Question

A company has five applicants for two positions: two women and three men. Suppose that the five applicants are equally qualified and that no preference is given for choosing either gender. Let \(x\) equal the number of women chosen to fill the two positions. a. Write the formula for \(p(x)\), the probability distribution of \(x\). b. What are the mean and variance of this distribution? c. Construct a probability histogram for \(x\).

Short answer

Answer: The mean of the discrete probability distribution is 0.8, and the variance is 0.56.

Step-by-step solution

Obtaining the sample space for 'x'

For this probability distribution problem, we consider the possible values of 'x', which is the number of women chosen. Since there are only two positions, the possible values for 'x' are 0, 1, and 2.

Calculating the probabilities for each value of 'x'

In this step, we will calculate the probabilities for the three possible values of 'x' (0, 1, and 2). Since the applicants are equally qualified and no preference is given to gender, the events for each case are equally likely. There are a total of 10 possible combinations of choosing the two applicants. - x = 0: This means no women are chosen, and both positions are filled by men (3 men). There are \(C(3,2)=3\) combinations in this case. So, the probability of this event is \(P(x=0)=\frac{3}{10}\). - x = 1: This means one woman and one man are chosen. There are \(C(2,1) \cdot C(3,1) = 2 \cdot 3=6\) combinations in this case. So, the probability of this event is \(P(x=1)=\frac{6}{10}\). - x = 2: This means both women are chosen, and no men are chosen. There is \(C(2,2)=1\) combination in this case. So, the probability of this event is \(P(x=2)=\frac{1}{10}\).

Deriving the formula for p(x), the probability distribution of 'x'

We can now form the probability distribution function for 'x': \(p(x)= \left\{ \begin{array}{ll} \frac{3}{10}, & x=0 \\ \frac{6}{10}, & x=1 \\ \frac{1}{10}, & x=2 \end{array} \right.\)

Calculating the mean and variance of this distribution

To calculate the mean and variance of the distribution, we will use the formulas for the mean (\(E(x)\)) and variance (\(Var(x)\)) of a discrete random variable: \(E(x) = \sum_{i=1}^{n} x_i \cdot p(x_i)\) \(Var(x) = \sum_{i=1}^{n} (x_i - E(x))^2 \cdot p(x_i)\) Mean, E(x): Mean = \((0)\left(\frac{3}{10}\right)+(1)\left(\frac{6}{10}\right)+(2)\left(\frac{1}{10}\right)=\frac{8}{10}=0.8\) Variance, Var(x): Variance = \((0-0.8)^2\left(\frac{3}{10}\right)+(1-0.8)^2\left(\frac{6}{10}\right)+(2-0.8)^2\left(\frac{1}{10}\right)=0.56\) So, the mean and variance of this distribution are 0.8 and 0.56, respectively.

Constructing a probability histogram for 'x'

A probability histogram is a visualization of the probability distribution. In this case, there are three possible values for 'x' (0, 1, and 2) and their corresponding probabilities (\(\frac{3}{10}\), \(\frac{6}{10}\), \(\frac{1}{10}\)). To construct the probability histogram, create a bar graph with 'x' on the horizontal axis and 'p(x)' on the vertical axis. Mark the values of 'x' as 0, 1, and 2. For each value of 'x', draw a rectangle with a width of 1 and height equal to the probability for that value. The resulting histogram should have three bars with heights of \(\frac{3}{10}\), \(\frac{6}{10}\), and \(\frac{1}{10}\), respectively.

Key concepts

Probability Calculation

Probability calculation is a fundamental concept in statistics that helps us understand the likelihood of different outcomes. In this exercise, we are looking at a situation where there are five equally qualified applicants (two women and three men) for two job positions. The goal is to find the probability of choosing a certain number of women, represented by the variable \(x\).

To find the probability of each possible outcome for \(x\), we calculate how many ways we can achieve each situation, then divide it by the total number of possibilities. There are ten total combinations for picking two people out of five.

• When \(x = 0\): No women are selected, both positions filled by men. There are 3 combinations, so the probability is \(P(x=0) = \frac{3}{10}\).
• When \(x = 1\): One woman and one man are chosen. There are 6 combinations, so \(P(x=1) = \frac{6}{10}\).
• When \(x = 2\): Both women are chosen, zero men. There is only 1 combination, so \(P(x=2) = \frac{1}{10}\).

The probabilities add up to 1, ensuring our distribution is valid. The probability function \(p(x)\) succinctly represents these probabilities for each value of \(x\).

Mean and Variance

The mean and variance are critical concepts in understanding the properties of a probability distribution. The mean, or expected value, is a measure of the central tendency of the distribution. It tells us the average outcome we can expect when repeating the experiment many times.

To calculate the mean \(E(x)\) for our distribution, we multiply each value of \(x\) by its probability and sum these products: \[E(x) = 0 \left(\frac{3}{10}\right) + 1 \left(\frac{6}{10}\right) + 2 \left(\frac{1}{10}\right) = 0.8\]

Variance indicates how much the values in the distribution deviate from the mean. It provides a measure of the spread or dispersion: \[Var(x) = (0 - 0.8)^2 \left(\frac{3}{10}\right) + (1 - 0.8)^2 \left(\frac{6}{10}\right) + (2 - 0.8)^2 \left(\frac{1}{10}\right) = 0.56\]

In this context, a mean of 0.8 suggests that, on average, 0.8 women are chosen for the positions. A variance of 0.56 indicates moderate spread around this mean, suggesting that while deviations occur, outcomes close to the mean are slightly more common.

Probability Histogram

A probability histogram is a graphical representation of a probability distribution, helping us visualize the likelihood of each outcome.

Constructing a probability histogram involves creating a bar graph where the x-axis represents the possible number of women chosen \(x\), and the y-axis represents the probability associated with each value \(p(x)\).

• For \(x = 0\), draw a bar with height \(\frac{3}{10}\).
• For \(x = 1\), the bar height is \(\frac{6}{10}\).
• And for \(x = 2\), the bar height is \(\frac{1}{10}\).

Each bar is placed above its corresponding value of \(x\), with all bars having the same width.

This histogram's visual representation makes it easier to see that the probability of selecting one woman is higher compared to selecting none or selecting both. Such visual tools are particularly useful for interpreting the distribution and gaining insights at a glance.