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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 mean but different standard deviations.

Short Answer

Expert verified
Creating two tables with different probability distributions for the same events (number of items). For the first distribution, assign the same probability for each outcome (0 to 5). For the second distribution, make the probabilities for the outcomes closer to the mean higher while decreasing the probabilities for the outcomes farther from the mean. By doing this, the second table will have the same mean but a smaller standard deviation.

Step by step solution

01

Constructing the first table

The first table can have equal probabilities for each outcome. This would mean all outcomes (0 to 5 items) are equally likely.
02

Calculating mean and standard deviation for Table 1

The mean (\(μ\)) of a distribution can be calculated as \(μ=E[X]=x_{1}p_{1}+x_{2}p_{2}+...+x_{n}p_{n}\). Here, each \(x_{i}\) is the number of items, and each \(p_{i}\) is the probability of that number of items. Similarly, the standard deviation (\(σ\)) can be calculated as \(σ=\sqrt{E[(X − μ)^2]}\), where \(E[(X − μ)^2] = x_{1}^2p_{1} + x_{2}^2p_{2} +... + x_{n}^2p_{n} - μ^2)\). Calculate the mean and standard deviation for the first table using these formulas.
03

Constructing the second table

Now, the second table can be made by changing the distribution of probabilities keeping the mean the same as the previous one. This can be accomplished by increasing the probabilities for the outcomes close to the mean and decreasing the probabilities for the outcomes farther from the mean.
04

Calculating mean and standard deviation for Table 2

Calculate the mean and standard deviation for the second table just like before. Now this table should have a smaller standard deviation since the probabilities have been adjusted to be closer to the mean. Ensure the mean matches with the previous table's mean while the standard deviation is different.

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Key Concepts

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

Mean
In the context of probability distributions, the mean is a measure that indicates the average outcome you can expect from a set of possibilities. It's calculated by multiplying each possible outcome by the probability of its occurrence and then summing those values. In our grocery store express line example, the mean would give us an idea of the average number of items that customers purchase.

For instance, if you were to randomly select a number of customers and note the amount of items each of them purchased, by multiplying those amounts by how likely they are (the probability), and adding them up, you get what is called the expected value or the average. This number can be crucial for the store to plan their checkout resources efficiently. Remember to consider each possible outcome and its associated probability; not doing so could lead to an incorrect calculation of the mean, which could skew the store's resource planning and lead to inefficiencies.
Standard Deviation
The standard deviation is a statistic that measures the amount of variation or dispersion from the mean within a set of data. In simpler terms, it tells us how spread out the numbers in our distribution are. A high standard deviation indicates that the data points are spread out over a wide range of values, while a low standard deviation indicates that they are clustered closely around the mean.

In our problem, we're looking at two probability distributions with the same mean but different standard deviations. Let's imagine that the first distribution represents a scenario where customers are just as likely to buy 1, 2, 3, 4, or 5 items, leading to a relatively high standard deviation. However, for the second distribution, customers may be more inclined to buy around the average number, causing a lower standard deviation. This concept is essential for making informed decisions based on variability in customer behavior.
Random Variable
A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. In our grocery store example, the random variable 'x' defines the possible outcomes for the number of items purchased. It's what we're observing and trying to understand in terms of probability distributions.

Think of it like this: each customer approaches the line is like rolling a die, but instead of numbers, each side represents the number of items they end up buying. Termed a discrete random variable because it can only take on a finite number of values, 'x' helps us represent and analyze processes that involve uncertainty and randomness, making it a cornerstone of probability theory.
Probability
Probability is a measure of how likely an event is to occur. It ranges from 0 (the event will never occur) to 1 (the event will always occur). So when we talk about probability distributions, we're really referring to a mapping of all the possible outcomes of our random variable to the likelihood of each outcome.

For example, if it's equally likely for a customer to purchase any number of items from 0 to 5, each of those outcomes would have a probability of 1/6. However, shopping habits can vary, and it's possible that more customers tend to buy certain amounts of items more than others. By adjusting the probabilities based on observed behaviors, we can create different distributions that still reflect the same average number of items bought (mean) but show a different amount of variation (standard deviation) among customers. Understanding the probability of diverse scenarios can help businesses forecast demand, manage stock, and customize customer service.

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Most popular questions from this chapter

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?

A machine producing vitamin \(\mathrm{E}\) capsules operates so that the actual amount of vitamin \(\mathrm{E}\) in each capsule is normally distributed with a mean of \(5 \mathrm{mg}\) and a standard deviation of \(0.05 \mathrm{mg}\). What is the probability that a randomly selected capsule contains less than \(4.9 \mathrm{mg}\) of vitamin \(\mathrm{E}\) ? At least \(5.2 \mathrm{mg}\) of vitamin \(\mathrm{E}\) ?

Of all airline flight requests received by a certain ticket broker, \(70 \%\) are for domestic travel \((\mathrm{D})\) and \(30 \%\) are for international flights (I). Define \(x\) to be the number that are for domestic flights among the next three requests received. Assuming independence of successive requests, determine the probability distribution of \(x\). (Hint: One possible outcome is DID, with the probability \((0.7)(0.3)(0.7)=0.147 .)\)

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. $$ \begin{array}{lcccclcccc} \boldsymbol{x} & 0 & 1 & 2 & 3 & \boldsymbol{y} & 0 & 1 & 2 & 3 \\ \boldsymbol{p}(\boldsymbol{x}) & 0.4 & 0.3 & 0.2 & 0.1 & \boldsymbol{p}(\boldsymbol{y}) & 0.2 & 0.6 & 0.2 & 0 \end{array} $$

State whether each of the following random variables is discrete or continuous: a. The number of defective tires on a car b. The body temperature of a hospital patient c. The number of pages in a book d. The number of draws (with replacement) from a deck of cards until a heart is selected e. The lifetime of a light bulb

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