Question

Your flight has been delayed: At Denver International Airport, 85% of recent flights have arrived on time. A sample of 14 flights is studied. Round the probabilities to at least four decimal places.(a) Find the probability that all 12 of the flights were on time.(b) Find the probability that exactly 10 of the flights were on time.(c) Find the probability that 10 or more of the flights were on time.(d) Would it be unusual for 11 or more of the flights to be on time?

Answer 1

Answer:

a. p=1.000

b. p=0.2924

c. p=0.7358

d. No

Step-by-step explanation:

a. This problem satisfies all the criteria for a binomial experiment expressed as:

$P(X=x){n\choose x}p^x(1-p)^{n-x}$

-Given that p=0.85, n=14, the probability that exactly all 14 were on time is calculated as:

$P(X\geq 1)=1-P(X=0)\\\\=1-{12\choose 0}0.85^0(1-0.85)^{12}\\\\=1-1.297\times 10^{-10}\\\\=1.0000$

Hence, the probability that all 12 flights are on time is 1.0000

b. Given that n=12, and p=0.85

-The probability that exactly 10 flights are on time is calculated as;

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\={12\choose 10}0.85^{10}(1-0.85)^{2}\\\\=0.2924$

Hence, the probability that exactly 10 flights are on time is 0.2924

c. Given that n=12, and p=0.85

-The probability that more of 10 or more flights are on time:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 10)=P(X=10)+P(X=11)+P(X=12)\\\\={12\choose 10}0.85^{10}(0.15)^{2}+{12\choose 11}0.85^{11}(0.15)^{1}+{12\choose 12}0.85^{12}(0.15)^{0}\\\\=0.2924+0.3012+0.1422\\\\=0.7358$

Hence, the probability of 10+ flights being on time is 0.7358

d. We first find the mean of the distribution:

$\mu=E(X)=0.85\times 14\\\\=11.9$

#We then find the probability of 11+=0.3012+0.1422=0.4434

-We compare the expectation to the probability of 11+ flights being on time.

No. Since the probability $P(X\geq 10)$=0.4434 < that the expectation, 11.9, it is not unusual  for 11+ flights to be on time.

*I have used a sample size of n=12 since there are two separate n values: