According to a flight statistics website, in 2009, a certain airline had the highest percentage of on-time flights in the airlines industry, which was 80.8 %. Assume this percentage still holds true for that airline. Use the normal approximation to the binomial distribution to complete parts a through c below. a. Determine the probability that, of the next 30 flights from this airline, less than 20 flights will arrive on time. P (xless than20 ) equals 0.0140 (Round to four decimal places as needed.) b. Determine the probability that, of the next 30 flights from this airline, exactly 24 flights will arrive on time. P (xequals24 ) equals 0.1822 (Round to four decimal places as needed.) c. Determine the probability that, of the next 30 flights from this airline, 25 , 26 , 27 , or 28 flights will arrive on time. P (25less than or equalsxless than or equals28 ) equals 0.4279 (Round to four decimal places as needed.)

Question

Cheyanne Blevins

Mathematics

2 February, 11:08

According to a flight statistics website, in 2009, a certain airline had the highest percentage of on-time flights in the airlines industry, which was 80.8 %. Assume this percentage still holds true for that airline. Use the normal approximation to the binomial distribution to complete parts a through c below. a. Determine the probability that, of the next 30 flights from this airline, less than 20 flights will arrive on time. P (xless than20 ) equals 0.0140 (Round to four decimal places as needed.) b. Determine the probability that, of the next 30 flights from this airline, exactly 24 flights will arrive on time. P (xequals24 ) equals 0.1822 (Round to four decimal places as needed.) c. Determine the probability that, of the next 30 flights from this airline, 25 , 26 , 27 , or 28 flights will arrive on time. P (25less than or equalsxless than or equals28 ) equals 0.4279 (Round to four decimal places as needed.)

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Answers (1)

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Wade Lopez · Answer 1

a) 0.0139; b) 0.1809; c) 0.4278

Step-by-step explanation:

We use a normal approximation to a binomial distribution for these problems.

The sample size, n, for each is 30; p, the probability of success, is 0.808. This makes the mean, μ = np = 30 (0.808) = 24.24. The standard deviation,

σ = √ (npq) = √ (30 (0.808) (1-0.808)) = √ (30 (0.808) (0.192)) = √4.65408 = 2.1573

For part a,

We are asked for P (X < 20). Using continuity correction to account for the discrete variable, we find

P (X < 19.5)

z = (19.5-24.24) / (2.1573) = - 4.74/2.1573 = - 2.20

Using a z table, we see that the area under the curve to the left of this is 0.0139.

For part b,

We are asked for P (X = 24). Using continuity correction, we find

P (23.5 < X < 24.5)

z = (23.5-24.24) / 2.1573 = - 0.74/2.1573 = - 0.34

z = (24.5-24.24) / 2.1573 = 0.26/2.1573 = 0.12

Using a z table, we see that the area under the curve to the left of z = - 0.34 is 0.3669. The area under the curve to the left of z = 0.12 is 0.5478. The area between them is then

0.5478-0.3669 = 0.1809.

For part c,

We are asked to find P (25 ≤ X ≤ 28). Using continuity correction, we find

P (24.5 < X < 28.5)

z = (24.5-24.24) / 2.1573 = 0.26/2.1573 = 0.12

z = (28.5-24.24) / 2.1573 = 4.26/2.1573 = 1.97

Using a z table, we see that the area under the curve to the left of z = 0.12 is 0.5478. The area under the curve to the left of z = 1.97 is 0.9756. The area between them is 0.9756 - 0.5478 = 0.4278.