Respuesta :
Answer:
a) 0.0668 = 6.68% probability that a domestic airfare is $550 or more
b) 0.1093 = 10.93% probability than a domestic airfare is $250 or less
c) 0.6313 = 63.13% probability that a domestic airfare is between $300 and $500
d) The cost for the 3% highest domestic airfares are $592 and higher.
Step-by-step explanation:
Problems of normally distributed samples are 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.
In this problem, we have that:
[tex]\mu = 385, \sigma = 110[/tex]
a. What is the probability that a domestic airfare is $550 or more (to 4 decimals)?
This is 1 subtracted by the pvalue of Z when X = 550. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{550 - 385}{110}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a pvalue of 0.9332
1 - 0.9332 = 0.0668
0.0668 = 6.68% probability that a domestic airfare is $550 or more
b. What is the probability than a domestic airfare is $250 or less (to 4 decimals)?
This is the pvalue of Z when X = 250. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{250 - 385}{110}[/tex]
[tex]Z = -1.23[/tex]
[tex]Z = -1.23[/tex] has a pvalue of 0.1093
0.1093 = 10.93% probability than a domestic airfare is $250 or less
c. What if the probability that a domestic airfare is between $300 and $500 (to 4 decimals)?
This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 300. So
X = 500
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{500 - 385}{110}[/tex]
[tex]Z = 1.045[/tex]
[tex]Z = 1.045[/tex] has a pvalue of 0.8519
X = 300
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{300 - 385}{110}[/tex]
[tex]Z = -0.77[/tex]
[tex]Z = -0.77[/tex] has a pvalue of 0.2206
0.8519 - 0.2206 = 0.6313
0.6313 = 63.13% probability that a domestic airfare is between $300 and $500
d. What is the cost for the 3% highest domestic airfares?
At least the 97th percentile, so at least X when Z has a pvalue of 0.97. So X when Z = 1.88.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.88 = \frac{X - 385}{110}[/tex]
[tex]X - 385 = 1.88*110[/tex]
[tex]X = 592[/tex]
The cost for the 3% highest domestic airfares are $592 and higher.
