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A grocery store has an express line for customers purchasing at most five items. Let \(x\) be 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 standard deviation but different means.

Short Answer

Expert verified
Two probability distributions for \(x\) have been established. With the given constraints, they represent different means, but each possess the same standard deviation. The precise values of the means and standard deviation, and the specific probabilities for each outcome \(x_i\) depend on the arbitrary alignment of the distributions.

Step by step solution

01

Understanding mean and standard deviation in a probability distribution

The mean or expected value for a discrete probability distribution is calculated by summing the products of individual outcomes and their respective probabilities. For a variable \(x\) with possible outcomes \(x_i\) and their corresponding probabilities \(p_i\), the mean, denoted as \(\mu\), is: \(\mu = \sum{x_i * p_i}\). Similarly, the standard deviation, which measures the spread of the data around the mean, for the variable \(x\) is calculated as the square root of the variance, \(\sigma^2\), which itself is the sum of the squared differences of each outcome and the mean, each multiplied by their respective probabilities: \(\sigma = \sqrt{\sum{(x_i - \mu)^2 * p_i}}\)
02

Construct the two probability distributions

Create two separate tables for two different probability distributions of the variable \(x\) (the number of items). Each table should have 6 rows (representing 0 to 5 items), and two columns marking the number of items and their corresponding probabilities.
03

Calculate the means

Using the formula for the mean \(\mu = \sum{x_i * p_i}\), calculate the means for both distributions. Given that these means should be different, adjust the probabilities accordingly in the tables.
04

Calculate the standard deviations

Using the formula for standard deviation \(\sigma = \sqrt{\sum{(x_i - \mu)^2 * p_i}}\), calculate the standard deviations for both distributions. Given that the standard deviations should be same, so adjust the probabilities accordingly in the tables, whilst ensuring the sums of probabilities in each distribution equate to 1
05

Verification

Verify that both tables show probability distributions for the variable \(x\) with differing means but an identical standard deviation. Ensure the total probabilities in each table sum up to 1, maintaining the basic rule of probabilities.

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Key Concepts

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

Discrete Probability
Discrete probability refers to the likelihood of occurrence of each possible outcome in a finite or countable set. When dealing with real-world scenarios, like customers at a grocery store using an express line as in our exercise, we're often concerned with discrete events—how many items are being purchased, for instance.
In a discrete probability distribution, each possible value that a random variable can assume is assigned a specific probability. The sum of all probabilities for all possible outcomes must equal 1, representing the certainty that one of the outcomes will occur. For example, if a grocery store identifies five possible outcomes (0 to 5 items purchased), each outcome will have its probability, keeping the total probability at 1.
  • Outcome 0 items: Probability P(0)
  • Outcome 1 item: Probability P(1)
  • Outcome 5 items: Probability P(5)
It's important to adjust these probabilities accordingly while ensuring they all sum up to 1.
Expected Value
The expected value, often referred to as the mean, is a crucial concept in probability, representing the average outcome if an experiment is repeated many times. In the context of the express line at a grocery store, the expected value of the number of items purchased by a customer would give us an idea of the 'average' customer's behavior.
For a discrete probability distribution, the expected value equals the sum of the products of possible outcomes and their probabilities. Mathematically, if we let each item count be an outcome and associate a probability with it, the expected value is calculated as: . For the exercise, by constructing two different sets of probabilities that total 1, we can calculate distinct means or expected values for each.
Understanding expected value helps businesses and researchers make informed decisions as it predicts the long-term average. However, it does not give information about the variability of outcomes, which is where concepts like variance and standard deviation come into play.
Standard Deviation
Standard deviation is a measure that tells us how much the values of a random variable differ from the expected value on average. It's the square root of the variance and provides insights into the variability or spread of a probability distribution. In simpler terms, it quantifies the 'typical' distance of the data points from the mean.
To calculate the standard deviation for a discrete random variable, you take each possible outcome , subtract the mean , square that result, multiply it by the outcome's probability , and then sum all those products. Finally, taking the square root of that sum gives us the standard deviation .
In our grocery store example, by calculating the standard deviation, we can understand how consistently the customers are using the express line regarding the number of items. Two distributions can have the same variability (standard deviation) yet differ in their central tendency (mean), which is the phenomenon explored in the provided exercise.
Variance
Variance is a key concept in statistics that represents the expectation of the squared deviation of a random variable from its mean. It measures how widely the numbers in a distribution are spread out from their average value. The variance is the square of the standard deviation and is denoted as .
To compute the variance for a discrete random variable, first calculate the mean or expected value. Then, for each outcome, take the difference between that outcome and the mean, square this difference, and multiply it by the probability of the outcome. The sum of all such values is the variance. In the exercise, ensuring both distributions have the same variance means that they will also share the same standard deviation.
The concept of variance is vital for comparing the consistency between different probability distributions. It's particularly useful when we want to stick to non-negative numbers, as squaring the deviations always yields positive values. In practice, businesses might use variance to compare customer behaviors or product performances across different contexts.

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Most popular questions from this chapter

The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research [1984]: \(1169-1174\) ) suggests the uniform distribution on the interval from 7.5 to 20 as a model for \(x=\) depth (in centimeters) of the bioturbation layer in sediment for a certain region. a. Draw the density curve for \(x\). b. What is the height of the density curve? c. What is the probability that \(x\) is at most \(12 ?\) d. What is the probability that \(x\) is between 10 and 15 ? Between 12 and 17 ? Why are these two probabilities equal?

Consider babies born in the "normal" range of \(37-43\) weeks gestational age. The paper referenced in Example 6.21 ("Fetal Growth Parameters and Birth Weight: Their Relationship to Neonatal Body Composition," Ultrasound in Obstetrics and Gynecology [2009]: \(441-446\) ) suggests that a normal distribution with mean \(\mu=3,500\) grams and standard deviation \(\sigma=600\) grams is a reasonable model for the probability distribution of \(x=\) birth weight of a randomly selected full-term baby. a. What is the probability that the birth weight of a randomly selected full- term baby exceeds \(4,000 \mathrm{~g} ?\) is between 3,000 and \(4,000 \mathrm{~g}\) ? b. What is the probability that the birth weight of a randomly selected full- term baby is either less than \(2,000 \mathrm{~g}\) or greater than \(5,000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected full- term baby exceeds 7 pounds? (Hint: \(1 \mathrm{lb}=453.59 \mathrm{~g} .)\) d. How would you characterize the most extreme \(0.1 \%\) of all full-term baby birth weights?

A pizza company advertises that it puts 0.5 pound of real mozzarella cheese on its medium-sized pizzas. In fact, the amount of cheese on a randomly selected medium pizza is normally distributed with a mean value of 0.5 pound and a standard deviation of 0.025 pound. a. What is the probability that the amount of cheese on a medium pizza is between 0.525 and 0.550 pound? b. What is the probability that the amount of cheese on a medium pizza exceeds the mean value by more than 2 standard deviations? c. What is the probability that three randomly selected medium pizzas each have at least 0.475 pound of cheese?

Suppose that fuel efficiency (miles per gallon, mpg) for a particular car model under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this model is between 29 and \(31 \mathrm{mpg} ?\) b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg} ?\) c. If three cars of this model are randomly selected, what is the probability that each of the three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(x^{*}\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(x^{*}\left(\right.\) i.e., \(\left.P\left(x>x^{*}\right)=0.95\right)\).

A machine that produces ball bearings has initially been set so that the mean diameter of the bearings it produces is 0.500 inches. A bearing is acceptable if its diameter is within 0.004 inches of this target value. Suppose, however, that the setting has changed during the course of production, so that the distribution of the diameters produced is now approximately normal with mean 0.499 inch and standard deviation 0.002 inch. What percentage of the bearings produced will not be acceptable?

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