Respuesta :
Answer:
Approximately nineteen percent of the service calls take more than 92.6 minutes.
Step-by-step explanation:
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.
In this problem, we have that:
[tex]\mu = 75, \sigma = 20[/tex]
Approximately nineteen percent of the service calls take more than how many minutes?
This is the value of X when Z has a pvalue of 1 - 0.19 = 0.81. So it is X when Z = 0.88.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.88 = \frac{X - 75}{20}[/tex]
[tex]X - 75 = 20*0.88[/tex]
[tex]X = 92.6[/tex]
Approximately nineteen percent of the service calls take more than 92.6 minutes.
