Answer: a) 0.2073
Step-by-step explanation:
Binomial distribution formula :
[tex]P(X=x)=^nC_xp^x q^{n-x}[/tex] ,
where n= total number of trials.
x= number of successes.
p= probability of getting success in each trial.
q= 1-p = probability of getting failure in each trial.
For the given question , let the flight are on time be success.
Let x denote the number of flight are on time .
Given : Proportion of domestic flights were delayed in 2007 at JFK airport : q=0.27
Proportion of domestic flights were on time in 2007 at JFK airport :
p=1-0.27=0.73
Five flights are randomly selected at JFK.
i.e. n=5
Now, the probability that all five flights are on time will be :-
[tex]P(X=5)=^5C_5(0.73)^5 (0.27)^{5-5}\\\\=(1)(0.73)^5=0.2073071593\approx0.2073[/tex]
Hence, the probability that all five flights are on time = 0.2073
