/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Exercise 4.36 states that the di... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Exercise 4.36 states that the distribution of speeds of cars traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. The speed limit on this stretch of the I-5 is 70 miles/hour. (a) A highway patrol officer is hidden on the side of the freeway. What is the probability that 5 cars pass and none are speeding? Assume that the speeds of the cars are independent of each other. (b) On average, how many cars would the highway patrol officer expect to watch until the first car that is speeding? What is the standard deviation of the number of cars he would expect to watch?

Short Answer

Expert verified
(a) Probability is 0.0024; (b) Expected cars is 1.414, standard deviation is 0.766.

Step by step solution

01

Calculate Probability of a Car Not Speeding

First, we need to find the probability that a single car is not speeding. "Not speeding" means the car's speed is 70 miles/hour or less. To find this, we need to calculate the z-score for 70 miles/hour using the formula: \[ z = \frac{X - \mu}{\sigma} \] where \( X = 70 \) miles/hour, \( \mu = 72.6 \) miles/hour, and \( \sigma = 4.78 \) miles/hour.Substituting the values, we get: \[ z = \frac{70 - 72.6}{4.78} = \frac{-2.6}{4.78} \approx -0.543 \]Using a standard normal distribution table or a calculator, we find that the probability \( P(Z \leq -0.543) \approx 0.293 \). Thus, the probability that a single car is not speeding is about 0.293.
02

Calculate Probability That None of the Cars are Speeding

We now use the probability that a single car is not speeding to find the probability that all 5 cars are not speeding. Since the cars are independent, we multiply the probability of a single event: \[ P(\text{none speeding}) = (0.293)^5 \]Calculating this gives \( (0.293)^5 \approx 0.002375 \). Therefore, the probability that none of the 5 cars are speeding is approximately 0.0024.
03

Define and Use Geometric Distribution for First Speeding Car

The problem of finding how many cars the officer expects to see until the first speeding car can be modeled by a geometric distribution. The probability of observing a car that is speeding (i.e., faster than 70 mph) is the complement of the probability of not speeding, which is \( 1 - 0.293 = 0.707 \).The expected number of trials until the first success in a geometric distribution is given by \( E(X) = \frac{1}{p} \), where \( p \) is the probability of speeding. So, \( E(X) = \frac{1}{0.707} \approx 1.414 \).Thus, the highway officer expects to see about 1.414 cars on average until the first car that is speeding.
04

Calculate Standard Deviation for Number of Trials

The standard deviation for a geometric distribution is given by the formula: \( \sigma = \sqrt{\frac{1-p}{p^2}} \).Substitute \( p = 0.707 \) into the formula: \[ \sigma = \sqrt{\frac{1-0.707}{0.707^2}} \approx \sqrt{\frac{0.293}{0.499849}} \approx \sqrt{0.586805} \approx 0.766 \].Hence, the standard deviation of the number of cars until the first speeding car is about 0.766.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Normal Distribution
In statistics, a normal distribution is a continuous probability distribution that is symmetrical around its mean. Most values cluster around a central point and probabilities for values further away from the mean taper off equally in both directions.
The normal distribution is often depicted as a bell curve due to its shape. The key characteristics of this distribution are its mean, median, mode and its standard deviation.
  • The mean (average) indicates the peak of the bell curve, representing the most probable value.
  • The spread of data, or the amount of variation or dispersion of values, around the mean is defined by the standard deviation.
  • In a standard normal distribution, the mean is 0 and the standard deviation is 1.
In the provided exercise, the speed of cars follows a normal distribution, with a mean speed of 72.6 miles per hour and a standard deviation of 4.78 miles per hour. This helps in calculating probabilities of different speed ranges using the properties of the normal distribution.
Z-Score Calculation
A z-score is a numerical measurement that describes a value's relationship to the mean of a group of values. A z-score of 0 indicates the value is identical to the mean. A positive z-score indicates the value is above the mean, while a negative z-score indicates it is below the mean.
  • The z-score formula is given by: \[ z = \frac{X - \mu}{\sigma} \]where:
    • \(X\) is the value for which you want to find the z-score
    • \(\mu\) is the mean of the distribution
    • \(\sigma\) is the standard deviation
In the exercise, to find the probability that a car is not speeding (70 miles/hour or less), the z-score was calculated as \(-0.543\). This tells us how many standard deviations below the mean the speed limit is and helps find probabilities using the standard normal distribution table.
Geometric Distribution
The geometric distribution is a probability distribution that models the number of trials needed to get the first success in a series of independent Bernoulli trials – each having the same probability of success.
  • The probability \(p\) of getting the first success and \(1-p\) for failure are constant.
  • The expected number of trials until the first success is given by the formula: \[ E(X) = \frac{1}{p} \]
  • The standard deviation is expressed as: \[ \sigma = \sqrt{\frac{1-p}{p^2}} \]
In the scenario, speeding is considered a 'success' (though undesirable in real life), and this setup helps determine the number of cars expected before observing the first speeding car. With a speeding probability of approximately 0.707, the calculation aids in understanding real-world probabilities.
Expected Value
The expected value is a central concept in probability theory, providing a measure of the center of a probability distribution. It represents the average outcome one would anticipate from an experiment if it were repeated under the same conditions.
  • In equations, the expected value is often represented by \(E(X)\), where \(X\) is the random variable.
  • For a geometric distribution, the expected value formula is: \[ E(X) = \frac{1}{p} \]indicating how many trials to expect before the first "success".
In the exercise, the expected number of cars until one is speeding is about 1.414. This means the highway patrol officer would, on average, expect to see slightly more than one car before witnessing speeding.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means the values tend to be close to the mean, while a high standard deviation indicates a wider spread of values.
  • Standard deviation is symbolized by \(\sigma\) in equations.
  • For a normal distribution, it quantifies how spread out the values are around the mean.
  • In the context of a geometric distribution, it's calculated using the formula: \[ \sigma = \sqrt{\frac{1-p}{p^2}} \]
In the current problem, the standard deviation for the number of cars expected until a speeding one is observed is calculated as approximately 0.766. This provides insight into the variability of the number of cars the officer may expect to monitor before catching a speeder.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Defective rate. A machine that produces a special type of transistor (a component of computers) has a \(2 \%\) defective rate. The production is considered a random process where each transistor is independent of the others. (a) What is the probability that the \(10^{\text {th }}\) transistor produced is the first with a defect? (b) What is the probability that the machine produces no defective transistors in a batch of \(100 ?\) (c) On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation? (d) Another machine that also produces transistors has a \(5 \%\) defective rate where each transistor is produced independent of the others. On average how many transistors would you expect to be produced with this machine before the first with a defect? What is the standard deviation? (e) Based on your answers to parts (c) and (d), how does increasing the probability of an event affect the mean and standard deviation of the wait time until success?4.14 Defective rate. A machine that produces a special type of transistor (a component of computers) has a \(2 \%\) defective rate. The production is considered a random process where each transistor is independent of the others. (a) What is the probability that the \(10^{\text {th }}\) transistor produced is the first with a defect? (b) What is the probability that the machine produces no defective transistors in a batch of \(100 ?\) (c) On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation? (d) Another machine that also produces transistors has a \(5 \%\) defective rate where each transistor is produced independent of the others. On average how many transistors would you expect to be produced with this machine before the first with a defect? What is the standard deviation? (e) Based on your answers to parts (c) and (d), how does increasing the probability of an event affect the mean and standard deviation of the wait time until success?

Sampling at school. For a sociology class project you are asked to conduct a survey on 20 students at your school. You decide to stand outside of your dorm's cafeteria and conduct the survey on a random sample of 20 students leaving the cafeteria after dinner one evening. Your dorm is comprised of \(45 \%\) males and \(55 \%\) females. (a) Which probability model is most appropriate for calculating the probability that the \(4^{\text {th }}\) person you survey is the \(2^{\text {nd }}\) female? Explain. (b) Compute the probability from part (a). (c) The three possible scenarios that lead to \(4^{\text {th }}\) person you survey being the \(2^{\text {nd }}\) female are $$ \\{M, M, F, F\\},\\{M, F, M, F\\},\\{F, M, M, F\\} $$ One common feature among these scenarios is that the last trial is always female. In the first three trials there are 2 males and 1 female. Use the binomial coefficient to confirm that there are 3 ways of ordering 2 males and 1 female. (d) Use the findings presented in part (c) to explain why the formula for the coefficient for the negative binomial is \(\left(\begin{array}{c}n-1 \\\ k=1\end{array}\right)\) while the formula for the binomial coefficient is \(\left(\begin{array}{l}n \\ k\end{array}\right)\).

4.17 Underage drinking, Part I. Data collected by the Substance Abuse and Mental Health Services Administration (SAMSHA) suggests that \(69.7 \%\) of \(18-20\) year olds consumed alcoholic beverages in any given year. (a) Suppose a random sample of ten \(18-20\) year olds is taken. Is the use of the binomial distribution appropriate for calculating the probability that exactly six consumed alcoholic beverages? Explain. (b) Calculate the probability that exactly 6 out of 10 randomly sampled 18- 20 year olds consumed an alcoholic drink. (c) What is the probability that exactly four out of ten \(18-20\) year olds have not consumed an alcoholic beverage? (d) What is the probability that at most 2 out of 5 randomly sampled \(18-20\) year olds have consumed alcoholic beverages? (e) What is the probability that at least 1 out of 5 randomly sampled 18-20 year olds have consumed alcoholic beverages?

LA weather, Part II. Exercise 4.7 states that average daily high temperature in June in LA is \(77^{\circ} \mathrm{F}\) with a standard deviation of \(5^{\circ} \mathrm{F}\), and it can be assumed that they to follow a normal distribution. We use the following equation to convert \({ }^{\circ} \mathrm{F}\) (Fahrenheit) to \({ }^{\circ} \mathrm{C}\) (Celsius): $$ C=(F-32) \times \frac{5}{9} $$ (a) Write the probability model for the distribution of temperature in \({ }^{\circ} \mathrm{C}\) in June in LA. (b) What is the probability of observing a \(28^{\circ} \mathrm{C}\) (which roughly corresponds to \(83^{\circ} \mathrm{F}\) ) temperature or higher in June in LA? Calculate using the \({ }^{\circ} \mathrm{C}\) model from part (a). (c) Did you get the same answer or different answers in part (b) of this question and part (a) of Exercise \(4.7 ?\) Are you surprised? Explain. (d) Estimate the IQR of the temperatures (in \({ }^{\circ} \mathrm{C}\) ) in June in LA.

Exploring permutations. The formula for the number of ways to arrange \(n\) objects is \(n !=n \times(n-\) 1) \(\times \cdots \times 2 \times 1\). This exercise walks you through the derivation of this formula for a couple of special cases. A small company has five employees: Anna, Ben, Carl, Damian, and Eddy. There are five parking spots in a row at the company, none of which are assigned, and each day the employees pull into a random parking spot. That is, all possible orderings of the cars in the row of spots are equally likely. (a) On a given day, what is the probability that the employees park in alphabetical order? (b) If the alphabetical order has an equal chance of occurring relative to all other possible orderings, how many ways must there be to arrange the five cars? (c) Now consider a sample of 8 employees instead. How many possible ways are there to order these 8 employees' cars?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.