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2 years ago

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Mathematics

1 answer:

Ahat [919]2 years ago

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3 years ago

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Your flight has been delayed: At Denver International Airport, 85% of recent flights have arrived on time. A sample of 14 flight

Bezzdna [24]

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:

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3 years ago