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Chapter 3: Problem 23

A consumer organization that evaluates new automobiles customarily reports the number of major defects in each car examined. Let \(X\) denote the number of major defects in a randomly selected car of a certain type. The cdf of \(X\) is as follows: $$ F(x)= \begin{cases}0 & x<0 \\ .06 & 0 \leq x<1 \\ .19 & 1 \leq x<2 \\ .39 & 2 \leq x<3 \\ .67 & 3 \leq x<4 \\ .92 & 4 \leq x<5 \\ .97 & 5 \leq x<6 \\\ 1 & 6 \leq x\end{cases} $$ Calculate the following probabilities directly from the cdf: a. \(p(2)\), that is, \(P(X=2)\) b. \(P(X>3)\) c. \(P(2 \leq X \leq 5)\) d. \(P(2

Short Answer

Expert verified
a. 0.20, b. 0.33, c. 0.58, d. 0.53

Step by step solution

01

Calculate P(X=2)

The probability of a discrete random variable taking a specific value is the difference between the cumulative distribution function (CDF) values at that value and the previous integer. Therefore, the probability that a car has exactly 2 major defects, \( P(X=2) \), is calculated by finding \( F(2) - F(1) \).From the given CDF, \( F(2) = 0.39 \) and \( F(1) = 0.19 \).So, \[ P(X=2) = F(2) - F(1) = 0.39 - 0.19 = 0.20 \].
02

Calculate P(X>3)

To find the probability that a random variable is greater than a certain value, we subtract the CDF value at that point from 1. Here, we want \( P(X>3) \).We know \( F(3) = 0.67 \), so:\[ P(X > 3) = 1 - F(3) = 1 - 0.67 = 0.33 \]
03

Calculate P(2 鈮� X 鈮� 5)

The probability that \( X \) lies in an interval \( [a, b] \) is given by the difference \( F(b+1) - F(a) \). Here, we need \( P(2 \leq X \leq 5) \).Since \( F(5) = 0.97 \) and \( F(2) = 0.39 \), we calculate:\[ P(2 \leq X \leq 5) = F(5+1) - F(2) = F(6) - F(2) = 1 - 0.39 = 0.58 \]
04

Calculate P(2 < X < 5)

For the probability of \( X \) between non-inclusive boundaries \( (a, b) \), we use \( F(b) - F(a+1) \). Therefore, the probability \( P(2 < X < 5) \) is found as follows.Given \( F(4) = 0.92 \) and \( F(3) = 0.67 \):\[ P(2 < X < 5) = F(5) - F(3) = 0.92 - 0.39 = 0.53 \]

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Key Concepts

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

Probability Calculations
Understanding probability calculations is key to grasping how we analyze the likelihood of different outcomes of a random variable. When given the cumulative distribution function (CDF), like in our exercise, we can extract probabilities by focusing on the difference in values between specific points on the CDF.

Let's explore this further:
  • For a discrete random variable, the probability of it taking a specific value, say at point 2, is calculated by finding the difference in the CDF values at that point and the previous integer. This is shown in the calculation of \( P(X=2) \) where \( F(2) - F(1) \) gives us the probability of 0.20.
  • To calculate the probability of \( X \) being greater than a certain value, say 3, you subtract the CDF value at that point from 1. Therefore, \( P(X>3) \) is calculated as \( 1 - F(3) \), resulting in 0.33.
  • Probabilities for intervals, like \( P(2 \leq X \leq 5) \), use the difference between distributing values at the endpoints, resulting in 0.58 in our example.
These techniques provide a systematic way to deduce probabilities from a CDF, enhancing our understanding of the behavior of discrete random variables.
Discrete Random Variable
A discrete random variable, such as the one in this exercise, is a variable that can take on a finite or countably infinite number of distinct values. In our context, it represents the number of major defects in a car. Each potential outcome, or number of defects, is associated with a probability derived from the CDF.

Here鈥檚 what you need to understand about discrete random variables:
  • The outcomes are countable. For instance, a car can have 0, 1, 2, etc., defects, but not 1.5 defects.
  • Each specific outcome has a specific probability, calculated by finding the change in CDF values at key integer points.
  • The sum of the probabilities over all possible outcomes is always 1.
In our exercise, the values of the CDF at specific points reveal the probabilities of a car having a precise number of defects. This helps in quality studies and predictions.
Probability Distribution
A probability distribution describes how the probabilities are distributed over the values that the random variable can take. For a discrete random variable like in our exercise, the CDF provides a way to map each potential outcome to its cumulative probability.

Key points about probability distributions include:
  • The cumulative distribution function (CDF) aids in evaluating the likelihood that a random variable will assume a value less than or equal to a certain level.
  • With it, we can determine any individual or range of outcomes' probabilities. For instance, using the CDF to find \( P(2 < X < 5) \) as 0.53.
  • Knowing how to translate CDF values into probability insights involves understanding that the steeper the inclination of the function at any point, the more concentrated the probability mass.
Understanding this distribution allows analysts to make informed predictions and decisions based on potential occurrences.

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Most popular questions from this chapter

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?

Suppose that trees are distributed in a forest according to a two-dimensional Poisson process with parameter \(\alpha\), the expected number of trees per acre, equal to 80 . a. What is the probability that in a certain quarter-acre plot, there will be at most 16 trees? b. If the forest covers 85,000 acres, what is the expected number of trees in the forest? c. Suppose you select a point in the forest and construct a circle of radius \(.1\) mile. Let \(X=\) the number of trees within that circular region. What is the pmf of \(X\) ? [Hint: 1 sq mile \(=640\) acres. \(]\)

An individual who has automobile insurance from a certain company is randomly selected. Let \(Y\) be the number of moving violations for which the individual was cited during the last 3 years. The pmf of \(Y\) is $$ \begin{array}{l|cccc} y & 0 & 1 & 2 & 3 \\ \hline p(y) & .60 & .25 & .10 & .05 \end{array} $$ a. Compute \(E(Y)\). b. Suppose an individual with \(Y\) violations incurs a surcharge of \(\$ 100 Y^{2}\). Calculate the expected amount of the surcharge.

A small market orders copies of a certain magazine for its magazine rack each week. Let \(X=\) demand for the magazine, with pmf $$ \begin{array}{c|cccccc} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline p(x) & \frac{1}{15} & \frac{2}{15} & \frac{3}{15} & \frac{4}{15} & \frac{3}{15} & \frac{2}{15} \end{array} $$ Suppose the store owner actually pays \(\$ 1.00\) for each copy of the magazine and the price to customers is \(\$ 2.00\). If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine?

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?
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