Question

An engineer travels daily from home tothelaboratory each morning with an average commutetime of 24.6 minutes with a standard deviation of 3.4 minutes. The travel timeapproximates very closely a Normal Distribution. a.What is the probability that the commute takes more than 32 minutes

Answer 1

Answer:

1.74% probability that the commute takes more than 32 minutes.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 24.6, \sigma = 3.5$

a.What is the probability that the commute takes more than 32 minutes?

This is 1 subtracted by the pvalue of Z when X = 32. So:

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

$Z = \frac{32 - 24.6}{3.5}$

$Z = 2.11$

$Z = 2.11$ has a pvalue of 0.9826.

So there is a 1-0.9826 = 0.0174 = 1.74% probability that the commute takes more than 32 minutes.