Question

For the random variables described, find and graph the probability distribution for \(x .\) Then calculate the mean, variance, and standard deviation. A piece of electronic equipment contains 6 computer chips, two of which are defective. Three chips are randomly selected and inspected, and \(x\), the number of defective chips in the selection is recorded.

Short answer

Answer: The probability distribution is given by \(P(x=0) = \frac{1}{5}\), \(P(x=1) = \frac{3}{5}\), and \(P(x=2) = \frac{1}{5}\). The mean is \(\mu = 1\), the variance is \(\sigma^2 = 1\), and the standard deviation is \(\sigma = 1\).

Step-by-step solution

Find the probabilities for each possible value of \(x\)

The random variable \(x\) can take on values of 0, 1, or 2 defective chips, as all 3 chips being defective is impossible in this scenario. For each possible value of \(x\), we need to calculate the probability by determining the number of ways to achieve that value and dividing by the total number of ways to select 3 chips from a set of 6.

Step 1a: Calculate the total number of ways to select 3 chips

Using the combination formula, we find the total number of ways to select 3 chips out of 6: $$C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$

Step 1b: Calculate the probability of x = 0 defective chips

To get 0 defective chips in the selection, we can only pick all 3 non-defectives. The number of ways to select 3 non-defectives from 4 is: $$C(4,3) = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4$$ Therefore, the probability of getting 0 defective chips is: $$P(x=0) = \frac{4}{20} = \frac{1}{5}$$

Step 1c: Calculate the probability of x = 1 defective chip

For 1 defective chip in the selection, we need to pick 1 defective chip from 2, and 2 non-defective chips from 4. The number of ways for this selection is: $$C(2,1) \times C(4,2) = 2 \times 6 = 12$$ The probability of getting 1 defective chip is: $$P(x=1) = \frac{12}{20} = \frac{3}{5}$$

Step 1d: Calculate the probability of x = 2 defective chips

For 2 defective chips in the selection, we need to pick both defective chips from 2, and 1 non-defective chip from 4. The number of ways for this selection is: $$C(2,2) \times C(4,1) = 1 \times 4 = 4$$ The probability of getting 2 defective chips is: $$P(x=2) = \frac{4}{20} = \frac{1}{5}$$ Now we have the probability distribution of \(x\): $$P(x=0) = \frac{1}{5}, \quad P(x=1) = \frac{3}{5}, \quad P(x=2) = \frac{1}{5}$$

Calculate the mean of the distribution

To calculate the mean, we multiply the value of \(x\) with its corresponding probability and sum the results: $$\mu = \sum xP(x) = 0 \times \frac{1}{5} + 1 \times \frac{3}{5} + 2 \times \frac{1}{5} = 1$$

Calculate the variance and standard deviation of the distribution

To calculate the variance, we first calculate \(E(x^2)\) and then use the formula \(\sigma^2 = E(x^2) - \mu^2\): $$E(x^2) = \sum x^2P(x) = 0^2 \times \frac{1}{5} + 1^2 \times \frac{3}{5} + 2^2 \times \frac{1}{5} = 2$$ Now, we can find the variance: $$\sigma^2 = E(x^2) - \mu^2 = 2 - 1^2 = 1$$ Finally, we find the standard deviation by taking the square root of the variance: $$\sigma = \sqrt{\sigma^2} = \sqrt{1} = 1$$

Create the graph of the probability distribution

To graph the probability distribution, create a bar chart with \(x\) values on the horizontal axis and their corresponding probabilities on the vertical axis. There will be 3 bars representing the probabilities of having 0, 1, and 2 defective chips in the selection with heights of \(\frac{1}{5}\), \(\frac{3}{5}\), and \(\frac{1}{5}\) respectively. In conclusion, the probability distribution for \(x\) is: $$P(x=0) = \frac{1}{5}, \quad P(x=1) = \frac{3}{5}, \quad P(x=2) = \frac{1}{5}$$ The mean, variance, and standard deviation are: $$\mu = 1, \quad \sigma^2 = 1, \quad \sigma = 1$$

Key concepts

Random Variable

A random variable is a fundamental concept in probability theory that essentially assigns numerical values to the outcomes of random phenomena. For instance, consider a situation where electronic equipment contains 3 computer chips, and we're interested in the amount of defective chips. Here, the random variable 'x' represents the number of defective chips found during inspection. This can lead to several discrete outcomes in which 'x' can take on specific values: 0, 1, or 2 defective chips. Such variables are discrete random variables because they arise from processes that can have countable outcomes.

When we analyze random variables, we develop a probability distribution, which, as you might suspect, is a table or an equation that links each possible value the variable can take with its probability of occurring. In the provided exercise, the solution involved finding the probability distribution by first determining all possible values of 'x' and then calculating the corresponding probabilities for those values.

Mean of Distribution

The mean of a distribution, often symbolized as \(\mu\) in the realm of probability and statistics, reflects the center or 'average' value expected for a random variable. To find the mean of the discrete probability distribution, we multiply each possible value of the random variable by its probability and then sum all these products. This process is called the expected value of a random variable.

In our example, we calculate the mean \(\mu\) by taking the sum of all the possible values of the random variable 'x' each weighted by their probability: \[\mu = \sum xP(x)\]. The computation of the expected value is a vital component in understanding the behavior of a random variable because it provides a measure of central tendency—a single value indicating where on the distribution's graph one might expect most values to cluster around.

Variance and Standard Deviation

Variance, represented by \(\sigma^2\), is a measure of how spread out the values of a random variable are from the mean. In other words, it quantifies the degree of variation or dispersion within a set of possible outcomes. The standard deviation \(\sigma\) is the square root of the variance and it gives us a measure of spread in the same units as the random variable.

To compute the variance, we first need to find the expected value of the square of 'x', denoted by \(E(x^2)\). We then subtract the square of the mean \(\mu^2\) from this expected value. The standard deviation is the square root of the variance, which provides an easily interpretable value of how much the outcomes deviate on average from the mean. In our exercise, the standard deviation came out to be 1, indicating a moderate level of variability relative to the mean value of 'x' which was also 1. Understanding the concepts of variance and standard deviation is crucial for students as these metrics are often used to describe the uncertainty and reliability of measurements in diverse fields.