Answer:
a) It is binomial distribution
b) 0.08224
c) 0.05719
d) 0.9428
e) 0.13344
Step-by-step explanation:
We are given the following information:
a) We treat flights on a certain route are on time as a success.
P(flights on a certain route are on time ) = 85% = 0.85
Then the number of flights on a certain route are on time follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
b) Now, we are given n = 24 and x = 18
P(exactly 18 flights are on time)
We have to evaluate:
[tex]P(x = 18) = \binom{24}{18}(0.85)^{18}(1-0.85)^{6} = 0.0978 + 0.0230 = 0.08224 = 8.2\%[/tex]
c) We have to evaluate:
P( fewer than 18 flights are on time)
[tex]P(x < 18) =\binom{24}{18}(0.85)^{0}(1-0.85)^{24} +...+ \binom{24}{18}(0.85)^{17}(1-0.85)^{7} = 0.05719= 5.7\%[/tex]
d) We have to evaluate:
P( at least 18 flights are on time.)
[tex]P(x \geq 18) =\binom{24}{18}(0.85)^{18}(1-0.85)^{6} +...+ \binom{24}{18}(0.85)^{24}(1-0.85)^{0} = 0.9428= 94.28\%[/tex]
e) We have to evaluate:
P(between 16 and 18 flights, inclusive, are on time.)
[tex]P(16 \leq x \leq 18) =\binom{24}{18}(0.85)^{18}(1-0.85)^{6} +\binom{24}{17}(0.85)^{17}(1-0.85)^{7}+ \binom{24}{16}(0.85)^{16}(1-0.85)^{8} =0.08224+ 0.0373+0.0139=0.13344= 13.34\%[/tex]