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Example 6.27 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 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 \(x\).

Short Answer

Expert verified
a. The possible values of \(x\) are 0, 1, 2, 3, 4, 5.\nb. For each possible value of \(x\), we can use the binomial probability formula to find the associated probability. Suggested practice is to compute these probabilities and create a table displaying each possible \(x\) value and its associated probability.\nc. A histogram can be created with the x-axis representing the possible values of \(x\) and the y-axis representing the associated probabilities. Remember the binomial shape while creating the histogram.

Step by step solution

01

Determining possible values of \(x\)

The possible values of \(x\) are all integer values from 0 to 5, inclusive. This is based on the person having no successful identifications (0) to having all successful identifications (5).
02

Calculating probability for each possible value of \(x\)

For each possible value of \(x\), we can use the binomial probability formula:\n\[p(x) = C(n, x) \cdot (p)^x \cdot (1-p)^{n-x}\]\n\nwhere \(C(n, x)\) is the number of combinations of \(n\) items taken \(x\) at a time. Here, \(p = \frac{1}{4}\) is the probability of success (correctly identifying the shirt), \(n = 5\) is the number of trials, and \(x\) is the number of successes we're finding the probability for.
03

Constructing a probability table for each possible value of \(x\)

Based on the calculated probabilities, we can create a table showing each possible value of \(x\) and its corresponding probability \(p(x)\). The table columns will be 'Value of \(x\)' and 'Probability \(p(x)\)'.
04

Constructing a histogram

A histogram can be made by taking each possible value of \(x\) as the x-axis and the corresponding probability \(p(x)\) as the height of the bars. The histogram will exhibit the shape of the binomial distribution.

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Key Concepts

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

Binomial Probability Formula
At the heart of understanding the scenario with the shirts is the binomial probability formula. This formula lets you calculate the likelihood of achieving a certain number of successes in a fixed number of trials, where each trial has two possible outcomes—success or failure—and each trial is independent of the others.

The formula is given as:
\[ p(x) = C(n, x) \cdot (p)^x \cdot (1-p)^{n-x} \]

Here, C(n, x) refers to the number of combinations, which is the count of different ways you can achieve x successes out of n trials; p is the probability of success on a single trial, and (1-p) is the probability of failure.

To apply this in our t-shirt scenario, where a person is randomly picking one correct shirt out of four, p equals \( \frac{1}{4} \), n is 5, and x can vary from 0 to 5. The formula helps us compute the probability for each possible outcome of correct identifications.
Probability Distribution
A probability distribution describes how probabilities are distributed over the values of the random variable. In a binomial distribution, which is applicable to our example, the random variable represents the number of successes (in our case, the number of times the person correctly identifies the t-shirt).

The probabilities associated with each value of the random variable must satisfy two conditions: the probability of each value must be between 0 and 1, and the sum of all probabilities must be 1. When you use the binomial probability formula to calculate the probabilities for each possible value of x, you essentially plot the probability distribution.

For the problem at hand, after using the binomial probability formula across all possible values of x, you would get a set of probabilities that create the distribution. For instance, the probability of getting no correct identifications (x=0) versus getting all five correct (x=5)—each of these scenarios has a probability that contributes to the overall distribution.
Histogram
A histogram is a graphical representation that organizes a group of data points into specified ranges. In the context of the binomial distribution from our exercise, a histogram can visually illustrate the probability of each number of correct identifications.

In creating the histogram, the x-axis would represent the number of successful identifications (x), and the y-axis would show the associated probability (p(x)). Each bar in the histogram corresponds to a possible value of x, and the height reflects its probability.

By constructing a histogram, students can better visualize the distribution; lower probabilities will have shorter bars, and higher probabilities will have taller bars. Observing the shape and height of the bars can also provide insights into the most and least likely outcomes—for example, getting exactly half of the identifications correct.
Combinatorics
In our exercise, combinatorics plays a crucial role. It is a field of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It is the core of calculating C(n, x) in the binomial probability formula, which determines the number of ways to choose x successes out of n trials.

The number of combinations is often written as:
\[ C(n, x) = \frac{n!}{x!(n-x)!} \]

where n! (n factorial) is the product of all positive integers up to n. For instance, to find out how many ways the person can correctly identify 3 shirts out of 5 attempts, you'd calculate C(5, 3) using the formula above.

In summary, combinatorics is fundamental in understanding and calculating the probabilities associated with binomial distributions. By looking into all possible combinations of outcomes, we can better understand the random process and predict the likelihood of various results.

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Most popular questions from this chapter

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccl} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ? \end{array} $$ a. Only \(y\) values of \(0,1,2,3,\) and 4 have probabilities greater than 0 . What is \(p(4)\) ? b. How would you interpret \(p(1)=0.20 ?\) c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. d. Calculate \(P(y<2),\) the probability that the carton contains fewer than two broken eggs. Why is this smaller than the probability in Part (c)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

The light bulbs used to provide exterior lighting for a large office building have an average lifetime of 700 hours. If lifetime is approximately normally distributed with a standard deviation of 50 hours, how often should all the bulbs be replaced so that no more than \(20 \%\) of the bulbs will have already burned out?

Example 6.14 gave the probability distributions shown below for \(x=\) number of flaws in a randomly selected glass panel from Supplier 1 \(y=\) number of flaws in a randomly selected glass panel from Supplier 2 for two suppliers of glass used in the manufacture of flat screen TVs. If the manufacturer wanted to select a single supplier for glass panels, which of these two suppliers would you recommend? Justify your choice based on consideration of both center and variability. $$ \begin{array}{lcccclcccc} \boldsymbol{x} & 0 & 1 & 2 & 3 & \boldsymbol{y} & 0 & 1 & 2 & 3 \\ \boldsymbol{p}(\boldsymbol{x}) & 0.4 & 0.3 & 0.2 & 0.1 & \boldsymbol{p}(\boldsymbol{y}) & 0.2 & 0.6 & 0.2 & 0 \end{array} $$

The paper "The Effect of Temperature and Humidity on Size of Segregated Traffic Exhaust Particle Emissions" (Atmospheric Environment [2008]: 2369-2382) gave the following summary quantities for a measure of traffic flow (vehicles/second) during peak traffic hours. Traffic flow was recorded daily at a particular location over a long sequence of days. Mean \(=0.41\) Standard Deviation \(=0.26\) Median \(=0.45\) 5th percentile \(=0.03 \quad\) Lower quartile \(=0.18\) \(\begin{array}{ll}\text { Upper quartile } & =0.57 & \text { 95th Percentile } & =0.86\end{array}\) Based on these summary quantities, do you think that the distribution of the measure of traffic flow is approximately normal? Explain your reasoning.

Let \(z\) denote a random variable that has a standard normal distribution. Determine each of the following probabilities: a. \(P(z<2.36)\) b. \(P(z \leq 2.36)\) c. \(P(z<-1.23)\) d. \(P(1.142)\) g. \(P(z \geq-3.38)\) h. \(P(z<4.98)\)

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