91Ó°ÊÓ

Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerators are 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.

Step by step solution

01

Identify the Distribution

The problem involves selecting items without replacement from a finite population. This is the setting for the hypergeometric distribution, characterized by three parameters: the population size (12), the number of success states in the population (7 defective compressors), and the sample size (6).

02

Calculate the Probability of Exactly Five Defective Compressors

Use the hypergeometric probability mass function to find \( P(X = 5) \). The formula is \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \], where \( N = 12 \), \( K = 7 \), \( n = 6 \), and \( k = 5 \). Thus, \[ P(X = 5) = \frac{\binom{7}{5} \binom{5}{1}}{\binom{12}{6}} = \frac{21 \times 5}{924} = \frac{105}{924} \approx 0.1137 \].

03

Calculate the Probability of at Most Four Defective Compressors

To find \( P(X \leq 4) \), sum the probabilities from 0 to 4 defective compressors: \[ P(X \leq 4) = \sum_{k=0}^{4} P(X = k) \]. You need to calculate \( P(X = 0), P(X = 1), P(X = 2), P(X = 3), \) and \( P(X = 4) \) using the hypergeometric formula and sum them.

04

Compute the Mean and Standard Deviation of X

For the hypergeometric distribution, the mean \( \mu \) is given by \( \mu = n \frac{K}{N} \) and the variance \( \sigma^2 \) is \( n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1} \). Here, \( n = 6, K = 7, N = 12 \). Calculate \( \mu = 6 \frac{7}{12} \approx 3.5 \) and \( \sigma^2 = 6 \cdot \frac{7}{12} \cdot \frac{5}{12} \cdot \frac{6}{11} \approx 1.1364 \) so \( \sigma \approx 1.066 \).

05

Calculate the Probability that X Exceeds its Mean by More Than One Standard Deviation

We want \( P(X > 3.5 + 1.066) \), which is approximately \( P(X > 4.566) \). As \( X \) is discrete, it means finding \( P(X \geq 5) \), which is \( P(X = 5) + P(X = 6) \). Use the earlier calculation for \( P(X = 5) \) and compute \( P(X = 6) \).

06

Large Sample Approximation

For the case of 400 refrigerators with 40 defective compressors, we have a sample of 15. This can be approximated using a binomial distribution due to the large population size. Use \( n = 15 \) and \( p = \frac{40}{400} = 0.1 \) for calculations. Calculate \( P(X \leq 5) \) using the cumulative distribution function of the binomial distribution.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Mass Function

The Probability Mass Function (PMF) is key when dealing with discrete probability distributions like the hypergeometric distribution. Here, it defines the probability of a specific number of successes, given certain conditions.
To calculate the PMF for the hypergeometric distribution, the formula is:
\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]
where:

• \(N\) is the size of the total population; in this case, 12 refrigerators.
• \(K\) is the number of success states (defective compressors), which is 7.
• \(n\) is the number of draws, which is 6.
• \(k\) is the number of successful draws, the value we want to find the probability for.

The PMF helps us understand just how likely it is to draw a certain number of defective compressors from our sample of refrigerators.

Mean and Standard Deviation

In probability distributions, the mean provides the average expected value. For the hypergeometric distribution, it's calculated using:
\[ \mu = n \frac{K}{N} \]
For our problem, substituting the values gives \( \mu = 6 \times \frac{7}{12} \approx 3.5 \). This means, on average, 3.5 defective compressors are expected in a sample of 6.

The standard deviation measures how spread out numbers are from the mean. It's calculated as:
\[ \sigma = \sqrt{n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1}} \]
For our exercise, substituting gives \( \sigma \approx 1.066 \). This indicates how much variation we might expect from the mean number of defective compressors.

Binomial Approximation

When dealing with large populations, as in the case of 400 refrigerators with 40 defective compressors, computing probabilities directly from the hypergeometric PMF can be cumbersome.

Instead, a binomial approximation provides a simpler approach. This is feasible because when the population is large, the hypergeometric distribution closely resembles the binomial distribution.

For this approximation, substitute:

• \(n\) as the sample size, 15.
• \(p\) as the probability of success in a single trial, \( \frac{40}{400} = 0.1 \).

Using the binomial cumulative distribution function, \(P(X \leq 5)\) can be computed more efficiently compared to hypergeometric calculations.

Combinatorics

Combinatorics is the branch of mathematics dealing with combinations and permutations of objects. It plays a crucial role in the hypergeometric distribution.
Combinatorial calculations help determine the number of ways to choose a subset of objects from a larger set.

In the formula:
\[ \binom{K}{k} \binom{N-K}{n-k} \]\

• \(\binom{K}{k}\) is the number of ways to choose \(k\) defective compressors from \(K\), the total defective ones.
• \(\binom{N-K}{n-k}\) is how we select the rest from non-defective compressors.

These are divided by \(\binom{N}{n}\), the number of ways to draw \(n\) from \(N\), ensuring the correct overall probability calculation.