Problem 12 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The minimum age requirement for a regular driver's license differs from state to state. The frequency distribution for this age requirement in the 50 states is given in the following table: $$\begin{array}{lllllll}\hline \text { Minimum } & & & & & & \\\\\text { Age } & 15 & 16 & 17 & 18 & 19 & 21 \\ \hline \text { Frequency of } & & & & & & \\\\\text { Occurrence } & 1 & 15 & 4 & 28 & 1 & 1 \\ \hline\end{array}$$ a. Describe a random variable \(X\) that is associated with these data. b. Find the probability distribution for the random variable \(X\). c. Compute the mean, variance, and standard deviation of \(X\).

Step-by-Step Solution

Answer

The random variable X represents the minimum age at which a person is allowed to get a regular driver's license in a state. The probability distribution of X is given as: $$\begin{array}{ccccccc} \hline X & 15 & 16 & 17 & 18 & 19 & 21 \\ \hline P(X) & \frac{1}{50} & \frac{15}{50} & \frac{4}{50} & \frac{28}{50} & \frac{1}{50} & \frac{1}{50}\\ \hline \end{array}$$ The mean (μ) is 17.18, the variance (σ²) is 1.0348, and the standard deviation (σ) is 1.0173.

Step 1: (Part a: Define the random variable X)

Let X represent the minimum age at which a person is allowed to get a regular driver's license in a state. In this case, X can take on the values 15, 16, 17, 18, 19, and 21.

Step 2: (Part b: Find the probability distribution for X)

To find the probability distribution, divide the frequency of occurrence for each age by the total number of states (50). These probabilities are listed below: For \[X = 15, P(X = 15) = \frac{1}{50} \\ X = 16, P(X = 16) = \frac{15}{50} \\ X = 17, P(X = 17) = \frac{4}{50} \\ X = 18, P(X = 18) = \frac{28}{50} \\ X = 19, P(X = 19) = \frac{1}{50} \\ X = 21, P(X = 21) = \frac{1}{50}\] The probability distribution, now expressed as a table, is given as: $$\begin{array}{ccccccc} \hline X & 15 & 16 & 17 & 18 & 19 & 21 \\ \hline P(X) & \frac{1}{50} & \frac{15}{50} & \frac{4}{50} & \frac{28}{50} & \frac{1}{50} & \frac{1}{50}\\ \hline \end{array}$$

Step 3: (Part c: Compute the mean, variance, and standard deviation of X)

To compute the mean, variance, and standard deviation of X, we'll use the following formulas: Mean (μ): \[\mathrm{E}(X) = \sum_{i=1}^{n} x_i P(x_i)\] Variance (σ²): \[\mathrm{Var}(X) = \sum_{i=1}^{n} (x_i - \mathrm{E}(X))^2 P(x_i)\] Standard deviation (σ): \[\mathrm{SD}(X) = \sqrt{\mathrm{Var}(X)}\] Mean (μ): \[\mathrm{E}(X) = 15\cdot\frac{1}{50} + 16\cdot\frac{15}{50} + 17\cdot\frac{4}{50} + 18\cdot\frac{28}{50} + 19\cdot\frac{1}{50} + 21\cdot\frac{1}{50} = 17.18\] Variance (σ²): \[\mathrm{Var}(X) = (15-17.18)^2\cdot\frac{1}{50} + (16-17.18)^2\cdot\frac{15}{50} + (17-17.18)^2\cdot\frac{4}{50} + (18-17.18)^2\cdot\frac{28}{50} + (19-17.18)^2\cdot\frac{1}{50} + (21-17.18)^2\cdot\frac{1}{50} = 1.0348\] Standard deviation (σ): \[\mathrm{SD}(X) = \sqrt{1.0348} = 1.0173\] The mean (μ) is 17.18, the variance (σ²) is 1.0348, and the standard deviation (σ) is 1.0173.

Key Concepts

Random Variable

In the context of probability and statistics, a random variable is essentially a way to represent the outcomes of a particular experiment or process. In our scenario, the random variable \( X \) is defined as the minimum age at which someone can obtain a driver's license across different states.
It can take on specific values, reflecting the varied age requirements, such as 15, 16, 17, 18, 19, and 21.
In this instance:

• Each possible age (15, 16, 17, 18, 19, 21) represents an outcome that \( X \) can take.
• The frequency with which each age appears in the data can help calculate probability.

Random variables are crucial as they allow us to apply mathematical and statistical techniques to analyze real-world processes by converting qualitative outcomes into quantitative data.

Mean and Variance

The mean and variance are statistical measures used to describe and analyze random variables. The mean, also known as expectation or expected value, is a measure of the 'central tendency' or 'average' of a set of values. For our random variable \( X \), the mean is computed using the formula: \[ \text{Mean} (\mu) = \sum (x_i \times P(x_i)) \]Here, \( x_i \) represents the values taken by the random variable, and \( P(x_i) \) is the probability of each value.
In this scenario, the mean is calculated as 17.18, suggesting that on average, the minimum age for obtaining a driver's license is approximately 17 years and 2 months.
The variance measures how much the values spread out from the mean. It is defined by the formula:\[ \text{Variance} (\sigma^2) = \sum ((x_i - \mu)^2 \times P(x_i)) \]Variances help in understanding the degree of variation and inconsistency within the dataset. Here, we find a variance of 1.0348, indicating moderate variability in the driver's license age requirements across states.

Standard Deviation

The standard deviation is one of the most widely utilized tools for understanding the measure of spread or dispersion within a set of data. It simply represents the square root of the variance. The formula is given by:\[ \text{Standard Deviation} (\sigma) = \sqrt{\text{Variance}} \]This measure gives you insight into how much the values of the random variable \( X \) deviate from the mean, on average. For our example of minimum age requirements, the standard deviation is calculated to be approximately 1.0173.
A smaller standard deviation signifies that the age requirements among different states tend to be closer to the mean age, while a larger standard deviation would suggest more variability. In this case, since the standard deviation is slightly above 1, it indicates that the ages do not deviate too greatly and tend to stay fairly close around the mean age of 17.18. This means most states have similar minimum age requirements for obtaining a driver’s license.