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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 \(0.0001 ?\)

Short Answer

Expert verified
a) Probability that the load exceeds 4300 pounds is 0.0228. b) Design limit exceeded with probability 0.0001 is approximately 4578 pounds.

Step by step solution

01

Understand the Problem

We need to determine the probability that the combined weight of 25 people exceeds 4300 pounds given that the weights are normally distributed with a mean of 160 pounds and a standard deviation of 30 pounds.
02

Define the Random Variable

Let \( X \) be the total weight of 25 people. Each person has a weight \( W_i \) that is normally distributed with mean \( \mu = 160 \) and standard deviation \( \sigma = 30 \). The total weight \( X \) is then \( X = \sum_{i=1}^{25} W_i \).
03

Calculate Mean and Standard Deviation for Total Weight

Since weights are independent, the mean of the total weight \( X \) is \( \mu_{X} = 25 \times 160 = 4000 \). The standard deviation of the total weight is \( \sigma_{X} = \sqrt{25} \times 30 = 150 \). Thus, \( X \sim N(4000, 150) \).
04

Find the Probability using Z-Score

We need to find \( P(X > 4300) \). The Z-score for 4300 is \( Z = \frac{4300 - 4000}{150} = \frac{300}{150} = 2 \).
05

Calculate Probability from Z-Table

Using a standard normal distribution table, find \( P(Z > 2) \). \( P(Z > 2) \approx 0.0228 \). Thus, \( P(X > 4300) \approx 0.0228 \).
06

Design Limit for Part (b)

We need to find the weight \( L \) such that \( P(X > L) = 0.0001 \). Using the Z-table, find the Z-score that corresponds to a cumulative probability of \( 1 - 0.0001 = 0.9999 \). This Z-score is approximately \( 3.719 \).
07

Solve for the Design Limit

Using the Z-score formula, \( Z = \frac{L - 4000}{150} \), rearrange to find \( L \): \( L = 3.719 \times 150 + 4000 = 4577.85 \). Therefore, the design limit that is exceeded with a probability of \( 0.0001 \) is approximately 4578 pounds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation
In probability calculation, we are often interested in determining the likelihood of a particular event occurring. This involves analyzing data and forming predictions based on statistical methods. In this context, calculating probability means understanding how likely it is for the total weight of people in an elevator to exceed a certain threshold—specifically, the elevator's weight limit.

To calculate the probability that a group of people's total weight surpasses a designed limit, one may use the properties of the normal distribution, which is a common technique when dealing with normally distributed data. In the case of normally distributed weights, knowing the mean and standard deviation allows us to predict how likely it is that a sum (like total weight) exceeds the limit.

This can be achieved by using the Z-score formula, which translates the data into standard normal distribution values. By determining the Z-score for a particular weight limit, we can then refer to a Z-table to find related probabilities, helping us understand how likely it is to exceed that weight.
  • The total weight is considered a random variable following a normal distribution.
  • The Z-score helps in comparing this weight against any threshold.
  • Using Z-score tables facilitates an easier way to pinpoint associated probabilities.
Z-score
The Z-score concept is crucial in probability calculation, especially when dealing with normally distributed data. A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It's expressed as the number of standard deviations a data point is from the mean.

The formula for calculating the Z-score is: \( Z = \frac{X - \mu}{\sigma} \) where \( X \) is the raw score (or observed value), \( \mu \) is the mean of the data set, and \( \sigma \) is the standard deviation. Z-scores can help standardize scores on different scales, making them easier to compare.

In context, for the probability of exceeding the elevator's weight limit, we calculate the Z-score to understand how far this particular weight (4300 pounds) is from the average total weight of 4000 pounds. By calculating the Z-score, which in this example turned out to be 2, we can use standard normal distribution tables to find the probability that the total weight exceeds 4300 pounds.
  • A positive Z-score indicates the observed value is above the mean.
  • A negative Z-score shows it’s below the mean.
  • The magnitude indicates how far it deviates from the average value.
Mean and Standard Deviation
Mean and standard deviation are fundamental concepts in statistics, especially when dealing with normal distributions. The mean, often symbolized as \( \mu \), is the average value of a data set, indicating the central point around which data points are distributed.

The standard deviation, denoted as \( \sigma \), measures how spread out the values are around the mean. It essentially tells us how much variation or dispersion exists in the data set. A low standard deviation means the data points tend to be close to the mean, while a high standard deviation indicates they are more spread out.

In the elevator problem, the mean weight of individuals is given as 160 pounds, and the standard deviation is 30 pounds. When considering a group of 25 people, we calculate the mean and standard deviation of their total weight.

The mean total weight becomes \( 25 \times 160 = 4000 \) pounds, and the standard deviation of the total weight is \( \sqrt{25} \times 30 = 150 \). Understanding these figures helps us use the normal distribution to predict the probability of different weight outcomes.
  • The mean provides a central value to gauge if data points are typical or unusual.
  • The standard deviation gives us an idea of the consistency or variability within a data set.
  • Both are essential for determining probabilities in normal distribution scenarios.
Independent Random Variables
Independent random variables are a key concept in probability and statistics. They refer to variables whose outcomes do not affect each other. In other words, the occurrence of one event does not change the probability distribution of another event. This concept is essential when calculating the total weight of individuals in a group, as we can consider each person's weight independently.

For instance, in the exercise, each person's weight is considered an independent random variable, following a normal distribution with a mean of 160 pounds and a standard deviation of 30 pounds. Because of this independence, we can easily calculate the mean and standard deviation of the total weight by aggregating 25 individuals.

The total weight is then treated as another normal distribution, dependent on the combination of these individual distributions. Understanding that these variables are independent simplifies the calculation of the total mean and standard deviation when dealing with a group.
  • Each individual's weight is considered separately, simplifying total weight calculations.
  • Independence means one person's weight does not influence another’s likely weight.
  • This concept is pivotal in summing random variables to form a new distribution.

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