Problem 21 - A First Course in Probability

Question

Suppose that the height, in inches, of a 25 -year-old man is a normal random variable with parameters \(\mu=71\) and \(\sigma^{2}=6.25 .\) What percentage of 25 -year-old men are taller than 6 feet, 2 inches? What percentage of men in the 6-footer club are taller than 6 feet, 5 inches?

Step-by-Step Solution

Answer

Approximately 11.51% of 25-year-old men are taller than 6 feet 2 inches (74 inches), and approximately 0.82% of men in the 6-footer club are taller than 6 feet 5 inches (77 inches).

Step 1: Convert given heights to inches

First, we will convert the given heights 6 feet 2 inches and 6 feet 5 inches to inches for easier comparison. 6 feet = 6 * 12 = 72 inches So, 6 feet 2 inches = 72 + 2 = 74 inches 6 feet 5 inches = 72 + 5 = 77 inches

Step 2: Calculate z-scores

The z-score is calculated as follows: \(z = \frac{x - \mu}{\sigma}\) Where \(x\) is the height in inches, \(\mu\) is the mean height, and \(\sigma\) is the standard deviation of the height. For 74 inches: \(z_1 = \frac{74 - 71}{\sqrt{6.25}}\) For 77 inches: \(z_2 = \frac{77 - 71}{\sqrt{6.25}}\) Let's calculate the z-scores.

Step 3: Calculate the z-scores for the given heights

First, we need to find the standard deviation \(\sigma\): \(\sigma = \sqrt{6.25} = 2.5\) Now calculate the z-scores for the heights: \(z_1 = \frac{74 - 71}{2.5} = 1.2\) \(z_2 = \frac{77 - 71}{2.5} = 2.4\)

Step 4: Utilize the standard normal distribution table to find the percentage

Now that we have the z-scores for the heights, we can use the standard normal distribution table (or a calculator that gives the area under the standard normal curve) to find the percentage of 25-year-old men taller than the given height. For \(z=1.2\), consulting the table or using a calculator, we get a value of 0.8849. To find the percentage of men taller than 6 feet 2 inches (74 inches), we subtract the value we found from 1: Percentage taller than 74 inches = \(1 - 0.8849 = 0.1151 = 11.51\%\) For \(z=2.4\), consulting the table or using a calculator, we get a value of 0.9918. To find the percentage of men taller than 6 feet 5 inches (77 inches), we subtract the value we found from 1: Percentage taller than 77 inches = \(1 - 0.9918 = 0.0082 = 0.82\%\)

Step 5: Answer the questions

Now that we have the percentages, we can answer the two questions: 1. What percentage of 25-year-old men are taller than 6 feet 2 inches (74 inches)? Answer: Approximately 11.51% 2. What percentage of men in the 6-footer club are taller than 6 feet 5 inches (77 inches)? Answer: Approximately 0.82%

Key Concepts

Z-Score Calculation

Understanding z-score calculations is pivotal for analyzing data relative to the mean of a group. A z-score, or standard score, converts individual data points into a universal metric that can be compared across different sets of data.

When dealing with normal random variables such as height, the z-score indicates how many standard deviations an element is from the mean. The formula utilized for z-score is as follows: \[ z = \frac{x - \mu}{\sigma} \]where \( x \) represents the value of interest, \( \mu \) is the mean, and \( \sigma \) denotes the standard deviation.

In the context of height, calculating a z-score allows us to determine how tall a person is compared to the average height. For instance, a positive z-score means a person is taller than average, while a negative one signifies the opposite. This method helps in transforming the diverse range of height measurements into a standardized format, simplifying the process of determining the percentage of the population that falls above or below a particular height.

Standard Normal Distribution

The standard normal distribution is a specialized case of the normal distribution where the mean (\( \mu \)) is 0 and the standard deviation (\( \sigma \)) is 1. This distribution forms the basis for z-score interpretation because it allows us to refer to a single, common probability distribution.

When we calculate a z-score, we are effectively translating our normal random variable into a standard normal variable. The beauty of this approach is that we can use pre-calculated values from the standard normal distribution table — or the Z-table — to find probabilities directly.

For example, a z-score of 1.2 or 2.4 would correspond to a specific percentage of the area under the curve of the standard normal distribution. These values allow us to make exact probability calculations about our data, just like the percentage of 25-year-old men who are taller than a given height in the given problem.

Probability Calculations

Probability calculations in statistics are essential for quantifying the likelihood of different outcomes. When using the standard normal distribution, these probabilities tell us the proportion of data points that fall within a particular range.

Once we have the z-score for a data point, we turn to the Z-table or use statistical software to find the corresponding probability. The table offers cumulative probabilities from the left tail of the distribution up to the z-score. In our example, we're looking for the probability of men being taller than specific heights, which requires the complementary probability – the area to the right of the z-score on the curve.

By subtracting the obtained probability from 1, we find the percentage of the normative population that exceeds the set height. The facility of this process eliminates complex integrations and derives accurate probabilistic interpretations rapidly, aiding in answering questions like the ones posed by the textbook exercise.

Normal Random Variables

Normal random variables are data points that are assumed to conform to the normal distribution, which is the bell-shaped curve frequently observed in natural and social phenomena. Such variables are characterized by their mean and variance, and height is a classic example.

In practical scenarios like our example with men's heights, the assumption of normality simplifies the estimation of probabilities. It allows statisticians and researchers to make educated guesses and predictions about the occurrence of events based on the properties of the normal distribution.

By using the measures of mean and standard deviation, one can describe the normal distribution of these variables. Standardization via z-scores then permits comparisons across different scales or populations. For educational purposes and ease of understanding, interpreting real-world data in terms of normal random variables streamlines the calculation of probabilities and offers illustrative insight into statistical concepts.