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Chapter 14: Problem 25

Hot Spot. Hot Spot is a California lottery game. Players pick 1 to 10 Spots (sets of numbers, each from 1 to 80 ) that they want to play per draw. For example, if you select a 4 Spot, you play four numbers. The lottery draws 20 numbers, each from 1 to 80 . Your prize is based on how many of the numbers you picked match one of those selected by the lottery. The odds of winning depend on the number of Spots you choose to play. For example, the overall odds of winning some prize in 4 Spot is approximately \(0.256\). You decide to play the 4 Spot game and buy 5 tickets. Let \(X\) be the number of tickets that win some prize. a. \(X\) has a binomial distribution. What are \(n\) and \(p\) ? b. What are the possible values that \(X\) can take? c. Find the probability of each value of \(X\). Draw a probability histogram for the distribution of \(X\). (See Figure \(14.2\) on page \(\underline{331}\) for an example of a probability histogram.) d. What are the mean and standard deviation of this distribution? Mark the location of the mean on your histogram.

Short Answer

Expert verified
a. \( n = 5 \), \( p = 0.256 \). b. Possible values: 0, 1, 2, 3, 4, 5. c. Calculate probabilities using binomial formula, draw histogram. d. Mean: 1.28, Std. Dev: 0.977.

Step by step solution

01

Determine the Binomial Distribution Parameters

In a binomial distribution, there are two parameters to determine: the number of trials, denoted by \( n \), and the probability of success on each trial, denoted by \( p \). A trial in this context is buying a ticket. Thus, \( n = 5 \). Given that the overall probability of winning a prize with a 4 Spot ticket is \( p = 0.256 \), we identify these parameters as: \( n = 5 \) and \( p = 0.256 \).
02

Identify Possible Values of X

\( X \) represents the number of tickets that win some prize out of the 5 bought tickets. Since each trial (buying a ticket) can either result in a win or a loss, \( X \) can take on integer values from 0 to 5, inclusive. Thus, the possible values of \( X \) are \( 0, 1, 2, 3, 4, 5 \).
03

Calculate Probability of Each Value of X

The probability of exactly \( k \) successes (winning tickets) in \( n \) trials can be calculated using the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]For each value \( k = 0 \) to \( 5 \), substitute \( n = 5 \), \( p = 0.256 \), and calculate: - \( P(X = 0) = \binom{5}{0} (0.256)^0 (0.744)^5 \)- \( P(X = 1) = \binom{5}{1} (0.256)^1 (0.744)^4 \)- \( P(X = 2) = \binom{5}{2} (0.256)^2 (0.744)^3 \)- \( P(X = 3) = \binom{5}{3} (0.256)^3 (0.744)^2 \)- \( P(X = 4) = \binom{5}{4} (0.256)^4 (0.744)^1 \)- \( P(X = 5) = \binom{5}{5} (0.256)^5 (0.744)^0 \).
04

Draw the Probability Histogram

A probability histogram displays the probability of each outcome on the vertical axis, with each bar's height representing the probability for each value of \( X = 0, 1, 2, 3, 4, 5 \). Use the probabilities calculated in Step 3 to draw the histogram. Ensure that the x-axis is labeled with the values \( 0, 1, 2, 3, 4, 5 \) and the y-axis shows probabilities.
05

Calculate the Mean and Standard Deviation

The mean (expected value) of a binomial distribution is given by \( \mu = n \times p \). Given \( n = 5 \) and \( p = 0.256 \), the mean is \( \mu = 5 \times 0.256 = 1.28 \).The standard deviation is calculated using the formula \( \sigma = \sqrt{n \times p \times (1-p)} \). Substitute the values to get \( \sigma = \sqrt{5 \times 0.256 \times 0.744} \approx 0.977 \).Mark the mean (\( X = 1.28 \)) on your histogram by placing a vertical line at \( X = 1.28 \).

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Key Concepts

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

Probability Histogram
A probability histogram is a visual representation that showcases the likelihood of different outcomes in a probabilistic scenario. In essence, it resembles a bar graph where each bar corresponds to a possible outcome of the random variable. The height of each bar indicates the probability of that specific outcome occurring.

In our exercise involving the lottery game, each bar in the histogram will represent the probability of getting 0, 1, 2, 3, 4, or 5 winning tickets out of the total tickets purchased.

This helps us see at a glance which outcomes are more likely and which are less so. Such visual tools are particularly useful as they facilitate understanding complex probability distributions by transforming numbers into an easily interpretable format.
Expected Value
The expected value, often known as the mean, of a binomial distribution is the average outcome you expect after many trials. Mathematically, it is calculated as the product of the number of trials and the probability of success on each trial.

In our scenario, we calculate the expected value by multiplying the numbers of tickets bought ( =5) with the probability of winning with one ticket ( p=0.256). This gives us an expected value of 1.28.

Simply put, if you were to play this lottery game numerous times, on average, you would expect about 1.28 winning tickets per five tickets purchased. This number provides a central value around which the number of winning tickets will typically scatter.
Standard Deviation
Standard deviation is a measure of the dispersion or spread of a set of values. In probability, it tells us how much the values of the random variable deviate from the mean or expected value.

For a binomial distribution, the standard deviation is calculated using the formula: \[ \sigma = \sqrt{n \times p \times (1-p)} \] where \( n \) is the number of trials and \( p \) is the probability of success.

For our lottery exercise, the calculations result in a standard deviation of approximately 0.977.

This means that the number of winning lottery tickets, in our case, usually deviates from the mean of 1.28 by about 0.977. Understanding this concept helps anticipate the range of possible outcomes you might encounter.
Probability Calculation
Calculating probabilities in a binomial distribution involves determining the likelihood of a specific number of successes in a set number of trials. This is done using the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( n \) is the number of trials, \( k \) is the number of successes, and \( p \) is the probability of success for each trial.

In our scenario, it requires calculating the probability for each number (0 to 5) of winning tickets. For instance, to find the probability of winning with exactly 2 tickets, you plug the values into the formula: \( n=5 \), \( k=2 \), and \( p=0.256 \).

This formula accounts for all possible combinations of winning tickets and uses the concept of success and failure to gauge each scenario. Completing these calculations gives us a comprehensive view of all potential outcomes and their respective probabilities.

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Most popular questions from this chapter

Larry reads that half of all super jumbo eggs contain double yolks. So he always buys super jumbo eggs and uses two whenever he cooks. If eggs do or don't contain two yolks independently of each other, the number of eggs with double yolks when Larry uses two chosen at random has the distribution a. binomial with \(n=2\) and \(p=1 / 2\). b. binomial with \(n=2\) and \(p=1 / 3\). c. binomial with \(n=3\) and \(p=1 / 2\).

In a group of 10 college students, three are psychology majors. You choose three of the 10 students at random and ask their major. The distribution of the number of psychology majors you choose is a. binomial with \(n=10\) and \(p=0.3\). b. binomial with \(n=3\) and \(p=0.3\). c. not binomial. In a test for ESP (extrasensory perception), a subject is told that cards that the experimenter can see, but that the subject cannot see, contain either a star, a circle, a wave, or a square. As the experimenter looks at each of 4 cards in turn, the subject names the shape on the card. A subject who is just guessing has probability \(0.25\) of guessing correctly on each card. Questions \(14.16\) to \(\underline{14.18}\) use this information.

Checking for Survey Errors. One way of checking the effect of undercoverage, nonresponse, and other sources of error in a sample survey is to compare the sample with known facts about the population. About \(24 \%\) of the Canadian population is first generation-that is, they were born outside Canada. \(\underline{6}\) The number \(X\) of first-generation Canadians in random samples of 1500 persons should therefore vary with the binomial ( \(n=1500, p=0.24\) ) distribution. a. What are the mean and standard deviation of \(X\) ? b. Use the Normal approximation to find the probability that a sample will contain between 340 and 390 firstgeneration Canadians. Check that you can safely use the approximation.

Multiple-Choice Tests. Here is a simple probability model for multiple-choice tests. Suppose each student has probability \(p\) of correctly answering a question chosen at random from a universe of possible questions. (A strong student has a higher \(p\) than a weak student.) Answers to different questions are independent. a. Stacey is a good student for whom \(p=0.75\). Use the Normal approximation to find the probability that Stacey scores between \(70 \%\) and \(80 \%\) on a 100 -question test. b. If the test contains 250 questions, what is the probability that Stacey will score between \(70 \%\) and \(80 \%\) ? You see that Stacey's score on the longer test is more likely to be close to her "true score."

Does the Wall Street Culture Corrupt Bankers? Bank employees from a large international bank were recruited, and 67 were randomly assigned to a control group and the remaining 61 to a treatment group. All subjects first completed a short online survey. After answering some general filler questions, the members of the treatment group were asked seven questions about their professional background, such as "At which bank are you currently employed?" or "What is your function at this bank?" These are referred to as "identity priming" questions. The members of the control group were asked seven innocuous questions unrelated to their profession, such as "How many hours a week, on average, do you watch television?" After the survey, all subjects performed a coin tossing task that required tossing any coin 10 times and reporting the results online. They were told they would win \(\$ 20\) for each head tossed, for a maximum payoff of \(\$ 200\). Subjects were unobserved during the task, making it impossible to tell if a particular subject cheated. If the banking culture favors dishonest behavior, it was conjectured that it should be possible to trigger this behavior by reminding subjects of their profession. \(-\) Here are the results. The first line gives the possible number of heads on 10 tosses, and the next two lines give the number of subjects that reported tossing this number of heads for the control and treatment groups, respectively (for example, 16 control subjects reported getting 4 heads). a. Suppose that a subject tosses a fair coin and truthfully reports the number of heads. What is the distribution of the number of heads reported? What is the probability of collecting \(\$ 160\) or more? b. For a truthful subject, what would be the probability of doing better than chance-that is, tossing 6 or more heads? What proportion of subjects in the two groups reported tossing 6 or more heads? c. What does your result in (b) suggest about cheating in the two groups? (In Chapter 23 we will return to this example with more formal tools for comparing the groups.)
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