />
(d) Probability that fewer than 18 flights are on time is given by = P(X<18)
P(X < 18) = 1 - P(X 18)
= 1 - 0.677 = 0.323
<em>Therefore, probability that fewer than 18 flights are on time is 32.3%.</em>
(e) Probability that between 17 and 19 flights, inclusive, are on time is given by = P(17 X
19)
P(17 X
19) = P(X = 17) + P(X = 18) + P(X = 19)
=
=
= 0.745
<em>Therefore, probability that between 17 and 19 flights, inclusive, are on time is 74.5%.</em>
