Question

#11 According to an​ airline, flights on a certain route are on time 75​% of the time. Suppose 15 flights are randomly selected and the number of​ on-time flights is recorded.
​(a) Explain why this is a binomial experiment.
​(b) Find and interpret the probability that exactly 9 flights are on time.
​(c) Find and interpret the probability that fewer than 9 flights are on time.
​(d) Find and interpret the probability that at least 9 flights are on time.
​(e) Find and interpret the probability that between 7 and 9 ​flights, inclusive, are on time.

Identify the statements that explain why this is a binomial experiment. Select all that apply.

A. There are two mutually exclusive​ outcomes, success or failure.

B. The experiment is performed until a desired number of successes is reached.

C. There are three mutually exclusive possibly​ outcomes, arriving​ on-time, arriving​ early, and arriving late.

D. Each trial depends on the previous trial.

E. The probability of success is the same for each trial of the experiment.

F. The experiment is performed a fixed number of times.

G. The trials are independent.

b) The probability that exactly 9 flights are on time is

In 100 trials of this​ experiment, it is expected about __ to result in exactly 9 flights being on time.

c) The probability that fewer than 9 flights are on time is ____

In 100 trials of this​ experiment, it is expected about ___ to result in at least 9 flights being on time.

e) The probability that between 7 and 9 ​flights, inclusive, are on time is ___

In 100 trials of this​ experiment, it is expected about ___ to result in between 7 and 9 ​flights, inclusive, being on time.​(Round to the nearest whole number as​ needed.)

Answer 1

a)

A. There are two mutually exclusive​ outcomes, success or failure.

E. The probability of success is the same for each trial of the experiment.

F. The experiment is performed a fixed number of times.

G. The trials are independent.

b)The probability that exactly 9 flights are on time is:

P(X=9)=

(nCx)px(1−p)(n-x)    =

0.0917

In 100 trials of this​ experiment, it is expected about 9 to result in exactly 9 flights being on time.

c)The probability that fewer than 9 flights are on time is:

P(X<=8)=

∑x=0a     (nCx)px(1−p)(n-x)    =

0.0566

In 100 trials of this​ experiment, it is expected about 6 to result in  fewer than 9 flights are on time

d)

P(X>=9)=1-P(X<=8)=

1-∑x=0x-1   (nCx)px(q)(n-x) =

0.9434

In 100 trials of this​ experiment, it is expected about 94 to result in at least 9 flights being on time.

e)

P(7<=X<=9)=

∑x=ab     (nCx)px(1−p)(n-x)    =

0.1442

In 100 trials of this​ experiment, it is expected about 14 to result in between 7 and 9 ​flights, inclusive, being on time