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Let \(x\) represent the number of times a customer visits a grocery store in a 1 -week period. Assume this is the probability distribution of \(x\) :$$\begin{array}{l|llll}x & 0 & 1 & 2 & 3 \\\\\hline p(x) & .1 & .4 & .4 & .1\end{array}$$ Find the expected value of \(x\), the average number of times a customer visits the store.

Short Answer

Expert verified
Answer: The expected value (average) of the number of times a customer visits a grocery store in a 1-week period is 1.5 times.

Step by step solution

01

Identify the values and probabilities

The values 0, 1, 2, 3 are given as the number of times a customer visits a grocery store in a 1-week period with the corresponding probabilities 0.1, 0.4, 0.4, and 0.1 respectively.
02

Apply the expected value formula

We need to plug in the values \(x_i\) and their respective probabilities \(p(x_i)\) into the expected value formula: $$E(X) = \sum_{i} x_i p(x_i)$$ For this specific example, our sum will look like this: $$E(X) = (0)(0.1) + (1)(0.4) + (2)(0.4) + (3)(0.1)$$ Multiplying each value \(x_i\) by its respective probability: $$E(X) = 0 + 0.4 + 0.8 + 0.3$$
03

Calculate the expected value

Now, we just need to add all these terms together to find the expected value of \(x\): $$E(X) = 0 + 0.4 + 0.8 + 0.3 = 1.5$$ Therefore, the expected value (average) of the number of times a customer visits the grocery store in a 1-week period is 1.5 times.

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Key Concepts

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

Probability Distribution
When we speak about the probability distribution of a random variable, we are referring to how that variable's possible values are spread out or dispersed. This representation tells us the likelihood, or probability, of the variable taking on a given value. In our exercise, we are considering the variable to represent the number of visits a customer makes to a grocery store over a week. The given table shows the distribution of this discrete variable, with the probabilities of each occurrence painstakingly detailed. Think of it as a map that gives the probabilities for every possible outcome.

For a customer's number of visits, the possible values are laid out as 0, 1, 2, and 3. Attached to these numbers are the probabilities of .1, .4, .4, and .1 respectively. This means there's a 10% chance a customer doesn't visit in the week, a 40% chance they visit once or twice, and a 10% chance they visit three times. A proper understanding of this distribution is crucial as it is the foundation upon which we calculate the expected value, a paramount piece of statistical information that can inform business decisions and forecasts.
Discrete Random Variable
A discrete random variable is a type of variable that can take on a countable number of distinct values. Unlike continuous random variables which can assume any value within a given range, discrete ones can be listed out individually. In our example, the discrete random variable is the number of visits (\( x \)) a customer makes to the grocery store in a week. The values 0, 1, 2, and 3 are the only outcomes here. This distinction is key when you're conceptualizing how to analyze and present statistical data. Noting that the variable is discrete informs us directly that we can apply certain formulas – such as those for expected value – and anticipate a specific set of possible outcomes to evaluate.
Expected Value Formula
The expected value formula is a weighted average that incorporates both the possible values of a random variable and the probabilities of those values occurring. It's represented mathematically as: \[E(X) = \sum_{i} x_i p(x_i)\] where:
  • \( E(X) \) is the expected value of the random variable \( X \)
  • \( x_i \) is a particular value that the random variable can take
  • \( p(x_i) \) is the probability associated with \( x_i \)

By using this formula, you identify each possible outcome, weight it by its probability, then sum up all these products to obtain the expected value. In the context of our exercise, the formula helps us calculate the average number of customer visits to a store in a week. The result, 1.5 visits, doesn't mean a customer can visit half a time; instead, it's an average over the long term. For the store's management, this value can be instrumental for inventory planning, staffing needs, and analysis of customer behavior over time.

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Most popular questions from this chapter

One professional golfer plays best on short-distance holes. Experience has shown that the numbers \(x\) of shots required for \(3-, 4-,\) and 5 -par holes have the probability distributions shown in the table: $$\begin{array}{lccccc}\text { Par-3 Holes } & \multicolumn{3}{c} {\text { Par-4 Holes }} & \multicolumn{2}{c} {\text { Par-5 Holes }} \\ \hline x & p(x) & x & p(x) & x & p(x) \\\\\hline 2 & .12 & 3 & .14 & 4 & .04 \\\3 & .80 & 4 & .80 & 5 & .80 \\\4 & .06 & 5 & .04 & 6 & .12 \\ 5 & .02 & 6 & .02 & 7 & .04\end{array}$$ What is the golfer's expected score on these holes? a. A par-3 hole b. A par- 4 hole c. A par-5 hole

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