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.