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A bookstore has 15 copies of a particular textbook, of which 6 are first printings and the other 9 are second printings (later printings provide an opportunity for authors to correct mistakes). Suppose that 5 of these copies are randomly selected, and let \(X\) be the number of first printings among the selected copies. 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
a. Hypergeometric, \(N=15, K=6, n=5\); b. \(P(X=2)\approx 0.420, P(X\leq2)\approx 0.642, P(X\geq2)\approx 0.778\); c. Mean \(=2\), SD \(\approx0.986\).

Step by step solution

01

Identify the distribution

In this scenario, we are dealing with a hypergeometric distribution. This is applicable when we are selecting objects without replacement from a finite population consisting of two types of objects. We have \(N = 15\) textbooks, with \(K = 6\) first printings, \(N-K = 9\) second printings, and we are selecting \(n = 5\) books.
02

Compute probabilities using the hypergeometric distribution

The probability mass function (PMF) for a hypergeometric distribution is given by: \[ P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}}{{\binom{N}{n}}} \]For \(P(X=2)\), we have \(k = 2\), so \[ P(X = 2) = \frac{{\binom{6}{2} \binom{9}{3}}}{{\binom{15}{5}}} = \frac{{15 \times 84}}{{3003}} \approx 0.420 \].For \(P(X \leq 2)\), we calculate \(P(X=0) + P(X=1) + P(X=2)\):- \(P(X=0) = \frac{{\binom{6}{0} \binom{9}{5}}}{{\binom{15}{5}}} = \frac{126}{3003} \approx 0.042\)- \(P(X=1) = \frac{{\binom{6}{1} \binom{9}{4}}}{{\binom{15}{5}}} = \frac{540}{3003} \approx 0.180\)- Add these: \(P(X \leq 2) \approx 0.042 + 0.180 + 0.420 = 0.642 \).For \(P(X \geq 2)\), we compute \(1 - P(X \leq 1)\):- \(P(X \leq 1) = P(X=0) + P(X=1) \approx 0.042 + 0.180 = 0.222\)- Thus, \(P(X \geq 2) = 1 - 0.222 = 0.778\).
03

Calculate the mean and standard deviation

The mean \(\mu\) of a hypergeometric distribution is given by \[ \mu = n \left( \frac{K}{N} \right) = 5 \times \frac{6}{15} = 2 \].The standard deviation \(\sigma\) is given by\[ \sigma = \sqrt{n \frac{K}{N} \left(1 - \frac{K}{N}\right) \frac{N-n}{N-1}} = \sqrt{5 \times \frac{6}{15} \times \frac{9}{15} \times \frac{10}{14}} \approx 0.986 \].

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Key Concepts

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

Probability Mass Function
When working with the hypergeometric distribution, the Probability Mass Function (PMF) is an essential concept. The PMF helps us determine the probability of a particular number of successes in a sample drawn without replacement from a finite population. The PMF is expressed mathematically as follows:
  • We identify the population size, which is the total number of items, denoted as \( N \).
  • The number of success states in the population is \( K \), which refers to the items of interest.
  • We select \( n \) objects without replacement.
  • Finally, \( k \) is the number of observed successes in your sample.
The PMF for the hypergeometric distribution is given by: \[P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}}{{\binom{N}{n}}}\]
This formula helps us calculate the probability of drawing exactly \( k \) successes from the population of \( N \) objects. In simpler terms, it tells us the likelihood of picking a specific number of desired items from a group without putting any back after each pick. This is vital when you are dealing with scenarios like quality checks or card games where the sequence or replacement of items does not occur.
Random Selection
Random selection is a fundamental concept when working with probability and distribution models. In the context of the hypergeometric distribution, it refers to choosing items from a known set without bias and without replacement. The essence of random selection includes:
  • Each item has an equal chance of being selected initially.
  • Once an item is chosen, it is not replaced, altering the probability of subsequent selections.
  • This type of selection helps model real-world scenarios more accurately than models with replacement, when trying to capture everyday events.

Random selection in this exercise involves selecting 5 copies of a textbook from a group of 15, where some are first printings, and others are second. The randomness ensures that any combination of selections can occur, distinguished by the count of first and second printings. In probability terms, this could influence events like lottery draws, card games, or any context where items are drawn from a finite set.
Mean and Standard Deviation
Calculating the mean and standard deviation is crucial to understanding the spread and average distribution of data in a hypergeometric context. For a hypergeometric distribution:
  • The mean \( \mu \) provides the expected number of successes and is calculated as \( \mu = n \left( \frac{K}{N} \right) \).
  • Standard deviation \( \sigma \) measures the variability or spread of the distribution. It's calculated as \[ \sigma = \sqrt{n \cdot \frac{K}{N} \cdot \left( 1 - \frac{K}{N} \right) \cdot \frac{N-n}{N-1}} \]

In this textbook example:
  • The mean or expected number of first printings among the selected copies is 2, meaning that, on average, out of the 5 books chosen, 2 will be first printings.
  • The standard deviation is approximately 0.986, indicating that the number of first printings can vary around this mean. This tells us how spread out the number of first printings is likely to be.
These metrics guide how much variation exists and what typical outcomes one might expect when an experiment is replicated multiple times.
Textbook Distribution
In probability theory, a textbook distribution, like a hypergeometric distribution, applies to specific scenarios characterized by selection without replacement. This distribution is descriptive of situations where each draw affects subsequent outcomes. Let’s summarize the key features tailored for our textbook example:
  • We start with 15 textbooks, 6 first printings, and 9 second printings.
  • We select 5 books at random, which means we are looking for the probability of drawing a specific count of first printings among those selections.
  • This relates closely to real-life textbook issues, such as quality control processes or inventory checks.

By understanding textbook distributions, one gains valuable insight into how the characteristics of the population (like count and type of books) directly influence the selection process and its outcomes. This concept helps us develop strategies in fields beyond textbooks, focusing on environments where different types of objects are present and decision-making follows specific constraints and limitations.

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Most popular questions from this chapter

Suppose that trees are distributed in a forest according to a two-dimensional Poisson process with parameter \(\alpha\), the expected number of trees per acre, equal to 80 . a. What is the probability that in a certain quarter-acre plot, there will be at most 16 trees? b. If the forest covers 85,000 acres, what is the expected number of trees in the forest? c. Suppose you select a point in the forest and construct a circle of radius \(1 \mathrm{mile}\). Let \(X=\) the number of trees within that circular region. What is the pmf of \(X\) ? [Hint: 1 sq mile \(=640\) acres.]

A trial has just resulted in a hung jury because eight members of the jury were in favor of a guilty verdict and the other four were for acquittal. If the jurors leave the jury room in random onder and each of the first four leaving the room is accosted by a reporter in quest of an interview, what is the pmf of \(X=\) the number of jurors favoring acquittal among those interviewed? How many of those favoring acquittal do you expect to be interviewed?

A second-stage smog alert has been called in an area of Los Angeles County in which there are 50 industrial firms. An inspector will visit 10 randomly selected firms to check for violations of regulations. a. If 15 of the firms are actually violating at least one regulation, what is the pmf of the number of firms visited by the inspector that are in violation of at least one regulation? b. If there are 500 firms in the area, of which 150 are in violation, approximate the pmf of part (a) by a simpler pmf. c. For \(X=\) the number among the 10 visited that are in violation, compute \(E(X)\) and \(V(X)\) both for the exact pmf and the approximating pmf in part (b).

Each of 12 refrigerators has been retumed to a distributor because of an audible, high-pitched, oscillating noise when the refrigerator is running. Suppose that 7 of these refrigerators have a defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order, let \(X\) be the number among the first 6 examined that have a defective compressor. Compute the following: a. \(P(X=5)\) b. \(P(X \leq 4)\) c. The probability that \(X\) exceeds its mean yalue by more than 1 standard deviation. d. Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If \(X\) is the number among 15 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) \(P(X \leq 5)\) than to use the hypergeometric pmf.

Suppose that only \(25 \%\) of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible. What is the probability that, of 20 randomly chosen drivers coming to an intersection under these conditions, a. At most 6 will come to a complete stop? b. Exactly 6 will come to a complete stop? c. At least 6 will come to a complete stop? d. How many of the next 20 drivers do you expect to come to a complete stop?

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