Question

In a study investigating the effect of car speed on accident severity, 5,000 reports of fatal automobile accidents were examined, and the vehicle speed at impact was recorded for each one. For these 5,000 accidents, the average speed was 46 mph and the standard deviation was 13 mph. A histogram revealed that the vehicle speed at impact distribution was approximately normal.

A. Roughly what proportion of vehicle speeds were between 27 and 57 mph?
B. Roughly what proportion of vehicle speeds exceeded 57 mph?

Answer 1

a) 73.02% of vehicle speeds were between 27 and 57 mph

b) 19.77% of vehicle speeds exceeded 57 mph

Explanation:

Problems of normally distributed distributions 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 = (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 = 46, \sigma = 13`

A. Roughly what proportion of vehicle speeds were between 27 and 57 mph?

This is the pvalue of Z when X = 57 subtracted by the pvalue of Z when X = 27. So

X = 57

`Z = (X - \mu)/(\sigma)`

`Z = (57 - 46)/(13)`

`Z = 0.85`

`Z = 0.85` has a pvalue of 0.8023.

X = 27

`Z = (X - \mu)/(\sigma)`

`Z = (27 - 46)/(13)`

`Z = -1.46`

`Z = -1.46` has a pvalue of 0.0721

0.8023 - 0.0721 = 0.7302

73.02% of vehicle speeds were between 27 and 57 mph

B. Roughly what proportion of vehicle speeds exceeded 57 mph?

This is 1 subtracted by the pvalue of Z when X = 57.

From a), when X = 57, Z = 0.85, which has a pvalue of 0.8023.

1 - 0.8023 = 0.1977

19.77% of vehicle speeds exceeded 57 mph