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Chapter 3: Problem 69

Each of 12 refrigerators of a certain type has been returned 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 value 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.

Short Answer

Expert verified
a. 0.1136, b. 0.5, c. 0.1212, d. Use binomial approximation.

Step by step solution

01

Understanding the Problem

There are 12 refrigerators in total. 7 have defective compressors and 5 do not. We are interested in the number among the first 6 examined that have defective compressors. This situation can be modeled with a hypergeometric distribution.
02

Calculating P(X=5)

The hypergeometric probability, where the number of successes in the population is 7, the population size is 12, the sample size is 6, and we want exactly 5 successes, can be calculated using the formula: \[ P(X=k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}} \]Plug in \(K=7\), \(N=12\), \(n=6\), \(k=5\):\[ P(X=5) = \frac{{\binom{7}{5} \cdot \binom{5}{1}}}{{\binom{12}{6}}} = \frac{{21 \cdot 5}}{{924}} = \frac{105}{924} \approx 0.1136 \].
03

Calculating P(X 鈮� 4)

To find \(P(X \leq 4)\), sum up the probabilities \(P(X=0)\), \(P(X=1)\), \(P(X=2)\), \(P(X=3)\), and \(P(X=4)\) using the hypergeometric formula:\[ P(X=0) = \frac{{\binom{7}{0} \cdot \binom{5}{6}}}{{\binom{12}{6}}} = 0 \]\[ P(X=1) = \frac{{\binom{7}{1} \cdot \binom{5}{5}}}{{\binom{12}{6}}} = \frac{7}{924} \]\[ P(X=2) = \frac{{\binom{7}{2} \cdot \binom{5}{4}}}{{\binom{12}{6}}} = \frac{105}{924} \]\[ P(X=3) = \frac{{\binom{7}{3} \cdot \binom{5}{3}}}{{\binom{12}{6}}} = \frac{175}{924} \]\[ P(X=4) = \frac{{\binom{7}{4} \cdot \binom{5}{2}}}{{\binom{12}{6}}} = \frac{175}{924} \]Sum these probabilities: \(0 + \frac{7}{924} + \frac{105}{924} + \frac{175}{924} + \frac{175}{924} = \frac{462}{924} = 0.5 \).
04

Calculating Expectation and Variance

The mean \(\mu\) of a hypergeometric distribution is \(\mu = \frac{nK}{N} = \frac{6 \times 7}{12} = 3.5 \).The variance \(\sigma^2\) is given by:\[ \sigma^2 = n \cdot \frac{K}{N} \cdot \frac{N-K}{N} \cdot \frac{N-n}{N-1} \]Substitute the values: \(\sigma^2 = 6 \cdot \frac{7}{12} \cdot \frac{5}{12} \cdot \frac{6}{11} = \frac{35}{22} \approx 1.59 \).
05

Calculating P(X > 渭 + 蟽)

The standard deviation \(\sigma\) is approximately \(\sqrt{1.59} \approx 1.26\). Therefore, \(\mu + \sigma \approx 3.5 + 1.26 = 4.76\).We're interested in \(P(X > 4.76)\), which translates to \(P(X \geq 5)\). Sum \(P(X=5)\) and \(P(X=6)\) using earlier computations:\[ P(X=5) \approx 0.1136, \quad P(X=6) = \frac{7}{924} \approx 0.0076 \]Sum: \(0.1136 + 0.0076 \approx 0.1212 \).
06

Alternative Method for Large Shipment

For the large shipment problem with 400 refrigerators and 40 defective, we can use the binomial approximation because \(N\) is much larger than the initial problem. Here, \(n=15\), \(p=\frac{40}{400}=0.1\).Use the binomial distribution:\[ P(X \leq 5) \approx \sum_{k=0}^{5} \binom{15}{k} (0.1)^k (0.9)^{15-k} \].Use computational tools or a statistical table to evaluate this sum, providing an approximate probability.

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Key Concepts

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

Probability Theory
Probability theory is the study of randomness and uncertainty. In situations where outcomes are uncertain, probability theory helps us determine the likelihood of various outcomes. This is crucial in fields like finance, science, and engineering.
Probabilistic models allow us to make predictions and decisions based on data. By understanding these models, like the hypergeometric distribution, one can compute specific probabilities for given scenarios, such as the likelihood of selecting a certain number of defective items out of a larger set.
To compute probabilities using probability theory, we use formulas that consider all possible outcomes and their chances of occurring. This enables us to estimate the probability of specific events happening, which can inform decision-making processes.
Statistical Distribution
Statistical distributions represent the possible values a random variable can take, along with their associated probabilities. Different distributions model different scenarios, making them crucial tools in statistics. The hypergeometric distribution is a specific type of statistical distribution used in situations where objects are chosen without replacement.
In our example, the hypergeometric distribution describes the probability of drawing a certain number of defective refrigerators from a batch, where the composition of defects and non-defects is known. This is different from a binomial distribution, which assumes replacement after each draw.
Understanding which distribution applies to a given problem is vital, as it dictates how you compute probabilities. For the refrigerators, the hypergeometric distribution was appropriate due to the finite sample size and the characteristic of drawing without replacement.
Mean and Variance Calculations
Mean and variance are fundamental components of statistical analysis. The mean provides a measure of the central tendency of a distribution, while the variance indicates how much the data are spread out.
For the hypergeometric distribution in our refrigerator problem, the mean is calculated by taking the product of the sample size, the number of successes in the population, and dividing by the total population size: \[ \mu = \frac{nK}{N} \]Substituting, we find \( \mu = \frac{6 \times 7}{12} = 3.5 \). This value represents the expected number of defective compressors.
The variance measures the distribution's spread, using the formula:\[ \sigma^2 = n \cdot \frac{K}{N} \cdot \frac{N-K}{N} \cdot \frac{N-n}{N-1} \]Substituting the values, the variance calculated is approximately 1.59. Understanding these calculations help us explore how much variation there is around the mean, and can inform expectations about future observations.
Binomial Approximation
The binomial approximation is a method used to simplify computations when dealing with large sample sizes. It is particularly useful when working with a large population where calculating exact probabilities with hypergeometric distribution is cumbersome.
In the case of the large shipment of 400 refrigerators where only a small fraction is defective (40 out of 400), using a binomial distribution serves as a practical shortcut. We approximate the hypergeometric distribution by assuming each selection is independent, which fits the conditions for a binomial distribution with parameters that represent probabilities of a defect and non-defect.The approximation involves substituting the probability of a defect \( p = \frac{40}{400} = 0.1 \), and calculating using the binomial formula. This ease of calculation can be completed with computational tools or statistical tables.
Recognizing when and how to deploy binomial approximation is a key statistical skill, improving efficiency and accuracy in scenarios requiring probability analysis.

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Most popular questions from this chapter

A \(k\)-out-of-n system is one that will function if and only if at least \(k\) of the \(n\) individual components in the system function. If individual components function independently of one another, each with probability \(.9\), what is the probability that a 3 -out-of-5 system functions?

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