Question

Is lack of sleep causing traffic fatalities? a study found that the average number of fatal crashes caused by drowsy drivers each year is 1555. assume that annual number of fatal crashes is normally distributed with a standard deviation of 300. what is the probability of fewer than 1000 fatal crashes a year due to drowsy drivers?

Answer 1

The probability of fewer than 1000 fatal crashes a year due to drowsy drivers is calculated by finding the Z-score and then looking up the corresponding probability in a normal distribution table. The Z-score here is -1.85, which corresponds to a probability of approximately 3.22%.

Step-by-step explanation:

The question is regarding the probability of fewer than 1000 fatal crashes a year due to drowsy drivers, assuming the average number is 1555 and the standard deviation is 300. This is a normal distribution probability question. The first step is to calculate the Z-score using the formula:

Z = (X - μ) / σ

where X is the value of interest (1000 fatal crashes), μ is the mean (1555 fatal crashes), and σ is the standard deviation (300 crashes).

Z = (1000 - 1555) / 300 = -555 / 300 = -1.85

We then look up the Z-score in a standard normal distribution table or use a calculator to find the probability associated with a Z-score of -1.85. The probability (P) of Z < -1.85 is approximately 0.0322.

So, the probability of fewer than 1000 fatal crashes a year due to drowsy drivers is roughly 3.22%.