Respuesta :
Answer:
a) 22.66% of customers receive the service for half-price.
b) The guaranteed time limit should be of 25.2 minutes.
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.
In this question, we have that:
[tex]\mu = 17, \sigma = 4[/tex]
(a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price?
Longer than 20 minutes is 1 subtracted by the pvalue of Z when X = 20. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20 - 17}{4}[/tex]
[tex]Z = 0.75[/tex]
[tex]Z = 0.75[/tex] has a pvalue of 0.7734
1 - 0.7734 = 0.2266
22.66% of customers receive the service for half-price.
(b) If the automotive center does not want to give the discount to more than 2% of its customers, how long should it make the guaranteed time limit?
The time limit should be the 100 - 2 = 98th percentile, which is X when Z has a pvalue of 0.98. So X when Z = 2.054.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.054 = \frac{X - 17}{4}[/tex]
[tex]X - 17 = 4*2.054[/tex]
[tex]X = 25.2[/tex]
The guaranteed time limit should be of 25.2 minutes.
