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 probability distribution is as follows: - p(0) (no women chosen) = 9/25 ≈ 0.36 - p(1) (one woman chosen) = 12/25 ≈ 0.48 - p(2) (two women chosen) = 4/25 ≈ 0.16 The mean is 4/5 and the variance is 12/25. The probability histogram has three bars, with heights corresponding to the probability values for x=0 (0.36), x=1 (0.48), and x=2 (0.16).

Step-by-step solution

Identify the binomial distribution parameters (n, p)

To find the probability distribution of x, we first need to identify the number of trials (n) and the probability of success, or in this case, the probability of choosing a woman (p). In our case, n = 2 (two positions to fill) and p = 2/5 (two women out of five applicants). The probability of failure, or choosing a man, q = 1-p = 3/5.

Write the formula for the binomial distribution p(x)

The binomial probability distribution formula is: \(p(x) = \binom{n}{x} \cdot p^x \cdot q^{n-x}\), where \(x\) can take the values 0, 1, or 2 (no women, one woman, or two women chosen), \(n\) is the number of trials (positions to fill), \(p\) is the probability of success (choosing a woman), and \(q\) is the probability of failure (choosing a man).

Calculate p(x) for different values of x

Now, we will calculate p(x) for x=0, x=1, and x=2. \(p(0) = \binom{2}{0} \cdot \left(\frac{2}{5}\right)^0 \cdot \left(\frac{3}{5}\right)^2 = 1 \cdot 1 \cdot \frac{9}{25} = \frac{9}{25}\) \(p(1) = \binom{2}{1} \cdot \left(\frac{2}{5}\right)^1 \cdot \left(\frac{3}{5}\right)^1 = 2 \cdot \frac{2}{5} \cdot \frac{3}{5} = \frac{12}{25}\) \(p(2) = \binom{2}{2} \cdot \left(\frac{2}{5}\right)^2 \cdot \left(\frac{3}{5}\right)^0 = 1 \cdot \frac{4}{25} \cdot 1 = \frac{4}{25}\)

Calculate the mean and variance

The mean and variance of a binomial distribution can be calculated using the formulas: Mean, \(\mu = np\) Variance, \(\sigma^2 = npq\) Using these formulas, we can find the mean and variance for our distribution: \(\mu = 2 \cdot \frac{2}{5} = \frac{4}{5}\) \(\sigma^2 = 2 \cdot \frac{2}{5} \cdot \frac{3}{5} = \frac{12}{25}\)

Construct the probability histogram for x

To construct a probability histogram for x, we will plot the value of x on the x-axis and the probability of each outcome, p(x), on the y-axis. - For x=0 (no women chosen), the probability is \(\frac{9}{25} \approx 0.36\) - For x=1 (one woman chosen), the probability is \(\frac{12}{25} \approx 0.48\) - For x=2 (two women chosen), the probability is \(\frac{4}{25} \approx 0.16\) Now, plot the probabilities as a histogram with x on the x-axis and probability (p(x)) on the y-axis, creating bars with heights corresponding to the probability values for x=0, x=1, and x=2.

Key concepts

Probability Histogram

A probability histogram is a graphical representation that shows the probability of outcomes in a discrete probability distribution. Each bar in the histogram corresponds to an outcome of the random variable and the height of the bar represents the probability of that outcome.

In the case of the company selecting applicants for positions, a probability histogram would effectively illustrate the likelihood of choosing zero, one, or two women. To visualize this, you would place 'number of women chosen' along the x-axis and 'probability' along the y-axis. Taller bars indicate a higher probability of that number of women being selected. For example, the tallest bar would be at x=1 in our case, signifying that it is the most likely scenario.

For exercise optimization, ensuring that each bar is correctly scaled to reflect probabilities will help students better understand the likelihood of each possible outcome in the binomial distribution.

Mean and Variance of Distribution

The mean and variance are fundamental statistical measures that describe the central tendency and the spread of a distribution, respectively. In a binomial probability distribution, the mean tells us the expected average number of successes, while the variance measures the amount of variability around this mean.

For our example, the mean \( \mu \) is calculated by multiplying the number of trials \( n \) by the probability of success \( p \) and is given by \( \mu = np \) which equals \( \frac{4}{5} \) women expected to be chosen. The variance \( \sigma^2 \) is found by multiplying the mean by the probability of failure \( q \) and equals \( \frac{12}{25} \) in our case. This implies a moderately low spread, suggesting that the outcomes are relatively close to the mean number of women selected.

Understanding the mean and variance allows students to grasp the behavior of the distribution beyond just knowing the individual probabilities; it helps them understand the general pattern the distribution follows and how much the outcomes can vary.

Binomial Distribution Parameters

The parameters of a binomial distribution define its shape and probabilities. The number of trials \( n \) and the probability of success on a single trial \( p \) are critical for understanding this type of distribution.

In our applicants' exercise, \( n = 2 \) representing the two available positions, and \( p = \frac{2}{5} \) denotes the probability of choosing a woman, given that there are two women out of five applicants. The parameter \( q \) signifies the probability of not choosing a woman, which is \( q = 1 - p = \frac{3}{5} \). Together, these parameters shape the binomial distribution, allowing us to calculate the probabilities for different values of \( x \), the number of women chosen.

For students to fully understand the binomial distribution, it is crucial that they learn to accurately identify and calculate these parameters, as this will aid them in constructing the probability distribution for various scenarios