Question

A study measured the speeds at which cars pass through a checkpoint. Assume the speeds are normally distributed such that the average is 61 miles per hour and the standard deviation is 4 miles per hour. Calculate the probability that the next car will be traveling less than 59 miles per hour.

Answer 1

Answer:

0.3085 = 30.85% probability that the next car will be traveling less than 59 miles per hour.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 61, \sigma = 4$

Calculate the probability that the next car will be traveling less than 59 miles per hour.

This is the pvalue of Z when X = 59. So

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

$Z = \frac{59 - 61}{4}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3085

0.3085 = 30.85% probability that the next car will be traveling less than 59 miles per hour.