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Chapter 6: Problem 6

Selling houses Let \(X\) represent the number of homes a real estate agent sells during a given month. Based on previous sales records, she estimates that \(\mathrm{P}(0)=0.68\), \(\mathrm{P}(1)=0.19, \mathrm{P}(2)=0.09, \mathrm{P}(3)=0.03, \mathrm{P}(4)=0.01\) with negligible probability for higher values of \(x\). a. Explain why it does not make sense to compute the mean of this probability distribution as \((0+1+2+3+4) / 5=2.0\) and claim that, on average, she expects to sell 2 homes. b. Find the correct mean and interpret.

Short Answer

Expert verified
The mean should be calculated using probabilities, resulting in an average of 0.5 homes sold per month.

Step by step solution

01

Understand the Problem

The problem involves understanding why calculating the mean of a probability distribution as an arithmetic average of outcomes, such as \((0+1+2+3+4)/5\), is incorrect. Instead, we should calculate the mean using the weighted average principle based on the probability distribution of the random variable \(X\).
02

Definition of Expected Value

In a probability distribution, the mean, also known as the expected value \(E(X)\), is calculated by summing the products of each possible outcome and its respective probability: \[ E(X) = \sum (x_i \cdot P(x_i)) \] where \(x_i\) are the possible outcomes, and \(P(x_i)\) are their probabilities.
03

Apply the Formula to Calculate the Correct Mean

Substitute the given probabilities and outcomes into the expected value formula: \[ E(X) = (0 \cdot 0.68) + (1 \cdot 0.19) + (2 \cdot 0.09) + (3 \cdot 0.03) + (4 \cdot 0.01) \] Compute each term: \[ 0 \cdot 0.68 = 0 \] \[ 1 \cdot 0.19 = 0.19 \] \[ 2 \cdot 0.09 = 0.18 \] \[ 3 \cdot 0.03 = 0.09 \] \[ 4 \cdot 0.01 = 0.04 \] Add these values together: \[ E(X) = 0 + 0.19 + 0.18 + 0.09 + 0.04 = 0.50 \]
04

Interpretation of the Mean

The computed mean \(E(X) = 0.50\) reflects the weighted average number of houses sold per month, considering the likelihood of each sales outcome. Therefore, on average, the real estate agent expects to sell 0.5 homes per month, which incorporates the lower probability of higher sales numbers.

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Key Concepts

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

Probability Distribution
In a probability distribution, each possible outcome of a random event is assigned a probability, which shows the likelihood of that outcome occurring. For the real estate agent selling houses, this is represented by a list of probabilities, \(P(0)=0.68\), \(P(1)=0.19\), \(P(2)=0.09\), \(P(3)=0.03\), and \(P(4)=0.01\). These probabilities describe the chance of selling 0, 1, 2, 3, and 4 houses respectively.Key characteristics of probability distributions include:
  • The sum of all probabilities must equal 1, ensuring they cover every possible outcome.
  • Each probability is always between 0 and 1.
  • Outcomes with a probability closer to 1 are more likely to occur compared to those closer to 0.
In this specific example, the highest probability 0.68 is associated with selling no houses, making it the most likely outcome each month. Understanding the probability distribution helps in predicting events based on their likelihood.
Weighted Average
A weighted average is used when different outcomes have varied levels of importance, indicated by their probabilities. Instead of treating each outcome equally, a weighted average takes into consideration how frequently each outcome is expected to occur.In calculating the expected number of homes sold per month, we multiply each number of houses (the outcome) by its probability to determine the contribution of each outcome to the average. For this problem:
  • For selling 0 houses: \(0 \cdot 0.68 = 0\).
  • For selling 1 house: \(1 \cdot 0.19 = 0.19\).
  • For selling 2 houses: \(2 \cdot 0.09 = 0.18\).
  • For selling 3 houses: \(3 \cdot 0.03 = 0.09\).
  • For selling 4 houses: \(4 \cdot 0.01 = 0.04\).
Weighted averages provide a more accurate representation of everyday scenarios than simple arithmetic averages, as they consider the probabilities associated with each potential outcome.
Mean Calculation
The mean, or expected value, of a probability distribution is a summary measure that offers insight into which outcomes are most likely when considering their probabilities. The formula for calculating the mean of a probability distribution is:\[ E(X) = \sum x_i \cdot P(x_i) \]Here, \(x_i\) are the possible outcomes, while \(P(x_i)\) are their respective probabilities. For our problem, using the outcomes and probabilities:\[ E(X) = (0 \cdot 0.68) + (1 \cdot 0.19) + (2 \cdot 0.09) + (3 \cdot 0.03) + (4 \cdot 0.01) \]After calculating each term, we find:\[ E(X) = 0 + 0.19 + 0.18 + 0.09 + 0.04 = 0.50 \]This means that, on average, the real estate agent will sell about 0.5 houses per month. This calculation considers all sales possibilities, weighting them by likelihood, to provide a realistic expectation rather than a simple count. It's a powerful tool for understanding real-world phenomena where different outcomes have different chances of occurring.

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Most popular questions from this chapter

Selling at the right price An insurance company wants to examine the views of its clients about the prices of three car insurance plans launched last year. It conducts a survey with two sets of plans with different prices and finds that: \- If plan A is sold for \(\$ 150\), plan \(\mathrm{B}\) for \(\$ 250\), and plan \(\mathrm{C}\) for \(\$ 350,\) then \(45 \%\) of the customers would be interested in plan \(\mathrm{A}, 15 \%\) in plan \(\mathrm{B},\) and \(40 \%\) in plan \(\mathrm{C}\). \- If plans \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are sold for \(\$ 170, \$ 250\), and \(\$ 310\) respectively, then \(15 \%\) of the customers would be interested in plan \(\mathrm{A}, 40 \%\) in plan \(\mathrm{B},\) and \(45 \%\) in plan \(\mathrm{C}\). a. For the first pricing set, construct the probability distribution of \(X=\) selling price for the sale of a car insurance plan, find its mean, and interpret. b. For the second pricing set, construct the probability distribution of \(X,\) find its mean, and interpret. c. Which pricing set is more profitable to the company? Explain.

Verify the empirical rule by using Table A, software, or a calculator to show that for a normal distribution, the probability (rounded to two decimal places) within a. 1 standard deviation of the mean equals 0.68 . b. 2 standard deviations of the mean equals 0.95 . c. 3 standard deviations of the mean is very close to 1.00 .

Binomial needs fixed \(n\) For the binomial distribution, the number of trials \(n\) is a fixed number. Let \(X\) denote the number of girls in a randomly selected family in Canada that has three children. Let \(Y\) denote the number of girls in a randomly selected family in Canada (that is, the number of children could be any number). A binomial distribution approximates well the probability distribution for one of \(X\) and \(Y\), but not for the other. a. Explain why. b. Identify the case for which the binomial applies and identify \(n\) and \(p\).

In the United States, the mean birth weight for boys is \(3.41 \mathrm{~kg}\), with a standard deviation of \(0.55 \mathrm{~kg}\). (Source: cdc.com.) Assuming that the distribution of birth weight is approximately normal, find the following using a table, calculator, or software. a. A baby is considered of low birth weight if it weighs less than \(2.5 \mathrm{~kg}\). What proportion of baby boys in the United States are born with low birth weight? b. What is the \(z\) -score for a baby boy that weighs \(1.5 \mathrm{~kg}\) (defined as extremely low birth weight)? c. Typically, birth weight is between \(2.5 \mathrm{~kg}\) and \(4.0 \mathrm{~kg}\). Find the probability a baby boy is born with typical birth weight. d. Matteo weighs \(3.6 \mathrm{~kg}\) at birth. He falls at what percentile? e. Max's parents are told that their newborn son falls at the 96 th percentile. How much does Max weigh?

In the National Basketball Association, the top free throw shooters usually have probability of about 0.90 of making any given free throw. a. During a game, one such player (Dirk Nowitzki) shot 10 free throws. Let \(X=\) number of free throws made. What must you assume for \(X\) to have a binomial distribution? (Studies have shown that such assumptions are well satisfied for this sport.) b. Specify the values of \(n\) and \(p\) for the binomial distribution of \(X\) in part a. c. Find the probability that he made (i) all 10 free throws (ii) 9 free throws and (iii) more than 7 free throws.
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