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Chapter 5: Problem 67

One piece of PVC pipe is to be inserted inside another piece. The length of the first piece is normally distributed with mean value \(20 \mathrm{in}\). and standard deviation \(5 \mathrm{in}\). The length of the second piece is a normal rv with mean and standard deviation \(15 \mathrm{in}\). and \(.4 \mathrm{in}\)., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation \(1 \mathrm{in}\). Assuming that the lengths and amount of overlap are independent of one another, what is the probability that the total length after insertion is between \(34.5 \mathrm{in}\). and 35 in.?

Short Answer

Expert verified
The probability is approximately 0.038.

Step by step solution

01

Define the Random Variables

Let's designate the length of the first piece of PVC pipe as \(L_1\) and the length of the second piece as \(L_2\). The overlap is denoted by \(O\). Then, \(L_1 \sim N(20, 5^2)\) and \(L_2 \sim N(15, 0.4^2)\), and \(O \sim N(1, 1^2)\).
02

Determine the Total Length RV

The total length after insertion, \(T\), is given by \(T = L_1 + L_2 - O\). Since the lengths and overlap are independent, the distribution of \(T\) is also normal. The mean is \(\mu_T = \mu_1 + \mu_2 - \mu_O = 20 + 15 - 1 = 34\).
03

Calculate the Standard Deviation of Total Length

For the standard deviation of \(T\), we use the formula for the sum of independent random variables: \(\sigma_T = \sqrt{\sigma_1^2 + \sigma_2^2 + \sigma_O^2} = \sqrt{5^2 + 0.4^2 + 1^2} = \sqrt{26.16} \approx 5.115\).
04

Calculate the Z-scores

To find the probability that \(34.5 \leq T \leq 35\), calculate the z-scores for 34.5 and 35: \(z_{34.5} = \frac{34.5 - 34}{\sigma_T} = \frac{34.5 - 34}{5.115} \approx 0.0978\)\(z_{35} = \frac{35 - 34}{\sigma_T} = \frac{35 - 34}{5.115} \approx 0.1956\).
05

Find the Probability

Using a standard normal distribution table, find the probabilities for the z-scores:\(P(Z < 0.0978) \approx 0.5390\)\(P(Z < 0.1956) \approx 0.5770\).Thus, the probability that \(34.5 \leq T \leq 35\) is \(P(34.5 \leq T \leq 35) = P(Z < 0.1956) - P(Z < 0.0978) = 0.5770 - 0.5390 = 0.038\).

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Key Concepts

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

Understanding Normal Distribution
The normal distribution is a foundational concept in statistics that describes how data points cluster around a mean. Imagine a bell curve, where most values land near the center or peak, known as the mean. In a normal distribution, data shows a symmetric distribution, with no skew. This allows us to calculate probabilities using the properties of the normal curve, which is characterized by its mean and standard deviation.
  • The curve is symmetrical around the mean.
  • About 68% of values lie within one standard deviation of the mean.
  • Approximately 95% within two standard deviations.
  • Almost 99.7% within three standard deviations.
The concept applies to many natural phenomena, making it useful to predict outcomes in various fields like finance, science, and engineering.
Exploring Random Variables
Random variables (RVs) are the foundation of probability and statistics. They help in modeling and understanding different outcomes of random processes. In the context of our exercise, we had to deal with three random variables: the lengths of two PVC pipes and their overlap. Each of these was described by a probability distribution, which in this case, was normal.
  • A random variable can be discrete (e.g., the roll of a die) or continuous (e.g., the length of a pipe).
  • Every random variable is associated with a set of probabilities.
  • The probability distribution provides a complete characterization of the variable.
Understanding random variables allows us to quantify uncertainty and make informed decisions based on probable outcomes.
Calculating Z-Scores
Z-scores are a statistical measurement that quantifies the distance of a data point from the mean of a distribution in terms of standard deviations. It's a way to compare different data points within a normal distribution. In our example, calculating z-scores allowed us to find probabilities using the standard normal distribution table.
  • Z-score formula: \( z = \frac{x - \mu}{\sigma} \)
  • \( x \) is the value we are examining.
  • \( \mu \) is the mean of the distribution.
  • \( \sigma \) is the standard deviation of the distribution.
Calculating z-scores is crucial when determining the probability of a value falling within a certain range in a normal distribution.
The Role of Standard Deviation
Standard deviation is a statistical measure that quantifies the amount of variation or dispersion in a set of values. In a normal distribution, it is a key parameter because it determines the spread of the data around the mean.
  • Larger standard deviation means data points are spread out over a wider range of values.
  • Smaller standard deviation indicates that data points are closer to the mean.
  • Calculating standard deviation involves using the variance, which is the average of the squared differences from the mean.
Understanding the standard deviation helps in predicting the likelihood of outcomes and is crucial for the accurate application of the normal distribution in problem-solving scenarios like our PVC pipe exercise.

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Most popular questions from this chapter

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