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Chapter 6: Problem 34

A grocery store has an express line for customers purchasing at most five items. Consider the random variable \(x=\) the number of items purchased by a randomly selected customer using this line. Make two tables that represent two different possible probability distributions for \(x\) that have the same mean but different standard deviations.

Short Answer

Expert verified
We have created two probability distributions with the same mean of 3 but different standard deviations. The first table has a standard deviation of 1.1, and the second table has a standard deviation of approximately 1.02. The tables are as follows: Table 1: | x | P(x) | |-----|------| | 1 | 0.1 | | 2 | 0.2 | | 3 | 0.4 | | 4 | 0.2 | | 5 | 0.1 | Table 2: | x | P(x) | |-----|--------| | 1 | 0.05 | | 2 | 0.35 | | 3 | 0.2 | | 4 | 0.35 | | 5 | 0.05 |

Step by step solution

01

Table 1

Assuming a simple probability distribution for x: | x | P(x) | |-----|------| | 1 | 0.1 | | 2 | 0.2 | | 3 | 0.4 | | 4 | 0.2 | | 5 | 0.1 | Now, let's calculate the mean (碌) and standard deviation (蟽) for Table 1. Mean for Table 1: \(碌 = 危[xP(x)] = (1\times0.1)+(2\times0.2)+(3\times0.4)+(4\times0.2)+(5\times0.1)= 3\) Standard Deviation for Table 1: \(\sigma = \sqrt{危[(x-碌)^2P(x)]} = \sqrt{(1-3)^2\times0.1 + (2-3)^2\times0.2 + (3-3)^2\times0.4 + (4-3)^2\times0.2 + (5-3)^2\times0.1} = 1.1\)
02

Table 2

Now we create another probability distribution for x with the same mean (碌 = 3) but a different standard deviation. | x | P(x) | |-----|--------| | 1 | 0.05 | | 2 | 0.35 | | 3 | 0.2 | | 4 | 0.35 | | 5 | 0.05 | Mean for Table 2 (already known as same mean as Table 1): \(碌 = 危[xP(x)] = 3\) Standard Deviation for Table 2: \(\sigma = \sqrt{危[(x-碌)^2P(x)]} = \sqrt{(1-3)^2\times0.05 + (2-3)^2\times0.35 + (3-3)^2\times0.2 + (4-3)^2\times0.35 + (5-3)^2\times0.05} \approx 1.02\) Now we have created two different probability distributions for the random variable x (number of items purchased by a customer using the express line) that have the same mean but different standard deviations.

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Key Concepts

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

Understanding Random Variables
When delving into the world of probability, the concept of a random variable is fundamental. A random variable is a numerical value determined by the outcome of a random phenomenon. Imagine flipping a coin; the outcome can be heads or tails. If we assign '1' to heads and '0' to tails, we have defined a random variable that represents the coin toss outcome.

Getting back to our example of the grocery story express line: the random variable 'x' represents the number of items a customer purchases. Since the express line is limited to at most five items, 'x' can take on one of five values (1 through 5). Each value has a probability associated with it, creating what is known as a 'probability distribution'.

A probability distribution lists all possible values of the random variable and the likelihood that each will occur. It succinctly summarizes the possible outcomes and how frequently we can expect to see them, based on our observation or assumptions about the random process in question.
Deciphering Probability Distribution Mean
The mean, or expected value, of a probability distribution is a measure of the 'center' of the distribution, weighted by the likelihood of each outcome. It's like an average where each outcome is considered as important as its probability. Mathematically, it is computed as the sum of each value of the random variable multiplied by its probability, succinctly represented as \( \mu = \Sigma[xP(x)] \).

In the exercise solution, the mean for both tables is calculated to be '3'. This doesn't necessarily mean that most customers buy exactly three items. Instead, it represents the average number of items that would be purchased if you observed a very large number of customers. It is the balancing point of the distribution and is crucial for understanding the overall behavior of the random variable 'x' in our grocery store scenario.
Exploring Probability Distribution Standard Deviation
Variability is just as crucial as the mean in understanding probability distributions. This is where standard deviation comes into play. It measures how spread out the values of the random variable are from the mean. A lower standard deviation implies that the values are clustered more closely around the mean, while a higher standard deviation indicates a more spread-out distribution.

The standard deviation is computed by taking the root of the sum of the squared differences between the values and the mean, each multiplied by their probability: \( \sigma = \sqrt{\Sigma[(x-\mu)^2P(x)]} \). The steps in our exercise show how different probability distributions can have the same mean yet varying standard deviations, reflecting different levels of predictability and consistency in the number of items purchased by customers. These varying standard deviations lead to different interpretations and expectations of the customer's behavior on the express line. For instance, Table 1, with a higher standard deviation, suggests a greater variety in the number of items customers are likely to purchase compared to Table 2.

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Most popular questions from this chapter

The probability distribution of \(x\), the number of tires needing replacement on a randomly selected automobile checked at a certain inspection station, is given in the following table: 0 \(\begin{array}{cccc}1 & 2 & 3 & 4 \\ 0.16 & 0.06 & 0.04 & 0.20\end{array}\) $$ p(x) \quad 0.54 $$ The mean value of \(x\) is \(\mu_{x}=1.2\). Calculate the values of \(\sigma_{x}^{2}\) and \(\sigma\).

Suppose that \(20 \%\) of all homeowners in an earthquakeprone area of California are insured against earthquake damage. Four homeowners are selected at random. Define the random variable \(x\) as the number among the four who have earthquake insurance. a. Find the probability distribution of \(x\). (Hint: Let \(S\) denote a homeowner who has insurance and \(\mathrm{F}\) one who does not. Then one possible outcome is SFSS, with probability (0.2)(0.8)(0.2)(0.2) and associated \(x\) value of 3 . There are 15 other outcomes.) b. What is the most likely value of \(x ?\) c. What is the probability that at least two of the four selected homeowners have earthquake insurance?

The size of the left upper chamber of the heart is one measure of cardiovascular health. When the upper left chamber is enlarged, the risk of heart problems is increased. The paper "Left Atrial Size Increases with Body Mass Index in Children" (International Journal of Cardiology [2009]: 1-7) described a study in which the left atrial size was measured for a large number of children ages 5 to 15 years. Based on these data, the authors concluded that for healthy children, left atrial diameter was approximately normally distributed with a mean of \(26.4 \mathrm{~mm}\) and a standard deviation of \(4.2 \mathrm{~mm}\). a. Approximately what proportion of healthy children have left atrial diameters less than \(24 \mathrm{~mm}\) ? b. Approximately what proportion of healthy children have left atrial diameters greater than \(32 \mathrm{~mm} ?\) c. Approximately what proportion of healthy children have left atrial diameters between 25 and \(30 \mathrm{~mm} ?\) d. For healthy children, what is the value for which only about \(20 \%\) have a larger left atrial diameter?

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean of 120 seconds and a standard deviation of 20 seconds. The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

Consider the population of all one-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal distribution with mean \(\mu=5 \mathrm{ml}\) and standard deviation \(\sigma=0.2 \mathrm{ml}\) is a reasonable model for the distribution of the variable \(x=\) amount of red dye in the paint mixture Use the normal distribution to calculate the following probabilities. (Hint: See Example \(6.21 .\) ) a. \(P(x<5.0)\) b. \(P(x<5.4)\) c. \(P(x \leq 5.4)\) d. \(P(4.64.5)\) f. \(P(x>4.0)\)
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