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.

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:

The standard deviation of the binomial distribution is:

Normal probability distribution

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

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

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 , .

In this problem, we have that:

So

(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

has a pvalue of 0.9418

X = 146.5

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.

, with continuity correction is

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

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.

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

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.

Using continuity correction

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

X = 154.5

has a pvalue of 0.9981

X = 144.5

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.

2 5/12

Step-by-step explanation:

(5 + 1/4) - (2 - 5/6)

5 - 2 = 3

(1/4 x 3/3) - (5/6 x 2/2) = 3/12 - 10/12 = -7/12

3 - 7/12 = 2 5/12