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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

The authors of the paper "Development of Nutritionally At-Risk Young Children Is Predicted by Malaria, Anemia, and Stunting in Pemba, Zanzibar" (The Journal of Nutrition [2009]: 763-772) studied factors that might be related to dietary deficiencies in children. Children were observed for a fixed period of time, and the amount of time spent in various activities was recorded. One variable of interest was the amount of time (in minutes) a child spent fussing. Data for 15 children, consistent with summary quantities in the paper, are given in the accompanying table. Normal scores for a sample size of 15 are also given. $$ \begin{array}{|cc|} \hline \text { Fussing Time }(x) & \text { Normal Score } \\ \hline 0.05 & -1.739 \\ 0.10 & -1.245 \\ 0.15 & -0.946 \\ 0.40 & -0.714 \\ 0.70 & -0.515 \\ 1.05 & -0.335 \\ 1.95 & -0.165 \\ 2.15 & 0.000 \\ 3.70 & 0.165 \\ 3.90 & 0.335 \\ 4.50 & 0.515 \\ 6.00 & 0.714 \\ 8.00 & 0.946 \\ 11.00 & 1.245 \\ 14.00 & 1.739 \\ \hline \end{array} $$ a. Construct a normal probability plot for the fussing time data. Does the plot look linear? Do you agree with the authors of the paper that the fussing time distribution is not normal? b. Calculate the correlation coefficient for the (normal score, \(x\) ) pairs. Compare this value to the appropriate critical \(r\) value from Table 6.2 to determine if it is reasonable to think that the fussing time distribution is approximately normal.

Consider the population of all one-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal distribution with mean \(\mu=5 \mathrm{ml}\) and standard deviation \(\sigma=0.2 \mathrm{ml}\) is a reasonable model for the distribution of the variable \(x=\) amount of red dye in the paint mixture Use the normal distribution to calculate the following probabilities. (Hint: See Example \(6.21 .\) ) a. \(P(x<5.0)\) b. \(P(x<5.4)\) c. \(P(x \leq 5.4)\) d. \(P(4.64.5)\) f. \(P(x>4.0)\)

Suppose a playlist on an MP3 music player consisting of 100 songs includes 8 by a particular artist. Suppose that songs are played by selecting a song at random (with replacement) from the playlist. The random variable \(x\) represents the number of songs until a song by this artist is played. a. Explain why the probability distribution of \(x\) is not binomial. b. Find the following probabilities. (Hint: See Example \(6.31 .\) ) i. \(p(4)\) ii. \(P(x \leq 4)\) iii. \(P(x>4)\) iv. \(P(x \geq 4)\) c. Interpret each of the probabilities in Part (b) and explain the difference between them.

Consider the following sample of 25 observations on \(x=\) diameter (in centimeters) of CD disks produced by a particular manufacturer: $$ \begin{array}{lccccccc} 15.66 & 15.78 & 15.82 & 15.84 & 15.89 & 15.92 & 15.94 & 15.95 \\ 15.99 & 16.01 & 16.04 & 16.05 & 16.06 & 16.07 & 16.08 & 16.10 \\ 16.11 & 16.13 & 16.13 & 16.15 & 16.15 & 16.19 & 16.22 & 16.27 \end{array} $$ 16.11 16.29 The 13 largest normal scores for a sample of size 25 are 1.964,1.519,1.259,1.064,0.903,0.763,0.636,0.519,0.408 \(0.302,0.200,0.099,\) and \(0 .\) The 12 smallest scores result from placing a negative sign in front of each of the given nonzero scores. Construct a normal probability plot. Is it reasonable to think that the disk diameter distribution is approximately normal? Explain.

The accompanying data on \(x=\) student-teacher ratio is for a random sample of 20 high schools in Maine selected from a population of 85 high schools. The data are consistent with summary values for the state of Maine that appeared in an article in the Bangor Daily News (September \(22,2016,\) bangordailynews.com/2016/09/22/mainefocus/we-discovered-a-surprise-when-we- looked-deeper-into-our-survey-of-maine-principals/?ref=morelnmidcoast, retrieved May 2, 2017). The corresponding normal scores are also shown. $$ \begin{array}{|cc|} \hline \text { Student-Teacher Ratio }(x) & \text { Normal Score } \\ \hline 9.0 & -1.868 \\ 10.0 & -1.403 \\ 11.0 & -1.128 \\ 11.2 & -0.919 \\ 11.6 & -0.744 \\ 11.7 & -0.589 \\ 11.8 & -0.448 \\ 11.9 & -0.315 \\ 12.0 & -0.187 \\ 12.1 & -0.062 \\ 12.5 & 0.062 \\ 12.6 & 0.187 \\ 13.0 & 0.315 \\ 13.2 & 0.448 \\ 13.6 & 0.589 \\ 13.7 & 0.744 \\ 14.0 & 0.919 \\ 14.5 & 1.128 \\ 14.9 & 1.403 \\ 15.0 & 1.868 \\ \hline \end{array} $$ a. Construct a normal probability plot. b. Calculate the correlation coefficient for the (normal score, \(x\) ) pairs. Compare this value to the appropriate critical \(r\) value from Table 6.2 to determine if it is reasonable to think that the distribution of student-teacher ratios for high schools in Maine is approximately normal.
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