Respuesta :
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;
[tex]P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......[/tex]
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) = [tex]\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}[/tex]
= [tex]190 \times 0.90^{18} \times 0.10^{2}[/tex]
= 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 [tex]\geq[/tex] 18)
P(X [tex]\geq[/tex] 18) = P(X = 18) + P(X = 19) + P(X = 20)
= [tex]\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}[/tex]
= [tex]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}[/tex]
= 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 [tex]\geq[/tex] 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 [tex]\leq[/tex] X [tex]\leq[/tex] 19)
P(17 [tex]\leq[/tex] X [tex]\leq[/tex] 19) = P(X = 17) + P(X = 18) + P(X = 19)
= [tex]\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}[/tex]
= [tex]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}[/tex]
= 0.745
Therefore, probability that between 17 and 19 flights, inclusive, are on time is 74.5%.
