Question

A certain flight arrives on time 8080 percent of the time. Suppose 174174 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that ​(a) exactly 147147 flights are on time. ​(b) at least 147147 flights are on time. ​(c) fewer than 145145 flights are on time. ​(d) between 145145 and 154154​, inclusive are on time.

Answer 1

Answer:

a) 2.56% probability that exactly 147 flights are on time.

b) 8.38% probability that at least 147 flights are on time.

c) 84.13% probability that fewer than 145 flights are on time.

d) 15.68% probability that between 145 and 154 flights, inclusive, are on time.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 174, p = 0.8$

So

$\mu = E(X) = np = 174*0.8 = 139.2$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{174*0.8*0.2} = 5.28$

(a) exactly 147 flights are on time.

This is P(X = 147)

Using continuity correction, this is P(146.5 < X < 147.5), which is the pvalue of Z when X = 147.5 subtracted by the pvalue of Z when X = 146.5.

X = 147.5

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

$Z = \frac{147.5 - 139.2}{5.28}$

$Z = 1.57$

$Z = 1.57$ has a pvalue of 0.9418

X = 146.5

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

$Z = \frac{146.5 - 139.2}{5.28}$

$Z = 1.38$

$Z = 1.38$ has a pvalue of 0.9162

0.9418 - 0.9162 = 0.0256

2.56% probability that exactly 147 flights are on time.

(b) at least 147 flights are on time.

$P(X \geq 147)$, with continuity correction is $X \geq 147-0.5 = 146.5$

So it is 1 subtracted by the pvalue of Z when X = 146.5

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

$Z = \frac{146.5 - 139.2}{5.28}$

$Z = 1.38$

$Z = 1.38$ has a pvalue of 0.9162

1 - 0.9162 = 0.0838

8.38% probability that at least 147 flights are on time.

(c) fewer than 145 flights are on time.

$P(X < 145)$, with continuity correction, is P(X < 145-0.5 = 144.5). So pvalue of Z when X = 144.5

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

$Z = \frac{144.5 - 139.2}{5.28}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

84.13% probability that fewer than 145 flights are on time.

d) between 145 and 154, inclusive are on time.

$P(145 \leq X \leq 154)$

Using continuity correction

$P(144.5 \leq X \leq 154.5)$

pvalue of Z when X = 154.5 subtracted by the pvalue of Z when X = 144.5

X = 154.5

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

$Z = \frac{154.5 - 139.2}{5.28}$

$Z = 2.9$

$Z = 2.9$ has a pvalue of 0.9981

X = 144.5

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

$Z = \frac{144.5 - 139.2}{5.28}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

0.9981 - 0.8413 = 0.1568

15.68% probability that between 145 and 154 flights, inclusive, are on time.