Problem 72 - Applied Statistics and Probability for Engineers

Question

Assume that the weights of individuals are independent and normally distributed with a mean of 160 pounds and a standard deviation of 30 pounds. Suppose that 25 people squeeze into an elevator that is designed to hold 4300 pounds. (a) What is the probability that the load (total weight) exceeds the design limit? (b) What design limit is exceeded by 25 occupants with probability \(0.0001 ?\)

Step-by-Step Solution

Answer

(a) Probability ≈ 0.0228. (b) Design limit is ≈ 4559 pounds.

Step 1: Identify the Variables

Let's denote the weights of the individuals as random variables. Since there are 25 people, the total weight can be expressed as \( X = X_1 + X_2 + \ldots + X_{25} \), where each \( X_i \) follows a normal distribution with mean \( \mu = 160 \) pounds and standard deviation \( \sigma = 30 \) pounds.

Step 2: Calculate Mean and Standard Deviation of Total Weight

The sum \( X = X_1 + X_2 + \ldots + X_{25} \) follows a normal distribution with mean \( \mu_X = 25 \times 160 = 4000 \) pounds and standard deviation \( \sigma_X = \sqrt{25} \times 30 = 150 \) pounds, because the variances add up.

Step 3: Find Probability for Part (a)

We need to find the probability that the total weight exceeds 4300 pounds. This can be expressed as \( P(X > 4300) \). First, standardize by calculating \( Z = \frac{X - \mu_X}{\sigma_X} = \frac{4300 - 4000}{150} = \frac{300}{150} = 2 \). Then, the probability \( P(X > 4300) \) is \( P(Z > 2) \). Looking at standard normal distribution tables or using a calculator, \( P(Z > 2) \approx 0.0228 \).

Step 4: Solve for Part (b) Design Limit

We need to find a weight \( W \) such that \( P(X > W) = 0.0001 \). Looking at standard normal tables, a Z-score corresponding to 0.0001 in the upper tail is approximately 3.719. We use the formula \( W = \mu_X + Z \cdot \sigma_X \), giving \( W = 4000 + 3.719 \times 150 \approx 4558.85 \) pounds.

Key Concepts

Normal Distribution

In statistics, the normal distribution is one of the most important probability distributions. It is often referred to as the bell curve due to its distinct shape, which is symmetrical and bell-shaped. The normal distribution is used to depict real-valued random variables whose distributions are not known, making it applicable in countless fields from natural and social sciences to business and economics.

Some key characteristics of a normal distribution include:

• The mean, median, and mode are all equal and located at the center of the distribution.
• It is defined by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)).
• The total area under the curve is 1, representing the total probability of all possible outcomes.

The weights of the individuals in our exercise follow a normal distribution with a mean of 160 pounds and a standard deviation of 30 pounds. Understanding this distribution allows us to predict and calculate various probabilities relating to the weights.

Random Variables

A random variable is a numerical outcome of a random phenomenon or experiment. It can take various values, each associated with a certain probability. Random variables can be either discrete (with a countable number of outcomes) or continuous (with an uncountable number of outcomes, such as weights or heights).

In the original exercise, each individual's weight is considered a random variable, denoted as \( X_i \). Since we examine 25 people, their total weight is represented by the sum of these random variables: \( X = X_1 + X_2 + \ldots + X_{25} \).
The randomness associated with these variables leads us to use probability theory to evaluate the probability of exceeding certain weight limits in the provided scenarios. Hence, the concept of random variables is crucial for solving these types of problems.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means the data points tend to be close to the mean of the set, while a high standard deviation indicates that the data points are spread out over a wider range of values.

For normally distributed data, about 68% of the data will fall within one standard deviation of the mean, 95% within two, and 99.7% within three. This is known as the empirical rule or 68-95-99.7 rule.

In the given problem, each person’s weight distribution has a standard deviation of 30 pounds. To find the standard deviation of the total weight of 25 people, we use \( \sigma_X = \sqrt{25} \times 30 = 150 \) pounds. This calculated value helps in determining probabilities related to the total weight.

Z-score

The Z-score is a numerical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. A Z-score of 0 indicates the value is exactly at the mean, while a Z-score of +2.5 or -2.5 shows the value is 2.5 standard deviations away from the mean.

Z-scores are a way to standardize scores across different normal distributions. It's particularly useful in calculating probabilities and comparing different data sets.

In our exercise, we calculate a Z-score to standardize the total weight exceeding the elevator's design limit. Specifically, we find \( Z = \frac{X - \mu_X}{\sigma_X} = \frac{4300 - 4000}{150} = 2 \), which reveals that a total weight of 4300 pounds is two standard deviations above the mean of the total weight.

Mean and Variance

The mean, often referred to as the average, is a measure of the central tendency of a set of values. The variance is a measure of how much the values in the dataset differ from the mean.

In probability theory, when dealing with a set of random variables that are normally distributed, the mean of these variables is crucial in determining their expected value. Meanwhile, variance describes how spread out those variables are around the mean, with variance being the square of the standard deviation.

In the elevator problem, the mean total weight (\( \mu_X \)) of 25 individuals is calculated as \( 25 \times 160 = 4000 \) pounds. The variance of individual weights is \( \sigma^2 = 900 \) (since \( 30^2 = 900 \)), and the variance of their combined weight becomes \( 25 \times 900 = 22500 \), leading to a standard deviation of 150 pounds. Together, these values allow us to make informed predictions about the total weight's behavior.