Question

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.)

Answer 1

Answer:

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.