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Chapter 6: Problem 63

For a car traveling 65 kilometers per hour \((\mathrm{km} / \mathrm{h}),\) the distance required to brake to a stop is normally distributed with a mean of 43 meters \(^{4}\) and a standard deviation of 5 meters. Suppose you are traveling \(65 \mathrm{~km} / \mathrm{h}\) in a residential area and a car moves abruptly into your path at a distance of 49 meters. a. If you apply your brakes, what is the probability that you will brake to a stop within 37 meters or less? Within 43 meters or less? b. If the only way to avoid a collision is to brake to a stop, what is the probability that you will avoid the collision?

Short Answer

Expert verified
Answer: The probability of stopping within 37 meters or less is 11.51%. The probability of stopping within 43 meters or less is 50%. The probability of stopping within 49 meters or less to avoid a collision is 88.49%.

Step by step solution

01

Problem Information

The braking distance of a car traveling at 65 km/h follows a normal distribution with a mean of 43 meters and a standard deviation of 5 meters. We are asked to find the probability of stopping within 37 meters or less, 43 meters or less, and if a car moves abruptly into the path at a distance of 49 meters.
02

Calculate the z-scores

To determine the probabilities of stopping within the given distances, we first need to calculate the z-scores. The z-score formula is: $$ z = \frac{X - \mu}{\sigma} $$ Where \(X\) is the value we want to find the probability for, \(\mu\) is the mean of the distribution (43 meters), and \(\sigma\) is the standard deviation (5 meters). Let's calculate the z-scores for 37 meters, 43 meters, and 49 meters. For 37 meters: $$ z_{37} = \frac{37 - 43}{5} = -1.2 $$ For 43 meters: $$ z_{43} = \frac{43 - 43}{5} = 0 $$ For 49 meters: $$ z_{49} = \frac{49 - 43}{5} = 1.2 $$
03

Find the probabilities

Now that we have the z-scores, we can use the standard normal distribution table or a calculator to find the corresponding probabilities. a. Probability of stopping within 37 meters or less is the area to the left of \(z_{37} = -1.2\). Checking the standard normal distribution table, we will have: $$ P(z \leq -1.2) = 0.1151 $$ Probability of stopping within 43 meters or less is the area to the left of \(z_{43} = 0\). By definition, the area to the left of \(z = 0\) is 0.5 or 50%: $$ P(z \leq 0) = 0.5 $$ b. The probability of stopping within 49 meters or less to avoid a collision is the area to the left of \(z_{49} = 1.2\). Checking the standard normal distribution table again, we find: $$ P(z \leq 1.2) = 0.8849 $$
04

Write the final answer

a. The probability of stopping within 37 meters or less is 0.1151 (11.51%) and within 43 meters or less is 0.5 (50%). b. The probability of stopping within 49 meters or less to avoid a collision is 0.8849 (88.49%).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution
The concept of a normal distribution is a foundational idea in statistics, used to describe how the values of a variable are distributed. Imagine a bell-shaped curve that symmetrical around the mean, with the majority of data points clustered around the center, and fewer data points appearing as you move away from the mean.

For our car braking distance example, the normal distribution indicates that most cars will stop within a distance close to the mean of 43 meters after applying brakes at 65 km/h. However, there will still be cars that stop much sooner or need more distance, represented by the tails of the distribution. This bell-shaped curve allows us to predict probabilities for any given distance, crucial for safety assessments and designing traffic regulations.
Z-score
The z-score is a measurement that indicates how many standard deviations an element is from the mean. In other words, it's a way to standardize individual data points against the whole distribution, which allows comparison across different datasets.

In the braking distance problem, we calculate z-scores to determine how likely it is for the car to stop within certain distances. A z-score of 0 means the value is exactly at the mean, while a positive z-score means the value is above the mean, and a negative z-score indicates it's below the mean. For instance, when the car stops precisely at the mean distance, the z-score is 0.
Standard Deviation
Standard deviation signifies the amount of variation or dispersion in a set of values. A low standard deviation means that values tend to be close to the mean, and a high standard deviation means values are spread out over a wider range.

In the context of braking distances, a standard deviation of 5 meters means that many of the stopping distances are within 5 meters of the mean, but there will still be some variation. To assess the risk of an accident, understanding this variability is essential, as it helps predict how likely it is for a car to exceed a certain braking distance and consequently, how much buffer space is advisable.

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Most popular questions from this chapter

Assume that the heights of American men are normally distributed with a mean of 176.5 centimeters and a standard deviation of 8.9 centimeters. Use this information to answer the questions. President Donald Trump is \(1.88 \mathrm{~m}\). Is this an unusual height?

Find the mean and standard deviation for the binomial random variable \(x\) using the information Then use the correction for continuity and approximate the probabilities using the normal approximation. $$ P(6 \leq x \leq 9) \text { when } n=25 \text { and } p=.3 $$

Let \(z\) be a standard normal random variable with mean \(\mu=0\) and standard deviation \(\sigma=1 .\) Find the value \(c\) that satisfies the inequalities given. The area to the left of \(c\) is \(.05 .\)

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Two of the biggest soft drink rivals, Pepsi and Coke, are very concerned about their market shares. The pie chart that follows claims that Coke's share of the beverage market is \(36 \% .^{11}\) Assume that this proportion will be close to the probability that a person selected at random indicates a preference for a Coke product when choosing a soft drink. A group of \(n=500\) consumers is selected and the number preferring a Coke product is recorded. Use the normal curve to approximate the following binomial probabilities. a. Exactly 200 consumers prefer a Coke product. b. Between 175 and 200 consumers (inclusive) prefer a Coke product. c. Fewer than 200 consumers prefer a Coke product. d. Would it be unusual to find that 232 of the 500 consumers preferred a Coke product? If this were to occur, what conclusions would you draw?
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