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

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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 [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 = 24.6, \sigma = 3.5[/tex]

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:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{32 - 24.6}{3.5}[/tex]

[tex]Z = 2.11[/tex]

[tex]Z = 2.11[/tex] 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.