Question

According to American Airlines, flight 71098 from New York to Los Angeles is on time 88.9% of the time. Assume that we randomly select 150 flights, use the normal approximation to the binomial to do the following:
a) approximately the probability that exactly 124 flights are on time.

b) approximate the probability that between 113 and 130 flights ,inclusive, are on time.

Answer 1

Using the normal approximation to the binomial, it is found that the probabilities are given as follows:

a) 0.0055 = 0.55%.

b) 0.2296 = 22.96%.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$.

The parameters of the binomial distribution are given as follows:

n = 150, p = 0.889.

Hence the mean and the standard deviation of the approximation are:

• $\mu = E(X) = np = 150 x 0.889 = 133.35$.

• $\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{150(0.889)(0.111)} = 3.8473$

Item a:

Using continuity correction, the probability is P(123.5 < X < 124.5), which is the p-value of Z when X = 124.5 subtracted by the p-value of Z when X = 123.5, hence:

X = 124.5:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{124.5 - 133.35}{3.8473}$

Z = -2.3

Z = -2.3 has a p-value of 0.0107.

X = 123.5:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{123.5 - 133.35}{3.8473}$

Z = -2.56

Z = -2.56 has a p-value of 0.0052.

Hence the probability is 0.0107 - 0.0052 = 0.0055 = 0.55%.

Item b:

The probability is P(112.5 < X < 130.5), which is the p-value of Z when X = 130.5 subtracted by the p-value of Z when X = 112.5, hence:

X = 130.5:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{130.5 - 133.35}{3.8473}$

Z = -0.74

Z = -0.74 has a p-value of 0.2296.

X = 112.5:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{112.5 - 133.35}{3.8473}$

Z = -5.42

Z = -5.42 has a p-value of 0.

Hence the probability is 0.2296 - 0 = 0.2296 = 22.96%.

More can be learned about the normal approximation to the binomial at brainly.com/question/14424710

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