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Chapter 6: Problem 30

A particular professor never dismisses class early. Let \(x\) denote the amount of additional time (in minutes) that elapses before the professor dismisses class. Suppose that \(x\) has a uniform distribution on the interval from 0 to 10 minutes. The density curve is shown in the following figure: a. What is the probability that at most 5 minutes elapse before dismissal? b. What is the probability that between 3 and 5 minutes elapse before dismissal?

Short Answer

Expert verified
The probability density function (pdf) for the uniform distribution in this case is \(f(x) = \frac{1}{10}\). Calculating the area under the pdf curve, we find that there is a 50% (\(\frac{1}{2}\)) probability that at most 5 minutes elapse before dismissal, and a 20% (\(\frac{1}{5}\)) probability that between 3 and 5 minutes elapse before dismissal.

Step by step solution

01

Understand the uniform distribution

A uniform distribution is a type of continuous probability distribution where all outcomes are equally likely within a given interval. In this case, the interval is from 0 to 10 minutes. For a uniform distribution, the probability density function (pdf) is given by: \[f(x) = \frac{1}{b - a}\] Where \(a\) and \(b\) are the endpoints of the interval, in this case, \(a = 0\) and \(b = 10\). Let's find the value of the pdf.
02

Calculate the probability density function (pdf)

Using the formula for the uniform distribution, we can calculate the pdf for the given interval: \[f(x) = \frac{1}{b - a} = \frac{1}{10 - 0} = \frac{1}{10}\]
03

Calculate the probability that at most 5 minutes elapse before dismissal

To find the probability that at most 5 minutes elapse before dismissal, we need to calculate the integral of the pdf from \(x = 0\) to \(x = 5\). For a uniform distribution, the integral of the pdf is equal to the area of the rectangle formed under the pdf curve from \(x = 0\) to \(x = 5\). The area of a rectangle is given by multiplying its width and height, so: \[P(x \le 5) = f(x) \cdot (5 - 0) = \frac{1}{10} \cdot 5 = \frac{1}{2}\] Thus, there is a 50% probability that at most 5 minutes elapse before dismissal.
04

Calculate the probability that between 3 and 5 minutes elapse before dismissal

To find the probability that between 3 and 5 minutes elapse before dismissal, we need to calculate the integral of the pdf from \(x = 3\) to \(x = 5\). For a uniform distribution, the integral of the pdf is equal to the area of the rectangle formed under the pdf curve from \(x = 3\) to \(x = 5\). The area of a rectangle is given by multiplying its width and height, so: \[P(3 \le x \le 5) = f(x) \cdot (5 - 3) = \frac{1}{10} \cdot 2 = \frac{1}{5}\] Thus, there is a 20% probability that between 3 and 5 minutes elapse before dismissal.

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Key Concepts

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

Probability Density Function
The probability density function (pdf) is a key concept when dealing with continuous probability distributions. A pdf helps to specify the likelihood of a random variable falling within a particular range of values, as opposed to taking on any one specific value. This is because, in a continuous distribution, the probability of the random variable exactly equaling any one particular value is zero. Instead, we look at intervals and the probability that a random variable falls within an interval.

The pdf is represented as a curve on a graph, and the area under the curve within a specified interval represents the probability of the variable falling within that interval. For a uniform distribution, like the one in the exercise where additional class time can be any value from 0 to 10 minutes, the pdf is a horizontal line since the probability is constant across the interval. Mathematically, a pdf must satisfy two conditions: the probability is always non-negative, and the total area under the curve must be equal to 1, indicating that the sum of all possible outcomes' probabilities is certain.
Continuous Probability Distribution
A continuous probability distribution differs significantly from a discrete probability distribution in that it deals with outcomes that can take on any value within an interval or collection of intervals. With continuous distributions, probabilities are assigned to ranges of values rather than specific outcomes. The uniform distribution described in the exercise is a classic example of a continuous distribution.

Continuous distributions are described by their probability density functions, such as the uniform probability density function in the exercise. The uniform distribution is quite straightforward since every outcome in the interval is equally likely, making it a useful starting point for understanding more complex distribution types. Continuity implies that between any two points within the distribution's range, no matter how close, there are infinitely many potential values the random variable might assume.
Integrating Probability Density Functions
Integrating the probability density function is the process of finding the total probability that a random variable falls within a certain range. This method is the continuous counterpart to summing probabilities in discrete distributions. The integral of the pdf over the entire range of possible values is 1, which corresponds to the certain event of the random variable taking on some value within that range.

To find the probability corresponding to a specific interval, as in the exercise scenarios for 0 to 5 minutes and 3 to 5 minutes, you calculate the integral over that interval. In certain cases, such as the uniform distribution seen here, integrating the pdf simplifies to calculating the area of a rectangle. However, with more complex distributions, integration might require applying more advanced calculus techniques. Understanding this process is crucial, for it allows us to make predictions and understand the likely outcomes of a continuous random variable.

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Most popular questions from this chapter

A point is randomly selected on the surface of a lake that has a maximum depth of 100 feet. Let \(x\) be the depth of the lake at the randomly chosen point. What are possible values of \(x\) ? Is \(x\) discrete or continuous?

Suppose that instead of three shirts, each participant was asked to choose among four shirts and that the process was performed five times. If a person can't identify her roommate by smell and is just picking a shirt at random, then \(x=\) number of correct identifications is a binomial random variable with \(n=5\) and \(p=\frac{1}{4}\). 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. (Hint: See Example 6.27.) c. Construct a histogram displaying the probability distribution of \(x\).

The paper "Risk Behavior, Decision Making, and Music Genre in Adolescent Males" (Marshall University, May 2009) examined the effect of type of music playing and performance on a risky, decision-making task. a. Participants in the study responded to a questionnaire used to assign a risk behavior score. Risk behavior scores (from a graph in the paper) for 15 participants follow. Use these data to construct a normal probability plot (the normal scores for a sample of size 15 appear in the previous exercise). \(\begin{array}{llllllll}102 & 105 & 113 & 120 & 125 & 127 & 134 & 135 \\ 139 & 141 & 144 & 145 & 149 & 150 & 160 & \end{array}\) b. Participants also completed a positive and negative affect scale (PANAS) designed to measure emotional response to music. PANAS values (from a graph in the paper) for 15 participants follow. Use these data to construct a normal probability plot (the normal scores for a sample of size 15 appear in the previous exercise). \(\begin{array}{llllllll}36 & 40 & 45 & 47 & 48 & 49 & 50 & 52 \\ 53 & 54 & 56 & 59 & 61 & 62 & 70 \end{array}\) c. The author of the paper stated that it was reasonable to consider both risk behavior scores and PANAS scores to be approximately normally distributed. Do the normal probability plots from Parts (a) and (b) support this conclusion? Explain.

6.99 A symptom validity test (SVT) is sometimes used to confirm diagnosis of psychiatric disorders. The paper "Developing a Symptom Validity Test for Posttraumatic Stress Disorder: Application of the Binomial Distribution" (Journal of Anxiety Disorders [2008]: 1297-1302) investigated the use of SVTs in the diagnosis of post-traumatic stress disorder. One SVT proposed is a 60 -item test (called the MENT test), where each item has only a correct or incorrect response. The MENT test is designed so that responses to the individual questions can be considered independent of one another. For this reason, the authors of the paper believe that the score on the MENT test can be viewed as a binomial random variable with \(n=60 .\) The MENT test is designed to help in distinguishing fictitious claims of post-traumatic stress disorder. The items on the test are written so that the correct response to an item should be relatively obvious, even to people suffering from stress disorders. Researchers have found that a patient with a fictitious claim of stress disorder will try to "fake" the test, and that the probability of a correct response to an item for these patients is 0.7 (compared to 0.96 for other patients). The authors used a normal approximation to the binomial distribution with \(n=60\) and \(p=0.7\) to calculate various probabilities of interest, where \(x=\) number of correct responses on the MENT test for a patient who is trying to fake the test. a. Verify that it is appropriate to use a normal approximation to the binomial distribution in this situation. b. Approximate the following probabilities: $$ \text { i. } \quad P(x=42) $$ ii. \(P(x<42)\) $$ \text { iii. } P(x \leq 42) $$ c. Explain why the probabilities calculated in Part (b) are not all equal. d. The authors calculated the exact binomial probability of a score of 42 or less for someone who is not faking the test. Using \(p=0.96,\) they found $$ P(x \leq 42)=.000000000013 $$ Explain why the authors calculated this probability using the binomial formula rather than using a normal approximation. e. The authors propose that someone who scores 42 or less on the MENT exam is faking the test. Explain why this is reasonable, using some of the probabilities from Parts (b) and (d) as justification.

Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboard can be approximated by a normal curve with mean 60 wpm and standard deviation 15 wpm ("The Effects of Split Keyboard Geometry on Upper Body Postures, Ergonomics [2009]: 104-111). a. What is the probability that a randomly selected typist's speed is at most 60 wpm? Less than 60 wpm? b. What is the probability that a randomly selected typist's speed is between 45 and 90 wpm? c. Would you be surprised to find a typist in this population whose speed exceeded 105 wpm? d. Suppose that two typists are independently selected. What is the probability that both their speeds exceed 75 wpm? e. Suppose that special training is to be made available to the slowest \(20 \%\) of the typists. What typing speeds would qualify individuals for this training? (Hint: See Example \(6.23 .)\)
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