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Chapter 3: Problem 11

An automobile service facility specializing in engine tune-ups knows that \(45 \%\) of all tune-ups are done on four-cylinder automobiles, \(40 \%\) on six- cylinder automobiles, and \(15 \%\) on eight-cylinder automobiles. Let \(X=\) the number of cylinders on the next car to be tuned. a. What is the pmf of \(X\) ? b. Draw both a line graph and a probability histogram for the pmf of part (a). c. What is the probability that the next car tuned has at least six cylinders? More than six cylinders?

Short Answer

Expert verified
a) The pmf is given by \(P(X = 4) = 0.45\), \(P(X = 6) = 0.40\), \(P(X = 8) = 0.15\). b) Use a line graph and histogram. c) \(P(X \geq 6) = 0.55\); \(P(X > 6) = 0.15\).

Step by step solution

01

Define the Probability Mass Function (pmf)

The problem gives us the probabilities for each type of automobile (four, six, and eight-cylinder cars). These probabilities form the pmf of the random variable \(X\). Let \(X\) be a random variable representing the number of cylinders on the next car. Based on the problem's data, we have:\[P(X = 4) = 0.45, \quad P(X = 6) = 0.40, \quad P(X = 8) = 0.15\]This defines the pmf: \(P_X(x) = \begin{cases} 0.45, & \text{if } x = 4 \0.40, & \text{if } x = 6 \0.15, & \text{if } x = 8 \end{cases}\)
02

Draw the Line Graph and Probability Histogram

To graph the pmf, we plot the probabilities for each value of \(X\). 1. **Line graph**: Plot points at \((4, 0.45)\), \((6, 0.40)\), and \((8, 0.15)\). Connect these points with straight lines.2. **Probability histogram**: Draw bars centered on each of \(4, 6, 8\) with heights corresponding to their probabilities. The width of each bar should be consistent (usually 1 unit wide).This visual representation helps in understanding the distribution of probabilities for the number of cylinders.
03

Calculate Probability for At Least Six Cylinders

To find the probability that the next car has at least six cylinders, we calculate the sum of probabilities for six and eight-cylinder cars:\[P(X \geq 6) = P(X = 6) + P(X = 8) = 0.40 + 0.15 = 0.55\]
04

Calculate Probability for More Than Six Cylinders

To find the probability of more than six cylinders, we consider only the eight-cylinder cars:\[P(X > 6) = P(X = 8) = 0.15\]

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Key Concepts

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

Discrete Probability Distribution
In statistics, a discrete probability distribution describes the probability of occurrence of each value of a discrete random variable. Unlike continuous distributions, discrete distributions only have specific, countable outcomes. In this problem, the discrete random variable \(X\) represents the number of cylinders in a car. Here, the possible values for \(X\) are 4, 6, and 8. These values directly correspond to the cylinder counts in the automobiles. Each specific count or outcome has an associated probability:
  • Four-cylinder: Probability \(P(X=4) = 0.45\)
  • Six-cylinder: Probability \(P(X=6) = 0.40\)
  • Eight-cylinder: Probability \(P(X=8) = 0.15\)
The probabilities within a discrete distribution must sum to 1, ensuring that they account for all possible outcomes. In this case, they do: \(0.45 + 0.40 + 0.15 = 1\). This distribution helps us understand the likelihood of encountering a certain type of engine in the next tune-up service.
Histograms in Probability
Histograms are one of the most common ways to visually represent a probability distribution. They are particularly useful in illustrating discrete probability distributions as they allow for an easy-to-understand visual snapshot of probabilities. For the given exercise, the histogram represents the pmf of the number of cylinders per automobile:
  • Each bar in the histogram is centered over each outcome (4, 6, and 8 in our case).
  • The height of each bar corresponds to the probability of each outcome. For example, a bar over 4 cylinders will have a height of 0.45.
This visual tool helps students more clearly understand the differences in probability between outcomes. By presenting probabilities as bar heights, it makes it obvious which outcomes are more likely or less likely compared to others. This complements the numerical pmf, offering an intuitive way to gauge data.
Probability Calculations
Probability calculations involve quantifying the likelihood of a particular event or set of events occurring, using the given probability distribution. In this problem, we were tasked to determine the probabilities of different cylinder conditions beyond the basic values provided.To find the probability of the next car having at least six cylinders, we summed the probabilities of six and eight-cylinder cars:\[P(X \geq 6) = P(X = 6) + P(X = 8) = 0.40 + 0.15 = 0.55\]Similarly, to calculate the probability of a car having more than six cylinders, we only consider the eight-cylinder vehicles:\[P(X > 6) = P(X = 8) = 0.15\]These calculations demonstrate how we can employ the given distribution to answer specific probability queries, enhancing the practical application of the probability mass function in real-world scenarios. This reinforces the concept that probabilities can be combined to solve broader questions beyond those initially provided in a pmf.

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Most popular questions from this chapter

An appliance dealer sells three different models of upright freezers having \(13.5,15.9\), and \(19.1\) cubic feet of storage space, respectively. Let \(X=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(X\) has pmf $$ \begin{array}{l|rrr} x & 13.5 & 15.9 & 19.1 \\ \hline p(x) & .2 & .5 & .3 \end{array} $$ a. Compute \(E(X), E\left(X^{2}\right)\), and \(V(X)\). b. If the price of a freezer having capacity \(X\) cubic feet is \(25 X-8.5\), what is the expected price paid by the next customer to buy a freezer? c. What is the variance of the price \(25 X-8.5\) paid by the next customer? d. Suppose that although the rated capacity of a freezer is \(X\), the actual capacity is \(h(X)=X-.01 X^{2}\). What is the expected actual capacity of the freezer purchased by the next customer?

Define a function \(p(x ; \lambda, \mu)\) by $$ p(x ; \lambda, \mu)=\left\\{\begin{array}{cl} \frac{1}{2} e^{-\lambda} \frac{\lambda^{x}}{x !}+\frac{1}{2} e^{-\mu \frac{\mu^{x}}{x !}} & x=0,1,2, \ldots \\ 0 & \text { otherwise } \end{array}\right. $$ a. Show that \(p(x ; \lambda, \mu)\) satisfies the two conditions necessary for specifying a pmf, [Note: If a firm employs two typists, one of whom makes typographical errors at the rate of \(\lambda\) per page and the other at rate \(\mu\) per page and they each do half the firm's typing, then \(p(x ; \lambda, \mu)\) is the pmf of \(X=\) the number of errors on a randomly chosen page.] b. If the first typist (rate \(\lambda\) ) types \(60 \%\) of all pages, what is the pmf of \(X\) of part (a)? c. What is \(E(X)\) for \(p(x ; \lambda, \mu)\) given by the displayed expression? d. What is \(\sigma^{2}\) for \(p(x ; \lambda, \mu)\) given by that expression?

Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerator is running. Suppose that 7 of these refrigerators have a defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order, let \(X\) be the number among the first 6 examined that have a defective compressor. Compute the following: a. \(P(X=5)\) b. \(P(X \leq 4)\) c. The probability that \(X\) exceeds its mean value by more than 1 standard deviation. d. Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If \(X\) is the number among 15 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) \(P(X \leq 5)\) than to use the hypergeometric pmf.

The article "Reliability-Based Service-Life Assessment of Aging Concrete Structures" (J. Structural Engr., 1993: \(1600-1621\) ) suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Suppose the mean time between occurrences of loads is .5 year. a. How many loads can be expected to occur during a 2year period? b. What is the probability that more than five loads occur during a 2-year period? c. How long must a time period be so that the probability of no loads occurring during that period is at most .1?

The College Board reports that \(2 \%\) of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities (Los Angeles Times, July 16, 2002). Consider a random sample of 25 students who have recently taken the test. a. What is the probability that exactly 1 received a special accommodation? b. What is the probability that at least 1 received a special accommodation? c. What is the probability that at least 2 received a special accommodation? d. What is the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated? e. Suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed \(4.5\) hours. What would you expect the average time allowed the 25 selected students to be?
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