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Suppose that \(X\) has a hypergeometric distribution with \(N=100, n=4,\) and \(K=20 .\) Determine the following: a. \(P(X=1)\) b. \(P(X=6)\) c. \(P(X=4)\) d. Mean and variance of \(X\)

Short Answer

Expert verified
a. \( P(X=1) \approx 0.418 \) b. \( P(X=6) = 0 \) c. \( P(X=4) \approx 0.001235 \) d. Mean = 0.8, Variance \( \approx 0.617 \).

Step by step solution

01

Understand the Hypergeometric Distribution

A hypergeometric distribution describes the probability of a certain number of successes from a finite population without replacement. The parameters for this distribution are:\[ N \] (the population size), \[ K \] (number of success states in the population), \[ n \] (number of draws), and \[ x \] (number of observed successes). The probability of observing exactly \( x \) successes is given by: \[ P(X = x) = \frac{\binom{K}{x} \binom{N-K}{n-x}}{\binom{N}{n}} \]
02

Calculate \(P(X=1)\)

To find the probability \( P(X=1) \), substitute \( x=1 \), \( N=100 \), \( n=4 \), and \( K=20 \) into the hypergeometric probability formula:\[ P(X = 1) = \frac{\binom{20}{1} \binom{80}{3}}{\binom{100}{4}} \]. Calculate each component:\[ \begin{align*} \binom{20}{1} &= 20, \ \binom{80}{3} &= 82160, \ \binom{100}{4} &= 3921225. \end{align*} \]Then the probability is \[ P(X = 1) = \frac{20 \times 82160}{3921225} \approx 0.418. \]
03

Calculate \(P(X=6)\)

For \( P(X=6) \), observe that this is not possible since you can only draw a maximum of 4 elements. Therefore, \( P(X=6) = 0. \)
04

Calculate \(P(X=4)\)

To find \( P(X=4) \), use the formula with \( x=4 \):\[ P(X = 4) = \frac{\binom{20}{4} \binom{80}{0}}{\binom{100}{4}} \]. Compute each:\[ \begin{align*} \binom{20}{4} &= 4845, \ \binom{80}{0} &= 1, \ \binom{100}{4} &= 3921225. \end{align*} \]Thus, \[ P(X = 4) = \frac{4845 \times 1}{3921225} \approx 0.001235. \]
05

Calculate Mean and Variance of \(X\)

The mean \( \mu \) of a hypergeometric distribution is given by \( \mu = n \frac{K}{N} \). The variance \( \sigma^2 \) is \( \sigma^2 = n \frac{K}{N} \left(1 - \frac{K}{N}\right) \frac{N-n}{N-1} \). Substitute \( K=20, N=100, n=4 \):\[ \begin{align*} \mu &= 4 \times \frac{20}{100} = 0.8, \ \sigma^2 &= 4 \times \frac{20}{100} \times \frac{80}{100} \times \frac{96}{99} \approx 0.617. \end{align*} \]

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Key Concepts

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

Probability Calculation
In the world of statistics, probability calculation is essential for predicting outcomes. For the hypergeometric distribution, calculating probabilities can be done with specific parameters. This distribution is used to determine the likelihood of a certain number of successes in draws without replacement from a finite population.
The formula to compute these probabilities is:
  • \( P(X = x) = \frac{\binom{K}{x} \binom{N-K}{n-x}}{\binom{N}{n}} \)
Where:
  • \( N \) is the total population size.
  • \( K \) is the number of success states in the population.
  • \( n \) is the number of draws.
  • \( x \) is the specific number of observed successes.
To calculate \( P(X=1) \), input the values: \( N=100, K=20, n=4, x=1 \) into the formula. Remember each part:
  • \( \binom{20}{1} = 20 \)
  • \( \binom{80}{3} = 82160 \)
  • \( \binom{100}{4} = 3921225 \)
Then \( P(X=1) \) calculates to approximately 0.418. Calculating \( P(X=6) \) yields 0 since the maximum successes in 4 draws can't exceed 4. And for \( P(X=4) \), the calculation gives approximately 0.001235. Understanding these calculations builds a solid foundation in probability.
Mean and Variance
The mean and variance are crucial concepts for understanding a distribution's characteristics. In a hypergeometric distribution, the mean (or expected value) indicates the average number of successes you can expect in your sample.
The formula to compute the mean, \( \mu \), is:
  • \( \mu = n \frac{K}{N} \)
Using our parameters, where \( K = 20 \), \( N = 100 \), and \( n = 4 \), plug them into the formula:
  • \( \mu = 4 \times \frac{20}{100} = 0.8 \)
For variance, which measures how much the number of successes varies from the mean, we use:
  • \( \sigma^2 = n \frac{K}{N} \left(1 - \frac{K}{N}\right) \frac{N-n}{N-1} \)
Substituting the values:
  • \( \sigma^2 = 4 \times \frac{20}{100} \times \frac{80}{100} \times \frac{96}{99} \approx 0.617 \)
Understanding mean and variance in this context helps predict and understand distribution behavior and variability.
Discrete Probability Distributions
Discrete probability distributions play an essential role in statistics when outcomes are distinct and separate. A hypergeometric distribution is one among several such types that specifically deals with instances where sampling is done without replacement, leading to changing probabilities with each draw.
Compared to other discrete distributions like the binomial distribution, where sampling is with replacement, hypergeometric distribution captures more complex real-life scenarios where the total population size matters significantly.
  • The outcomes in hypergeometric are finite, making it suitable for quality control or lottery type problems.
  • It is defined by a set number of trials, each of which results in one of two outcomes: success or failure.
Important points to remember include that as each draw from the population occurs, the makeup of the population changes, affecting future probabilities. This means that for hypergeometric distributions, probability calculi are more intricate but also more reflective of certain realistic settings.
Exploring discrete probability distributions like the hypergeometric distribution aids significantly in studying various practical situations where each sample affects the next.

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