/*! 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 104 A company that manufactures muff... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A company that manufactures mufflers for cars offers a lifetime warranty on its products, provided that ownership of the car does not change. Suppose that only \(20 \%\) of its mufflers are replaced under this warranty. a. In a random sample of 400 purchases, what is the approximate probability that between 75 and 100 (inclusive) mufflers are replaced under warranty? b. Among 400 randomly selected purchases, what is the probability that at most 70 mufflers are ultimately replaced under warranty? c. If you were told that fewer than 50 among 400 randomly selected purchases were ever replaced under warranty, would you question the \(20 \%\) figure? Explain.

Short Answer

Expert verified
a. The probability that between 75 and 100 mufflers are replaced under warranty is approximately 0.50. b. The probability that at most 70 mufflers are replaced under warranty is about 0.118. c. Yes, it would seem reasonable to question the \(20%\) figure if fewer than 50 mufflers out of 400 were replaced under warranty. This number of replacements is significantly less than the expected 80 replacements.

Step by step solution

01

Identify the Distribution

Identify that it's a problem related to Binomial Distribution, since there are only two possible outcomes (muffler getting replaced under warranty or not) and a fixed number of independent trials (400 in this case). In this problem, the success probability \(p\) is the chance of a muffler being replaced under warranty, which is \(0.20\), and our sample size \(n\) is 400.
02

Calculate Probability for Scenario (a)

Use the normal approximation to the binomial distribution, given that both \(np\) and \(n(1-p)\) are greater than 5, which is true in our case. We can approximate our binomial distribution to a normal distribution with mean \(μ = np = 400*0.20 = 80\) and standard deviation \(σ = \sqrt{np(1-p)} =\sqrt{400*0.20*0.80} = 8\). Now, compute the standardized scores (z-scores) for 74.5 (we subtract 0.5 due to continuity correction) and 100.5 (we add 0.5) and find the corresponding probabilities from the standard normal distribution: \(Z_{74.5} = \frac{74.5 - μ}{σ}=\frac{74.5 - 80}{8}=-0.6875\), \(Z_{100.5} = \frac{100.5 - 80}{8}=2.5625\). By looking up these z-scores in the standard normal table and subtracting the probabilities, the approximate probability is \(P(75 ≤ X ≤ 100) = P(Z_{74.5} ≤ Z ≤ Z_{100.5}) = P(-0.6875 ≤ Z ≤ 2.5625) \approx 0.7459 - 0.2461 = 0.4998\).
03

Calculate Probability for Scenario (b)

Calculate the z-score for 70.5 using the formula \(Z_{70.5} =\frac{70.5 - 80}{8}= -1.1875\) and look this up in the standard normal table.The probability that at most 70 mufflers get replaced under warranty is \(P(X ≤ 70) = P(Z ≤ Z_{70.5}) \approx 0.1175 \).
04

Analysis for Scenario (c)

The fact that fewer than 50 among 400 randomly selected purchases were ever replaced under warranty does seem to suggest questioning the \(20%\) figure. This is because the given 50 replacements are significantly less than the expected 80 replacements (\(20%\) of 400) under warranty.

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.

Understanding Probability
Probability is a measure of the likelihood that a particular event will occur. It's quantified as a number between 0 and 1, where 0 indicates the event is impossible to occur, and 1 denotes certainty. When dealing with probability, the sum of probabilities for all possible outcomes must equal 1.

In the exercise, the probability concept is applied to calculate the chances of a certain number of mufflers being replaced under warranty out of 400 purchases. When we talk about 'the probability that between 75 and 100 mufflers are replaced', we're seeking to find the likelihood that a random event falls within a certain range.

Furthermore, in this context, the terms 'at most' and 'fewer than' are used to denote probabilities of events not exceeding or falling short of a certain threshold. For instance, 'the probability that at most 70 mufflers are replaced' reflects how often 70 or fewer replacements could occur by random chance, according to our assumptions about muffler replacement rates.
Normal Approximation of the Binomial Distribution
The normal approximation to the binomial distribution is a method used when dealing with a large number of trials. It simplifies calculations considerably and is applicable when both the minimum expected successes (np) and failures (n(1-p)) are greater than 5.

In our exercise, because the sample size (n=400) is relatively large and the probability of success (p=0.20) meets the criteria, the binomial distribution can be approximated using a normal distribution. This means instead of dealing with discrete successful trials, we now treat the variable as continuous, allowing us to use z-scores and the properties of the standard normal distribution to find probabilities over a range.
Standard Deviation and Its Role
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that most numbers are close to the mean (average) of the dataset, while a high standard deviation indicates the numbers are more spread out.

The formula for the standard deviation of a binomial distribution is \( \sigma = \sqrt{np(1-p)} \), where \(n\) is the number of trials and \(p\) is the probability of success. In the context of the muffler scenario, it quantifies the variability in the number of mufflers getting replaced under warranty out of 400, giving us the spread of our bell-shaped normal approximation curve.
z-Score: A Key Concept in Statistics
A z-score indicates how many standard deviations an element is from the mean. It is a pivotal concept when analyzing data in statistics, especially under the normal distribution umbrella. To calculate a z-score, the formula is \( Z = \frac{X - \mu}{\sigma} \), where \(X\) is the value in question, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

In this exercise, we evaluate z-scores to find the probability that a value lies within a certain range. For example, the z-score calculation for 74.5 helps us understand how likely it is that at least 75 mufflers (after continuity correction) are replaced under warranty. The z-score transformation allows us to use standard normal distribution tables, simplifying the process of finding probabilities for complex binomial problems.

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

An author has written a book and submitted it to a publisher. The publisher offers to print the book and gives the author the choice between a flat payment of \(\$ 10,000\) and a royalty plan. Under the royalty plan the author would receive \(\$ 1\) for each copy of the book sold. The author thinks that the following table gives the probability distribution of the variable \(x=\) the number of books that will be sold: \(\begin{array}{lrrrr}x & 1000 & 5000 & 10,000 & 20,000 \\ p(x) & .05 & .30 & .40 & .25\end{array}\) Which payment plan should the author choose? Why?

A box contains five slips of paper, marked \(\$ 1, \$ 1\), \(\$ 1, \$ 10\), and \(\$ 25 .\) The winner of a contest selects two slips of paper at random and then gets the larger of the dollar amounts on the two slips. Define a random variable \(w\) by \(w=\) amount awarded. Determine the probability distribution of \(w\). (Hint: Think of the slips as numbered \(1,2,3,4\), and 5, so that an outcome of the experiment consists of two of these numbers.)

A multiple-choice exam consists of 50 questions. Each question has five choices, of which only one is correct. Suppose that the total score on the exam is computed as $$ y=x_{1}-\frac{1}{4} x_{2} $$ where \(x_{1}=\) number of correct responses and \(x_{2}=\) number of incorrect responses. (Calculating a total score by subtracting a term based on the number of incorrect responses is known as a correction for guessing and is designed to discourage test takers from choosing answers at random.) a, It can be shown that if a totally unprepared student answers all 50 questions by just selecting one of the five answers at random, then \(\mu_{x_{1}}=10\) and \(\mu_{x_{2}}=40\). What is the mean value of the total score, \(y\) ? Does this surprise you? Explain. b. Explain why it is unreasonable to use the formulas given in this section to compute the variance or standard deviation of \(y\).

Example \(7.18\) described a study in which a person was asked to determine which of three t-shirts had been worn by her roommate by smelling the shirts ("Sociochemosensory and Emotional Functions," Psychological Science [2009]: \(1118-1123\) ). Suppose that instead of three shirts, each participant was asked to choose among four shirts and that the process was repeated five times. Then, assuming that the participant is choosing at random, \(x=\) number of correct identifications is a binomial random variable with \(n=5\) and \(p=1 / 4\). a. What are the possible values of \(x\) ? b. For each possible value of \(x\), find the associated probability \(p(x)\) and display the possible \(x\) values and \(p(x)\) values in a table. c. Construct a probability histogram for the probability distribution of \(x\).

Emissions of nitrogen oxides, which are major constituents of \(\mathrm{smog}\), can be modeled using a normal distribution. Let \(x\) denote the amount of this pollutant emitted by a randomly selected vehicle (in parts per billion). The distribution of \(x\) can be described by a normal distribution with \(\mu=1.6\) and \(\sigma=0.4\). Suppose that the EPA wants to offer some sort of incentive to get the worst polluters off the road. What emission levels constitute the worst \(10 \%\) of the vehicles?

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.