### Solution Found! # Answer: On-Time Flights According to flightstats.com,

Chapter 11, Problem 35AYU

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QUESTION:

On-Time Flights According to flightstats.com, 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 is a binomial experiment.

(b) Find and interpret the probability that exactly 10 flights are on time.

(c) Find and interpret the probability that fewer than 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 between 8 and 10 flights, inclusive, are on time.

Step1 of 2

We have According to flightstats.com, 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).

This is a binomial experiment because:

1.  It is performed a fixed number of times, n = 15.

2.  The trials are independent.

3.  For each trial, there are two possible mutually exclusive outcomes: on time and not on time.

4.  The probability of “on time” is fixed at p = 0.80.

b).

We have X B(15, 0.80).

Probability mass function of binomial distribution is given by

P(X) = , x = 0,1,2,.....,15.

We need to find P(3) = ?.

Consider,

P(X)  = , x = 0,1,2,.....,15.

P(10)   = =  3003*(0.1073)*(0.00032)

= 0.103111 0.1032.

Therefore,the probability of x successes in the n independent trials of the experiment is P(38) = 0.1032.

In 100 trials of this experiment, we expect about 10 trials to result in exactly 10 flights being on time.

c).

The probability that fewer than 10 flights are on time is given by P(X < 10)

Probability mass function of binomial distribution is given by

P(X) = , x = 0,1,2,.....,15.

P(X) = , x = 0,1,2,.....,15.

Put x = 0,1,2,.....,9. In above equation we get  P(X < 10)

 x P(X) 0 3.2768E-11 1 1.9661E-09 2 5.505E-08 3 9.542E-07 4 1.145E-05 5 0.00010076 6 0.00067176 7 0.00345476 8 0.01381906 9 0.04299262 P(X < 10 ) = 0.06105143

Therefore,The probability that fewer than 10 flights are on time is P(X < 10 ) = 0.06105143.

In 100 trials of this experiment, we expect about 6 trials to result in fewer than 10 flights being on time.

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