Question

An engineer commutes daily from her suburban home to her midtown office. The average time for a one-way trip is 36 minutes, with a standard deviation of 4.9 minutes. Assume the distribution of trip times to be normally distributed. What is the probability that a trip will take at least 35 minutes

Answer 1

Answer:

57.93% probability that a trip will take at least 35 minutes.

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:

$\mu = 36, \sigma = 4.9$

What is the probability that a trip will take at least 35 minutes

This probability is 1 subtracted by the pvalue of Z when X = 35. So

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

$Z = \frac{35 - 36}{4.9}$

$Z = -0.2$

$Z = -0.2$ has a pvalue of 0.4207

1 - 0.4207 = 0.5793

57.93% probability that a trip will take at least 35 minutes.