Respuesta :
Answer:
The probability that at least 13 flights arrive late is 2.5196 [tex]\times 10^{-6}[/tex].
Step-by-step explanation:
We are given that Southwest Air had the best rate with 80 % of its flights arriving on time.
A test is conducted by randomly selecting 18 Southwest flights and observing whether they arrive on time.
The above situation can be represented through binomial distribution;
[tex]P(X = x) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.........[/tex]
where, n = number of trials (samples) taken = 18 Southwest flights
r = number of success = at least 13 flights arrive late
p = probability of success which in our question is probability that
flights arrive late, i.e. p = 1 - 0.80 = 20%
Let X = Number of flights that arrive late.
So, X ~ Binom(n = 18, p = 0.20)
Now, the probability that at least 13 flights arrive late is given by = P(X [tex]\geq[/tex] 13)
P(X [tex]\geq[/tex] 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)
= [tex]\binom{18}{13}\times 0.20^{13} \times (1-0.20)^{18-13}+ \binom{18}{14}\times 0.20^{14} \times (1-0.20)^{18-14}+ \binom{18}{15}\times 0.20^{15} \times (1-0.20)^{18-15}+ \binom{18}{16}\times 0.20^{16} \times (1-0.20)^{18-16}+ \binom{18}{17}\times 0.20^{17} \times (1-0.20)^{18-17}+ \binom{18}{18}\times 0.20^{18} \times (1-0.20)^{18-18}[/tex]
= [tex]\binom{18}{13}\times 0.20^{13} \times 0.80^{5}+ \binom{18}{14}\times 0.20^{14} \times 0.80^{4}+ \binom{18}{15}\times 0.20^{15} \times 0.80^{3}+ \binom{18}{16}\times 0.20^{16} \times 0.80^{2}+ \binom{18}{17}\times 0.20^{17} \times 0.80^{1}+ \binom{18}{18}\times 0.20^{18} \times 0.80^{0}[/tex]
= 2.5196 [tex]\times 10^{-6}[/tex].
