Question

Suppose that the travel time from your home to your office is normallydistributed with mean 40 minutes and standard deviation 7 minutes. 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?

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 40, \sigma = 7$

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

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 40}{7}$

$X - 40 = 7*1.645$

$X = 51.6$

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