Question

A certain flight arrives on time 80 percent of the time. Suppose 165 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that ​(a) exactly 135 flights are on time. ​(b) at least 135 flights are on time. ​(c) fewer than 146 flights are on time. ​(d) between 146 and 148 ​, inclusive are on time.

P(135)=

Answer 1

When using the normal approximation to the binomial, we need to find the mean and standard deviation for the normal distribution.

Mean () = n * p = 165 * 0.8 = 132

Standard Deviation () = Sqrt[n * p * (1 - p)] = Sqrt(165 * 0.8 * 0.2) = 5.138

To find the probability, we need to find the z scores.

We must also use the continuity correction factor, when using the normal approximation to the binomial. The table for the guidelines are as below.

Continuity Correction Factor Table

1) If P(X = n)       Use P(n – 0.5 < X < n+0.5)

2) If P(X > n)       Use P(X > n + 0.5)

3) If P(X < n)    Use P(X < n + 0.5)

4) If P(X < n)       Use P(X < n - 0.5)

5) If P(X > n)    Use P(X > n - 0.5)

6) If P(a < X < b), Use P(X < b + 0.5) - P(X < a - 0.5)

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(a) P(X = 135). From the continuity correction factor table, we need to find P(135 - 0.5 < X < 135 +0.5) = P(134.5 < X < 135.5)

P(134.5 < X < 135.5) = P(X < 135.5) - P(X < 134.5)

For P(X < 135.5) ; z = (135.5 - 132) / 5.138 = 0.68. The p value at this score is = 0.7517

For P(X < 200) ; z = (135.5 - 132) / 5.138 = 0.49. The p value at this score is = 0.6879

Therefore the required probability is 0.7517 – 0.6879 = 0.0638

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(b) P(At least 135) P(X > 135). From the continuity correction factor table, we need to find P(X > 135 - 0.5) = P(X > 134.5)

For P (X > 134.5) = 1 - P (X < 134.5), as the normal tables give us the left tailed probability only.

For P( X < 134.5)

Z = (134.5 – 132) / 5.138 = 0.49

The probability for P(X < 134.5) from the normal distribution tables is = 0.6879

Therefore the required probability = 1 – 0.6879 = 0.3121

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(c) P(X < 146). From the continuity correction factor table, we need to find P(X < 146 - 0.5) = P(X < 145.5)

Z = (145.5 – 132) / 5.138 = 2.63

The required probability from the normal distribution tables is = 0.9957

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(d) P(146 < X < 148). From the continuity correction factor table, we need to find P(X < 148 + 0.5) - P(X < 146 - 0.5)

= P(X < 148.5) - P(X < 145.5)

For P(X < 148.5)

Z = (148.5 – 132) / 5.138 = 3.21. The p value at this score is = 0.9993

For P(X < 145.5)

Z = (145.5 – 132) / 5.138 = 2.63. The p value at this score is = 0.9957

Therefore the required probability is 0.9993 – 0.9957 = 0.0036

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