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Chapter 3: Problem 68

An electronics store has received a shipment of 20 table radios that have connections for an iPod or iPhone. Twelve of these have two slots (so they can accommodate both devices), and the other eight have a single slot. Suppose that six of the 20 radios are randomly selected to be stored under a shelf where the radios are displayed, and the remaining ones are placed in a storeroom. Let \(X=\) the number among the radios stored under the display shelf that have two slots. a. What kind of a distribution does \(X\) have (name and values of all parameters)? b. Compute \(P(X=2), P(X \leq 2)\), and \(P(X \geq 2)\). c. Calculate the mean value and standard deviation of \(X\).

Short Answer

Expert verified
X follows a hypergeometric distribution; \(P(X=2)\approx0.1192\), \(P(X\leq2)\approx0.1373\), \(P(X\geq2)\approx0.982\). Mean=3.6, SD\(\approx1.063\).

Step by step solution

01

Identify Distribution Type

The problem describes a situation where we have a total of 20 radios, 12 with two slots and 8 with one slot. We are asked to find the distribution of the random variable \(X\), the number of radios with two slots selected. This follows the hypergeometric distribution since the selection is without replacement. In a hypergeometric distribution, considering:- Total population size \(N = 20\)- Number of items of desired type (two slots) \(K = 12\)- Number of draws \(n = 6\)Thus, \(X\) follows a hypergeometric distribution with parameters \(N = 20\), \(K = 12\), and \(n = 6\).
02

Calculate P(X=2)

Using the hypergeometric probability formula:\[P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{\binom{N}{n}}\]Plugging in the given numbers where \(k = 2\):\[P(X = 2) = \frac{{\binom{12}{2} \cdot \binom{8}{4}}}{\binom{20}{6}}\]Calculate each component:- \(\binom{12}{2} = 66\)- \(\binom{8}{4} = 70\)- \(\binom{20}{6} = 38760\)Thus:\[P(X = 2) = \frac{{66 \cdot 70}}{38760} = \frac{4620}{38760} \approx 0.1192\]
03

Calculate P(X 鈮� 2)

To find \(P(X \leq 2)\), sum up the probabilities for \(X = 0, 1, 2\):Using the hypergeometric formula again:- \(P(X = 0) = \frac{{\binom{12}{0} \cdot \binom{8}{6}}}{\binom{20}{6}} = \frac{1 \cdot 28}{38760} = \frac{28}{38760}\)- \(P(X = 1) = \frac{{\binom{12}{1} \cdot \binom{8}{5}}}{\binom{20}{6}} = \frac{12 \cdot 56}{38760} = \frac{672}{38760}\)Now summing these probabilities:\[P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \approx \frac{28}{38760} + \frac{672}{38760} + \frac{4620}{38760}\]\[= \frac{5320}{38760} \approx 0.1373\]
04

Calculate P(X 鈮� 2)

To find \(P(X \geq 2)\), we can either compute \(P(X = 2) + P(X = 3) + \cdots + P(X = 6)\) or use\[P(X \geq 2) = 1 - P(X < 2) = 1 - P(X \leq 1)\]We already calculated:- \(P(X \leq 1) = P(X = 0) + P(X = 1) \approx \frac{700}{38760} = \frac{700}{38760}\)Thus:\[P(X \geq 2) = 1 - \frac{700}{38760} = 1 - \approx 0.0181\]\[\approx 0.982\]
05

Calculate Mean and Standard Deviation

For a hypergeometric distribution, the mean \(\mu\) and variance \(\sigma^2\) are given by:\[\mu = n \left( \frac{K}{N} \right) = 6 \left( \frac{12}{20} \right) = 3.6\]The variance \(\sigma^2\) is:\[\sigma^2 = n \left( \frac{K}{N} \right) \left( \frac{N-K}{N} \right) \left( \frac{N-n}{N-1} \right)\]\[\sigma^2 = 6 \left( \frac{12}{20} \right) \left( \frac{8}{20} \right) \left( \frac{14}{19} \right)\]\[\sigma^2 \approx 1.1316\]Thus the standard deviation \(\sigma\) is:\[\sigma \approx \sqrt{1.1316} \approx 1.063\]So, the mean is 3.6 and the standard deviation is approximately 1.063.

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Key Concepts

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

Random Variable
In statistical terms, a random variable is a variable whose possible values are numerical outcomes of a random phenomenon. In this context, the random variable is denoted by \(X\), which represents the number of radios selected that have two slots.
To further understand this, consider the 20 radios being the population. Among these, 12 have two slots, considered a 'success' in this context. When selecting 6 radios without replacement, \(X\) becomes a hypergeometric random variable, counting the number of successes in the sample.
Unlike other distributions such as the binomial, the hypergeometric distribution does not assume that each draw is independent. This is a crucial aspect and makes it particularly useful in scenarios where the sampling is done without replacement.
Probability Calculation
Calculating probability involves determining the likelihood of various outcomes of a random variable. For a hypergeometric distribution like ours, this requires specific calculations.
In our scenario, if you want to find \(P(X=2)\), you use the hypergeometric probability formula:
\[P(X = k) = \frac{\binom{K}{k} \cdot \binom{N-K}{n-k}}{\binom{N}{n}}\] where \(\binom{a}{b}\) denotes a binomial coefficient. This notation represents the number of ways to choose \(b\) items from a total of \(a\) without regard to order.
Using the formula, you compute \(P(X=2)\) by calculating separate binomial coefficients for different components and then computing their product.
This approach extends to finding cumulative probabilities such as \(P(X \leq 2)\) and \(P(X \geq 2)\), which involve summing individual probabilities or employing complementary probabilities.
Mean and Standard Deviation
The mean and standard deviation of a distribution provide measures of central tendency and variability, respectively. For the hypergeometric distribution, they are calculated based on the parameters of the population and the sample.
The mean \(\mu\) reflects the expected number of two-slot radios among the selected ones:
\[\mu = n \left(\frac{K}{N}\right)\]This formula utilizes the total number of draws \(n\), the number of available two-slot radios \(K\), and the total radios in the shipment \(N\).
The standard deviation \(\sigma\) indicates how much variability or spread exists around the mean:
\[\sigma = \sqrt{n \left(\frac{K}{N}\right) \left(\frac{N-K}{N}\right) \left(\frac{N-n}{N-1}\right)}\]Both calculated values, a mean of 3.6 and a standard deviation of about 1.063, give us insights into what we can typically expect from such selections.

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Most popular questions from this chapter

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, \(60 \%\) can be repaired, whereas the other \(40 \%\) must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

A new battery's voltage may be acceptable \((A)\) or unacceptable \((U)\). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that \(90 \%\) of all batteries have acceptable voltages. Let \(Y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is, \(P(Y=2)\) ? b. What is \(p(3)\) ? [Hint: There are two different outcomes that result in \(Y=3\).] c. To have \(Y=5\), what must be true of the fifth battery selected? List the four outcomes for which \(Y=5\) and then determine \(p(5)\). d. Use the pattern in your answers for parts (a)-(c) to obtain a general formula for \(p(y)\).

Give three examples of Bernoulli rv's (other than those in the text).

According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Planning and Mgmnt., 2005: 383-393), the drought length \(Y\) is the number of consecutive time intervals in which the water supply remains below a critical value \(y_{0}\) (a deficit), preceded by and followed by periods in which the supply exceeds this critical value (a surplus). The cited paper proposes a geometric distribution with \(p=.409\) for this random variable. a. What is the probability that a drought lasts exactly 3 intervals? At most 3 intervals? b. What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

A small market orders copies of a certain magazine for its magazine rack each week. Let \(X=\) demand for the magazine, with pmf \begin{tabular}{l|llllll} \(x\) & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(\frac{1}{15}\) & \(\frac{2}{15}\) & \(\frac{3}{15}\) & \(\frac{4}{15}\) & \(\frac{3}{15}\) & \(\frac{2}{15}\) \end{tabular} Suppose the store owner actually pays \(\$ 2.00\) for each copy of the magazine and the price to customers is \(\$ 4.00\). If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine? [Hint: For both three and four copies ordered, express net revenue as a function of demand \(X\), and then compute the expected revenue.]
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