Respuesta :
Answer:
The latest time that you should leave home is 12:08 PM
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 = 40, \sigma = 7[/tex]
If you want tobe 95 percent certain that you will not be late for an office appointment at 1 P.M.,what is the latest time that you should leave home?
How many minutes is the 95th percentile of travel time?
it is X when Z has a pvalue of 0.95. So it is X when Z = 1.645. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 40}{7}[/tex]
[tex]X - 40 = 7*1.645[/tex]
[tex]X = 51.6[/tex]
Rouding up, the 95th percentile for travel time is 52 minutes.
52 minutes before 1PM is 12:08 PM.
The latest time that you should leave home is 12:08 PM
