Probability Distribution And Mean Of Defective Items In A Sample

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In this article, we delve into the fascinating world of probability distributions, specifically focusing on a scenario involving defective items in a sample. Imagine you have a lot of items, some of which are defective, and you randomly select a sample from this lot. The number of defective items you find in your sample is a random variable, and understanding its probability distribution is crucial in many quality control and risk assessment applications. We will explore how to calculate this probability distribution and then determine the mean, which represents the average number of defective items we expect to find in our sample.

In our specific case, we have a lot of 6 items, with 2 of them being defective. We draw a sample of 4 items at random, without replacement, meaning once an item is selected, it's not put back into the lot. Our random variable, denoted by X, represents the number of defective items in the sample. Our goal is two-fold: first, to find the probability distribution of X, which tells us the probability of observing each possible value of X; and second, to calculate the mean of X, also known as the expected value, which gives us a sense of the typical number of defective items we'd expect in a sample of 4.

This problem is a classic example of a hypergeometric distribution, which arises when we sample without replacement from a finite population containing two types of items (in our case, defective and non-defective). The principles and techniques we'll discuss here are applicable to a wide range of similar problems in various fields, including manufacturing, healthcare, and finance. Let's embark on this journey of discovery and unravel the mysteries of probability distributions and expected values.

(i) Probability Distribution of X

To determine the probability distribution of X, where X is the number of defective items in the sample, we need to calculate the probability of each possible value that X can take. Since we are drawing a sample of 4 items from a lot of 6 items containing 2 defective items, the possible values for X are 0, 1, and 2. We cannot have more than 2 defective items in the sample because there are only 2 defective items in the entire lot.

The hypergeometric distribution is the perfect tool for this scenario. It helps us calculate the probability of getting a specific number of successes (defective items in our case) in a sample drawn without replacement from a finite population. The formula for the hypergeometric probability mass function is:

P(X = k) = (C(K, k) * C(N - K, n - k)) / C(N, n)

Where:

  • N is the total number of items in the population (6 in our case).
  • K is the total number of defective items in the population (2 in our case).
  • n is the number of items drawn in the sample (4 in our case).
  • k is the number of defective items in the sample (0, 1, or 2).
  • C(a, b) represents the number of combinations of choosing b items from a items, also known as "a choose b", and is calculated as a! / (b! * (a - b)!).

Let's calculate the probabilities for each value of X:

  • P(X = 0): This is the probability of drawing 0 defective items (and therefore 4 non-defective items). Plugging the values into the formula:

    P(X = 0) = (C(2, 0) * C(6 - 2, 4 - 0)) / C(6, 4)
         = (C(2, 0) * C(4, 4)) / C(6, 4)
         = (1 * 1) / 15
         = 1/15

  • P(X = 1): This is the probability of drawing 1 defective item and 3 non-defective items:

    P(X = 1) = (C(2, 1) * C(6 - 2, 4 - 1)) / C(6, 4)
         = (C(2, 1) * C(4, 3)) / C(6, 4)
         = (2 * 4) / 15
         = 8/15

  • P(X = 2): This is the probability of drawing 2 defective items and 2 non-defective items:

    P(X = 2) = (C(2, 2) * C(6 - 2, 4 - 2)) / C(6, 4)
         = (C(2, 2) * C(4, 2)) / C(6, 4)
         = (1 * 6) / 15
         = 6/15
         = 2/5

Therefore, the probability distribution of X is:

  • P(X = 0) = 1/15
  • P(X = 1) = 8/15
  • P(X = 2) = 2/5

This distribution tells us the likelihood of observing each possible number of defective items in our sample. For instance, there's a 1/15 chance of finding no defective items, an 8/15 chance of finding one defective item, and a 2/5 chance of finding two defective items.

Verifying the Probability Distribution

It's always a good practice to verify that the probabilities in a probability distribution sum up to 1. This ensures that we have accounted for all possible outcomes. In our case:

P(X = 0) + P(X = 1) + P(X = 2) = 1/15 + 8/15 + 2/5 = 1/15 + 8/15 + 6/15 = 15/15 = 1

Since the probabilities sum to 1, our calculated probability distribution is valid.

(ii) Mean of X

The mean of a random variable, also known as the expected value, represents the average value we would expect to observe if we repeated the experiment many times. For a discrete random variable like X, the mean is calculated as the sum of each possible value multiplied by its probability:

E[X] = Σ [x * P(X = x)]

Where:

  • E[X] represents the expected value or mean of X.
  • x represents the possible values of X (0, 1, and 2 in our case).
  • P(X = x) represents the probability of X taking the value x.
  • Σ denotes the summation over all possible values of x.

Using the probability distribution we calculated earlier, we can find the mean of X:

E[X] = (0 * P(X = 0)) + (1 * P(X = 1)) + (2 * P(X = 2))
     = (0 * 1/15) + (1 * 8/15) + (2 * 2/5)
     = 0 + 8/15 + 4/5
     = 8/15 + 12/15
     = 20/15
     = 4/3

Therefore, the mean of X is 4/3, which is approximately 1.33. This means that, on average, we would expect to find about 1.33 defective items in a sample of 4 drawn from the lot. It's important to remember that we can't actually observe 1.33 defective items in a single sample, as the number of defective items must be a whole number. The mean represents the average number of defective items we would expect over many samples.

Alternative Formula for the Mean of a Hypergeometric Distribution

There's also a direct formula for calculating the mean of a hypergeometric distribution, which can be a handy shortcut:

E[X] = n * (K / N)

Where:

  • n is the sample size (4 in our case).
  • K is the number of defective items in the population (2 in our case).
  • N is the total number of items in the population (6 in our case).

Let's apply this formula to our problem:

E[X] = 4 * (2 / 6)
     = 4 * (1/3)
     = 4/3

As we can see, this formula gives us the same result as our previous calculation, confirming the accuracy of our probability distribution and mean.

Conclusion

In this article, we have successfully determined the probability distribution of the random variable X, representing the number of defective items in a sample drawn from a lot, and calculated its mean. We found that the probability distribution follows a hypergeometric distribution, and we calculated the probabilities for each possible value of X: 0, 1, and 2. We then calculated the mean of X, which represents the average number of defective items we would expect to find in a sample, using both the definition of the mean for a discrete random variable and the direct formula for the mean of a hypergeometric distribution. Both methods yielded the same result: 4/3, or approximately 1.33.

Understanding the probability distribution and mean of a random variable is crucial in various applications, particularly in quality control and risk management. By knowing the probability of observing different numbers of defective items, we can make informed decisions about the quality of a product or the risk associated with a particular process. The concepts and techniques we've explored here are applicable to a wide range of similar problems, making this a valuable skill for anyone working with data and statistics.

This example highlights the power of probability distributions in describing and predicting random phenomena. By understanding the underlying distribution, we can gain valuable insights into the behavior of the system we are studying and make more informed decisions. The hypergeometric distribution, in particular, is a powerful tool for analyzing situations involving sampling without replacement, and its applications extend far beyond the simple example we've discussed here.

As a final thought, remember that the mean is just one measure of the central tendency of a distribution. To get a more complete picture, it's often helpful to also consider other measures, such as the variance and standard deviation, which quantify the spread or variability of the distribution. Exploring these concepts further will deepen your understanding of probability and statistics and equip you with even more powerful tools for analyzing data and making decisions in the face of uncertainty.