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.

Respuesta :

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:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 174, p = 0.8[/tex]

So

[tex]\mu = E(X) = np = 174*0.8 = 139.2[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{174*0.8*0.2} = 5.28[/tex]

(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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{147.5 - 139.2}{5.28}[/tex]

[tex]Z = 1.57[/tex]

[tex]Z = 1.57[/tex] has a pvalue of 0.9418

X = 146.5

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{146.5 - 139.2}{5.28}[/tex]

[tex]Z = 1.38[/tex]

[tex]Z = 1.38[/tex] 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.

[tex]P(X \geq 147)[/tex], with continuity correction is [tex]X \geq 147-0.5 = 146.5[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{146.5 - 139.2}{5.28}[/tex]

[tex]Z = 1.38[/tex]

[tex]Z = 1.38[/tex] 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.

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{144.5 - 139.2}{5.28}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] 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.

[tex]P(145 \leq X \leq 154)[/tex]

Using continuity correction

[tex]P(144.5 \leq X \leq 154.5)[/tex]

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

X = 154.5

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{154.5 - 139.2}{5.28}[/tex]

[tex]Z = 2.9[/tex]

[tex]Z = 2.9[/tex] has a pvalue of 0.9981

X = 144.5

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{144.5 - 139.2}{5.28}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] 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.

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