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Chapter 5: Problem 37

A particular brand of dishwasher soap is sold in three sizes: \(25 \mathrm{oz}, 40 \mathrm{oz}\), and \(65 \mathrm{oz}\). Twenty percent of all purchasers select a 25 -oz box, \(50 \%\) select a 40 -oz box, and the remaining \(30 \%\) choose a \(65-\mathrm{oz}\) box. Let \(X_{1}\) and \(X_{2}\) denote the package sizes selected by two independently selected purchasers. a. Determine the sampling distribution of \(\bar{X}\), calculate \(E(\bar{X})\), and compare to \(\mu\). b. Determine the sampling distribution of the sample variance \(S^{2}\), calculate \(E\left(S^{2}\right)\), and compare to \(\sigma^{2}\).

Short Answer

Expert verified
a. The mean \(E(\bar{X})\) is 46.5 oz; it equals \(\mu\). b. The expected variance \(E(S^2)\) is 146.75 oz虏; it equals \(\sigma^2\).

Step by step solution

01

Identifying Probabilities and Values

First, we list the possible outcomes for the soap sizes and their respective probabilities: - Size of 25 oz with a probability of 0.2. - Size of 40 oz with a probability of 0.5. - Size of 65 oz with a probability of 0.3.
02

Calculating Expected Value (E(X))

Calculate the expected value (mean) of the distribution, \( \mu \):\[ \mu = (25)(0.2) + (40)(0.5) + (65)(0.3) = 46.5 \text{ oz} \].
03

Determine Sample Mean Distribution

Since \( X_1 \) and \( X_2 \) are selected independently, the sample mean \( \bar{X} = \frac{X_1 + X_2}{2} \). Calculate \( E(\bar{X}) = E\left(\frac{X_1 + X_2}{2}\right) = \frac{E(X_1) + E(X_2)}{2} = \mu = 46.5 \).
04

Comparing E(\bar{X}) to \(\mu\)

The expected value of the sample mean \(E(\bar{X})\) is the same as the population mean:\( \mu \), which is \(46.5 \text{ oz} \), confirming both are equal.
05

Determining Sample Variance \(S^2\)

Calculate the variance \( \sigma^2 \) of the distribution:\[ \sigma^2 = E(X^2) - \mu^2 \], where \(E(X^2) = (25^2)(0.2) + (40^2)(0.5) + (65^2)(0.3) = 2252.5\).Thus, \[ \sigma^2 = 2252.5 - 46.5^2 = 146.75 \].
06

Calculating E(S^2)

For two independent samples, \[ E(S^2) = \frac{n}{n-1}\sigma^2 = \sigma^2 = 146.75 \text{ oz}^2 \]. This implies \( E\left(S^2\right) \) should equal \( \sigma^2 \) when the sample size is very large.
07

Comparing E(S^2) to \(\sigma^2\)

Here, \( E(S^2) = \sigma^2 = 146.75 \text{ oz}^2 \). This equality holds because we used the unbiased formula for sample variance with enough samples.

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Key Concepts

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

Expected Value
When dealing with the expected value, or simply the mean, you're essentially looking at the "center" of a probability distribution. It's like asking "On average, where can we expect our outcomes to land?" Here, we examine the different sizes of soap packages and their probabilities.
The calculation of the expected value, denoted as \( E(X) \) or \( \mu \), involves multiplying each outcome by its probability and summing up all these products. In our scenario with soap sizes, we have:
  • 25 oz with 20% probability,
  • 40 oz with 50% probability,
  • 65 oz with 30% probability.
To find the expected size of the soap package, the formula is \( \mu = (25)(0.2) + (40)(0.5) + (65)(0.3) \). Simplify this to get \( 46.5 \) oz.
This means, on average, a randomly selected soap package will tend to be around 46.5 ounces. This value serves as a benchmark whenever you analyze other statistics related to the distribution.
Sample Variance
Sample variance might seem a bit tricky at first, but it really helps understand how spread out the data is around the expected value. It's like measuring how much soap sizes deviate from the average size. Variance tells us about the variability, or dispersion, in your data.
To calculate the variance of a distribution \( \sigma^2 \), you start with finding \( E(X^2) \), which involves squaring each size, multiplying by its probability, and adding the results. In this problem, it's \[ E(X^2) = (25^2)(0.2) + (40^2)(0.5) + (65^2)(0.3) = 2252.5 \].
Use the variance formula: \( \sigma^2 = E(X^2) - \mu^2 \), giving \( 146.75 \) oz虏.
Next, calculate the expected sample variance \( E(S^2) \), known as the unbiased estimator, which is \( \frac{n}{n-1}\sigma^2 \), crucial for estimating what variance looks like when we only have a few samples. When an appropriate number of samples are considered, \( E(S^2) \) becomes \( \sigma^2 \), indicating a similar variability structure to our original population just estimated differently.
Population Mean
The population mean is a foundational concept in statistics. It's the average of every possible observation in a whole population, basically if you thought of a giant "average" measuring stick. It's not just an educated guess like sample means, but rather the "true" average if you had unlimited resources and could actually measure everything.
In the exercise, population mean is calculated from the expected value of different soap sizes, giving us \( \mu = 46.5 \) oz. This represents the average size that anyone buying soap would end up with. It's the value that the sample mean \( E(\bar{X}) \) attempts to estimate.
Why is it important? Because it helps in understanding the overall behavior and expectation of data. Anytime you're trying to gauge what's normal in data, population mean helps set that baseline. It's an anchor in sea of data exploration, guiding you through averages, variances, and all sorts of statistical investigations.

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Most popular questions from this chapter

Suppose the proportion of rural voters in a certain state who favor a particular gubernatorial candidate is \(.45\) and the proportion of suburban and urban voters favoring the candidate is .60. If a sample of 200 rural voters and 300 urban and suburban voters is obtained, what is the approximate probability that at least 250 of these voters favor this candidate?

There are 40 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of \(6 \mathrm{~min}\) and a standard deviation of \(6 \mathrm{~min}\). a. If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins? b. If the sports report begins at \(11: 10\), what is the probability that he misses part of the report if he waits until grading is done before turning on the TV?

If the amount of soft drink that I consume on any given day is independent of consumption on any other day and is normally distributed with \(\mu=13 \mathrm{oz}\) and \(\sigma=2\) and if I currently have two six-packs of 16 -oz bottles, what is the probability that I still have some soft drink left at the end of 2 weeks (14 days)?

A shipping company handles containers in three different sizes: (1) \(27 \mathrm{ft}^{3}(3 \times 3 \times 3)\), (2) \(125 \mathrm{ft}^{3}\), and (3) \(512 \mathrm{ft}^{3}\). Let \(X_{i}(i=1,2,3)\) denote the number of type \(i\) containers shipped during a given week. With \(\mu_{i}=E\left(X_{i}\right)\) and \(\sigma_{i}^{2}=V\left(X_{i}\right)\), suppose that the mean values and standard deviations are as follows: $$ \begin{array}{lll} \mu_{1}=200 & \mu_{2}=250 & \mu_{3}=100 \\ \sigma_{1}=10 & \sigma_{2}=12 & \sigma_{3}=8 \end{array} $$ a. Assuming that \(X_{1}, X_{2}, X_{3}\) are independent, calculate the expected value and variance of the total volume shipped. b. Would your calculations necessarily be correct if the \(X_{i} \mathrm{~s}\) were not independent? Explain.

The mean weight of luggage checked by a randomly selected tourist-class passenger flying between two cities on a certain airline is \(40 \mathrm{lb}\), and the standard deviation is \(10 \mathrm{lb}\). The mean and standard deviation for a business-class passenger are \(30 \mathrm{lb}\) and \(6 \mathrm{lb}\), respectively. a. If there are 12 business-class passengers and 50 touristclass passengers on a particular flight, what are the expected value of total luggage weight and the standard deviation of total luggage weight? b. If individual luggage weights are independent, normally distributed rv's, what is the probability that total luggage weight is at most \(2500 \mathrm{lb}\) ?
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