For a specific truck traveling at 50 miles per hour (mph), the distance x required to brake to a complete stop is normally distributed with a mean of 120 feet and a standard deviation of 12 feet. Suppose that this truck is traveling at a constant speed of 50 mph and a car abruptly moves into the path of the truck and stops at a distance of 150 feet from the truck. Assuming that the only way to avoid a collision is for the truck to brake to a complete stop, what is the probability that there will be a collision? Round your answer to 4 decimal places. Remember to round all z values to two decimal places.

Respuesta :

Answer:

There is a 0.62% probability that there will be a collision.

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]

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

For a specific truck traveling at 50 miles per hour (mph), the distance x required to brake to a complete stop is normally distributed with a mean of 120 feet and a standard deviation of 12 feet. This means that [tex]\mu = 120, \sigma = 12[/tex].

Suppose that this truck is traveling at a constant speed of 50 mph and a car abruptly moves into the path of the truck and stops at a distance of 150 feet from the truck. Assuming that the only way to avoid a collision is for the truck to brake to a complete stop, what is the probability that there will be a collision?

There is going to be a collision if the car takes more than 150 feet to brake. So the probability of a collision is 1 subtracted by the pvalue of [tex]X = 150[/tex].

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

[tex]Z = \frac{150 - 120}{12}[/tex]

[tex]Z = 2.5[/tex]

[tex]Z = 2.5[/tex] has a pvalue of 0.9938.

This means that there is a 1-0.9938 = 0.0062 = 0.62% probability that there will be a collision.