Question

According to an​ airline, flights on a certain route are on time 90 90​% of the time. Suppose 20 20 flights are randomly selected and the number of on time flights is recorded. Use technology to find the probabilities. ​(a) Determine whether this is a binomial experiment. ​(b) Find and interpret the probability that exactly 18 18 flights are on time. ​(c) Find and interpret the probability that at least 18 18 flights are on time. ​(d) Find and interpret the probability that fewer than 18 18 flights are on time. ​(e) Find and interpret the probability that between 17 17 and 19 19 ​flights, inclusive, are on time.

Answer 1

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;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......$

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

Let X = Number of flights on a certain route that are on time

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) =  $\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}$

                              =  $190 \times 0.90^{18} \times 0.10^{2}$

                              =  0.285

Therefore, probability that exactly 18 flights are on time is 28.5%.

(c) Probability that at least 18 flights are on time is given by = P(X $\geq$ 18)

P(X $\geq$ 18) = P(X = 18) + P(X = 19) + P(X = 20)

=  $\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}+\binom{20}{19} \times 0.90^{19} \times (1-0.90)^{20-19}+\binom{20}{20} \times 0.90^{20} \times (1-0.90)^{20-20}$

=  $190 \times 0.90^{18} \times 0.10^{2}+20 \times 0.90^{19} \times 0.10^{1}+1 \times 0.90^{20} \times 0.10^{0}$

=  0.677

Therefore, probability that at least 18 flights are on time is 67.7%.

(d) Probability that fewer than 18 flights are on time is given by = P(X<18)  

             P(X < 18) = 1 - P(X $\geq$ 18)  

                            =  1 - 0.677 = 0.323

Therefore, 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 given by = P(17 $\leq$ X $\leq$ 19)

P(17 $\leq$ X $\leq$ 19) = P(X = 17) + P(X = 18) + P(X = 19)

=  $\binom{20}{17} \times 0.90^{17} \times (1-0.90)^{20-17}+\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}+\binom{20}{19} \times 0.90^{19} \times (1-0.90)^{20-19}$

=  $1140 \times 0.90^{17} \times 0.10^{3}+190 \times 0.90^{18} \times 0.10^{2}+20 \times 0.90^{19} \times 0.10^{1}$

=  0.745

Therefore, probability that between 17 and 19 ​flights, inclusive, are on time is 74.5%.