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Chapter 6: Problem 33

An appliance dealer sells three different models of freezers having 13.5,15.9 , and 19.1 cubic feet of storage space. Consider the random variable \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: \(x\) \(\begin{array}{lll}13.5 & 15.9 & 19.1\end{array}\) \(p(x)\) \(\begin{array}{lll}0.2 & 0.5 & 0.3\end{array}\) a. Calculate the mean and standard deviation of \(x\). (Hint: See Example \(6.15 .\) ) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.

Short Answer

Expert verified
. Using the calculated variance, we compute the standard deviation: Standard Deviation, \(SD(x) = \sqrt{Var(x)}\) \(SD(x) = \sqrt{2.88}\) \(SD(x) = 1.70\) The standard deviation of x is 1.70 cubic feet. #tag_title# Step 4: Interpret the mean and standard deviation #tag_content# The mean value of x, 16.38 cubic feet, represents the average storage space purchased by customers in this scenario. It gives us an idea of the typical storage size preferred by the customers. The standard deviation of x, 1.70 cubic feet, is a measure of the dispersion or variability in the storage space purchased by customers. A lower standard deviation would indicate that most customers' preferences are closely clustered around the mean, while a higher value would indicate a wider variety in their choices of storage size. In this case, a standard deviation of 1.70 cubic feet shows moderate variability in the customers鈥� preferences for freezer storage space.

Step by step solution

01

Calculate the mean of x

To calculate the mean, also known as expected value, of the discrete random variable x, we use the formula: \(E(x) = \sum_{i=1}^n x_i * p(x_i)\) Where \(x_i\) are the possible values of x and \(p(x_i)\) are their corresponding probabilities. Plugging in the given values, we get: Mean, \(E(x) = (13.5*0.2) + (15.9*0.5) + (19.1*0.3)\) \(E(x) = 2.7 + 7.95 + 5.73\) \(E(x) = 16.38\) The mean value of x is 16.38 cubic feet.
02

Calculate the variance of x

To calculate the variance, we use the formula: Var(x) = \(\sum_{i=1}^n (x_i - E(x))^2 * p(x_i)\) Plugging in the values, we get: Variance, Var(x) = \((13.5-16.38)^2*0.2 + (15.9-16.38)^2*0.5 + (19.1-16.38)^2*0.3\) Var(x) = \(2.88\) The variance of x is 2.88.
03

Calculate the standard deviation of x

The standard deviation is simply the square root of the variance

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Key Concepts

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

Expected Value
The concept of "Expected Value" in probability distribution is essentially the weighted average of all possible outcomes. In our freezer example, it gives us an estimate of the typical amount of storage space a customer is likely to purchase. Using the given values, the expected value, or mean, is calculated by multiplying each storage size with its probability and then summing these products. The formula used is:
  • \(E(x) = \sum_{i=1}^n x_i \times p(x_i)\)
For our specific example:
  • \(E(x) = (13.5\times0.2) + (15.9\times0.5) + (19.1\times0.3) = 16.38\)
This means, on average, a customer buys a freezer with about 16.38 cubic feet of storage. It gives dealers an estimate to optimize stock and understand customer preferences.
Variance
Variance is a measure of the spread between numbers in a probability distribution and indicates how much the values differ from the expected value. In simpler terms, it shows how much variation there is from our calculated average storage space of 16.38 cubic feet. For our dealer's data, variance is calculated by:
  • First, finding the difference of each storage space from the expected value, \(E(x)\).
  • Then, squaring each difference to eliminate negative values.
  • Lastly, multiplying each squared difference by its probability, and summing these results.\[Var(x) = \sum_{i=1}^n (x_i - E(x))^2 \times p(x_i) = 2.88\]
A variance of 2.88 tells us there's a moderate spread around the mean freezer size, indicating customers occasionally purchase significantly above or below the 16.38 average.
Standard Deviation
Standard deviation provides a more intuitive measure of variability, as it is in the same units as the data. It's calculated as the square root of the variance and indicates how much individual storage sizes deviate, on average, from the expected value. For our dataset:
  • \[\text{SD}(x) = \sqrt{\text{Var}(x)} = \sqrt{2.88}\approx 1.70\]
This tells the appliance dealer that typically, the amount of storage space purchased deviates by about 1.70 cubic feet from the average of 16.38 cubic feet. Consequently, knowing both the mean and the standard deviation helps in assessing whether most purchases are near the average size or vary widely.

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Most popular questions from this chapter

A box contains five slips of paper, marked \(\$ 1, \$ 1, \$ 1, \$ 10,\) and \(\$ 25 .\) The winner of a contest selects two slips of paper at random and then gets the larger of the dollar amounts on the two slips. Define a random variable \(w\) by \(w=\) amount awarded. Determine the probability distribution of \(w\). (Hint: Think of the slips as numbered \(1,2,3,4,\) and \(5 .\) An outcome of the experiment will consist of two of these numbers.)

Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboard can be approximated by a normal curve with mean 60 wpm and standard deviation 15 wpm ("The Effects of Split Keyboard Geometry on Upper Body Postures, Ergonomics [2009]: 104-111). a. What is the probability that a randomly selected typist's speed is at most 60 wpm? Less than 60 wpm? b. What is the probability that a randomly selected typist's speed is between 45 and 90 wpm? c. Would you be surprised to find a typist in this population whose speed exceeded 105 wpm? d. Suppose that two typists are independently selected. What is the probability that both their speeds exceed 75 wpm? e. Suppose that special training is to be made available to the slowest \(20 \%\) of the typists. What typing speeds would qualify individuals for this training? (Hint: See Example \(6.23 .)\)

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