Respuesta :
Answer:
85.99% of airline passengers incur this fee.
Step-by-step explanation:
When the distribution is normal, we use 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.
Mean 46 pounds and standard deviation 3.7 pounds.
This means that [tex]\mu = 46, \sigma = 3.7[/tex]
Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee.
As a proportion, this is 1 subtracted by the pvalue of Z when X = 50. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{50 - 46}{3.7}[/tex]
[tex]Z = 1.08[/tex]
[tex]Z = 1.08[/tex] has a pvalue of 0.8599
0.8599*100% = 85.99%
85.99% of airline passengers incur this fee.
