University:
University of Houston
Course:
MATH 2450 | Accelerated Calculus I
Academic year:

2023
Views:

315

Pages:

1
Author:

lo1kakr9cy

Example Problems 6.19 For a car traveling 30 mph, the distance required to brake to stop is normally distributed with a mean of 50 feet and a standard deviation of 8 feet. Suppose you are traveling 30 mph and a car moves abruptly into your path at a distance of 60 feet. If you apply your brakes, what is the probability that you will break to a stop within 40 ft or less? Let X = stopping distance in feet Given 1) the mean of X (i.e. the mean stopping distance): 2) the standard deviation of X: = 50 feet = 50 feet a) Find: P(X<40) [i.e. the probability that you will break to a stop within 40 ft or less] To find the required probability using the normal tables in the text, you must first convert X = 40 to a value of z. z X   = 40  50 = -1.25 8 A stopping distance of 40 feet is 1.25 standard deviations to the left of the mean of 50 feet. Using the table, find 1.25 to get a value of 0.3944. This value is P(-1.25 < z < 0). Since you want P(z < 40) = P(z< -1.25), subtract the value you found in the table from 0.5000. [0.5000 is the probability that a value is less than the mean when you have a normal curve.] P(z < 40) = P(z< -1.25) = 0.5 - .3944 = 0.1056 Thus, you will be able to stop in less than 40 feet about 10.56% of the time. b) To avoid a collision, find P(X < 60). z X   = 60  50 = +1.25 8 A stopping distance of 40 feet is 1.25 standard deviations to the right of the mean of 50 feet. P(z < 60) = P(z < 1.25) = 0.5 + .3944 = 0.8944 [Note since you want the probability of everything to the left of 60, you add the area between 0 and 1.25 to the area to the left of the mean.] 3360p619 1