Question

For the random variables described, find and graph the probability distribution for \(x .\) Then calculate the mean, variance, and standard deviation. Five applicants have applied for two positions: two women and three men. All are equally qualified and there is no preference for choosing either gender. Let \(x\) be the number of women chosen to fill the two positions.

Short answer

Answer: The probability distribution of \(x\) is: \(x=0\): \(P(x=0) = \frac{3}{10}\) \(x=1\): \(P(x=1) = \frac{6}{10}\) \(x=2\): \(P(x=2) = \frac{1}{10}\) The mean is \(\frac{8}{10}\), the variance is \(\frac{168}{100}\), and the standard deviation is approximately \(\sqrt{1.68}\).

Step-by-step solution

Identify possible values of \(x\)

We have two women and three men applying for two positions. The random variable \(x\) represents the number of women chosen. Since there are two positions, the range of possible values for \(x\) would be {0, 1, 2}, where 0 means no women chosen, 1 means one woman chosen, and 2 means both women chosen.

Calculate probabilities for each possible value of \(x\)

To calculate the probabilities for each possible value of \(x\), we can use the combinations formula. We have a total of 5 applicants and want to choose 2 positions: Total number of combinations = \({5 \choose 2} = \frac{5!}{2!(5-2)!} = 10\) Now, let's calculate the probabilities for each possible value of \(x\): 1. \(P(x=0)\): Since there are 3 men, and we want to choose both positions from these men, the number of combinations is \({3 \choose 2} = 3\). The probability is therefore \(\frac{3}{10}\). 2. \(P(x=1)\): We want to choose one position from the 2 women and one position from the 3 men. The number of combinations for choosing one woman and one man is \({2 \choose 1}{3 \choose 1} = 2\times 3 = 6\). The probability is therefore \(\frac{6}{10}\). 3. \(P(x=2)\): Since there are 2 women, and we want to choose both positions from these women, the number of combinations is \({2 \choose 2} = 1\). The probability is therefore \(\frac{1}{10}\). Now we have the probability distribution for \(x\): \(x = 0\): \(P(x=0) = \frac{3}{10}\) \(x = 1\): \(P(x=1) = \frac{6}{10}\) \(x = 2\): \(P(x=2) = \frac{1}{10}\)

Calculate the mean (expected value) of \(x\)

To find the mean of the probability distribution, we multiply each value of \(x\) by its corresponding probability and then sum the results: Mean \(= E(x) = 0 \times \frac{3}{10} + 1 \times \frac{6}{10} + 2 \times \frac{1}{10} = \frac{6}{10} + \frac{2}{10} = \frac{8}{10}\)

Calculate the variance of \(x\)

The variance is obtained by taking the weighted average of the squared differences from the mean: Variance \(= Var(x) = (0-\frac{8}{10})^2 \times \frac{3}{10} + (1-\frac{8}{10})^2 \times \frac{6}{10} + (2-\frac{8}{10})^2 \times \frac{1}{10} = \frac{168}{100}\)

Calculate the standard deviation of \(x\)

The standard deviation is simply the square root of the variance: Standard Deviation \(= SD(x) = \sqrt{\frac{168}{100}} = \sqrt{1.68}\) In conclusion, for this given scenario, the probability distribution of \(x\) is as follows: \(x=0\): \(P(x=0) = \frac{3}{10}\) \(x=1\): \(P(x=1) = \frac{6}{10}\) \(x=2\): \(P(x=2) = \frac{1}{10}\) And we also found the mean to be \(\frac{8}{10}\), the variance to be \(\frac{168}{100}\), and the standard deviation to be approximately \(\sqrt{1.68}\).

Key concepts

Combinations Formula

The combinations formula is crucial in calculating the number of ways a set number of items can be selected from a larger group, without taking order into account. It's represented by \[\begin{equation}{n \choose k} = \frac{n!}{k!(n-k)!},\end{equation}\]where \(n!\) is the factorial of \(n\), representing the product of all positive integers up to \(n\). The term \(k!\) is the factorial of \(k\), and \((n - k)!\) is the factorial of the difference between \(n\) and \(k\).For instance, in our example of selecting two positions from five applicants, the total number of combinations is calculated using the combinations formula shown in the step-by-step solution. This formula is essential for understanding the distribution of possibilities in probability and statistics.

Remembering that combinations do not consider the order of selection differentiates them from permutations, which do. Combinations are used in probability distributions, like the one in the exercise, to determine how likely different outcomes are when there is no preference among the outcomes.

Mean Calculation

The mean, or expected value, is a fundamental concept in statistics, representing the average outcome if an experiment were repeated many times. To calculate the mean of a probability distribution, you multiply each possible outcome by its probability and then add these products together. The formula is expressed as\[\begin{equation}E(x) = \sum (x_i \times P(x_i)),\end{equation}\]where \(E(x)\) represents the expected value or mean, \(x_i\) represents each possible outcome, and \(P(x_i)\) is the probability of that outcome occurring.

In simpler terms, you're calculating a weighted average, where each outcome is weighted by how likely it is to occur. This gives you a single number that summarizes the entire probability distribution. For example, in our problem, the mean is calculated as \(\frac{8}{10}\), reflecting the average number of women selected across all possible scenarios.

Variance and Standard Deviation

Variance and standard deviation are measures of dispersion in a probability distribution. They indicate how spread out the values are around the mean. Variance is defined as the average of the squared differences from the mean, calculated using the following formula:\[\begin{equation}Var(x) = \sum ((x_i - E(x))^2 \times P(x_i)),\end{equation}\]where \(Var(x)\) is the variance, \(x_i\) are the values the random variable can take, \(E(x)\) is the expected value or mean, and \(P(x_i)\) is the probability of \(x_i\).

Standard deviation, on the other hand, is just the square root of the variance, representing the average distance of the distribution's values from the mean. It is more commonly used because it is in the same units as the mean. For example, the solution showed us that the standard deviation of \(x\) for our probability distribution is approximately the square root of 1.68, which helps us understand the variability of the number of women chosen for the positions.

Together, variance and standard deviation give us a clear picture of variability, which can be especially useful when comparing different probability distributions or making informed decisions based on the spread of possible outcomes.