/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 One piece of PVC pipe is to be i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(.5\) in. The length of the second piece is a normal rv with mean and standard deviation 15 and \(.4\) in., respectively. The amount of overlap is normally distributed with mean value \(1 \mathrm{in}\). and standard deviation \(.1 \mathrm{in}\). Assuming that the lengths and amount of overlap are independent of each other, what is the probability that the total length after insertion is between \(34.5\) and 35 in.?

Short Answer

Expert verified
The probability is approximately 15.88%.

Step by step solution

01

Define Random Variables

Let \(L_1\) be the length of the first piece of PVC pipe, \(L_2\) the length of the second piece, and \(O\) the overlap between them. \(L_1 \sim N(20, 0.5^2)\), \(L_2 \sim N(15, 0.4^2)\), and \(O \sim N(1, 0.1^2)\).
02

Define Total Length Formula

The total length of the pipe after inserting the first pipe into the second is \(T = L_1 + L_2 - O\). According to the properties of normally distributed variables, \(T\) is also normally distributed.
03

Calculate Mean and Standard Deviation

Since means and variances can be added for independent random variables, calculate the mean and variance of \(T\):\[ E(T) = E(L_1) + E(L_2) - E(O) = 20 + 15 - 1 = 34 \]\[ \text{Var}(T) = \text{Var}(L_1) + \text{Var}(L_2) + \text{Var}(O) = 0.5^2 + 0.4^2 + 0.1^2 = 0.42 \]Thus, \(T \sim N(34, 0.42)\).
04

Standardize and Calculate Probability

We need to find \(P(34.5 < T < 35)\). First, standardize the distribution:\[ Z = \frac{T - 34}{\sqrt{0.42}} \]The standardized interval is:\[ \frac{34.5 - 34}{\sqrt{0.42}} < Z < \frac{35 - 34}{\sqrt{0.42}} \]Compute the standardized values:\[ Z_1 = \frac{0.5}{\sqrt{0.42}} \approx 0.77 \]\[ Z_2 = \frac{1}{\sqrt{0.42}} \approx 1.54 \]
05

Use Standard Normal Distribution Table

Using the standard normal distribution table, find the probabilities:\[ P(Z < 0.77) \approx 0.7794 \]\[ P(Z < 1.54) \approx 0.9382 \]Therefore:\[ P(0.77 < Z < 1.54) = P(Z < 1.54) - P(Z < 0.77) = 0.9382 - 0.7794 = 0.1588 \]
06

Interpret the Probability Result

Thus, the probability that the total length after insertion is between 34.5 and 35 inches is approximately 15.88%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Normal Distribution
The normal distribution is a probability distribution that is widely used in statistics and is often referred to as a Gaussian distribution. It is characterized by its bell-shaped curve, which is symmetric around its mean, indicating that data near the mean are more frequent in occurrence. The normal distribution is completely defined by two parameters:
  • The mean (\(\mu\)), which is the center of the distribution and the point of symmetry.
  • The standard deviation (\(\sigma\)), which determines the spread of the distribution. A larger standard deviation indicates a wider spread of data.
In practical applications, normal distributions are used to represent real-valued random variables whose distributions are unknown. In this particular exercise, the lengths of PVC pieces and the overlap are modeled using normal distributions. Each length is assumed to follow a normal distribution with given means and standard deviations, reflecting the variability in manufacturing or measuring processes.
Random Variables
Random variables are fundamental components of probability and statistics. A random variable represents numerical outcomes of a random phenomenon. There are two types of random variables:
  • Discrete random variables take on a countable number of distinct values, often integers, such as the number of heads in a series of coin flips.
  • Continuous random variables can assume an infinite number of values within a given range, such as lengths or weights. They often follow probability distributions like the normal distribution.
In the exercise, three random variables are involved:
  • \(L_1\): the length of the first PVC piece.
  • \(L_2\): the length of the second PVC piece.
  • \(O\): the overlap between the two pieces when combined.
These random variables are considered independent, meaning the length of each piece and the overlap are not influenced by one another. This assumption simplifies the analysis since it allows the use of properties like the addition of variances and means, forming the calculation for the total length distributed normally.
Standard Deviation
Standard deviation is a crucial concept in statistics that measures the amount of variation or dispersion in a set of values. It tells us how spread out these values are from the mean.Mathematically, standard deviation (\(\sigma\)) is defined as the square root of the variance (\(\sigma^2\)). In other words, it is:\[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} \]where:
  • \(N\) is the number of observations,
  • \(x_i\) are the individual observations,
  • \(\mu\) is the mean of the data set.
In the given exercise, different standard deviations are specified for each part, such as \(0.5\) inches for \(L_1\) and \(0.4\) inches for \(L_2\), illustrating the natural variability in the lengths. The standard deviation of the total length is determined in part by these individual standard deviations.
Expected Value
The expected value, also known as the mean, is a central concept in probability and statistics that provides the average outcome of a random variable if an experiment is repeated many times. It is denoted as \(E(X)\) for a random variable \(X\).For a continuous random variable, the expected value is computed using an integral of the function weighted by its probability:\[ E(X) = \int_{-\infty}^{\infty} x f(x) \, dx \]where \(f(x)\) is the probability density function.In the context of the exercise, the expected value represents the calculated means of the PVC lengths and the overlap:
  • The mean of \(L_1\) is 20 inches,
  • The mean of \(L_2\) is 15 inches,
  • The mean of \(O\) is 1 inch.
The expected value for the total length, \(T = L_1 + L_2 - O\), sums these means to provide an expected total length of 34 inches. This expected value is critical for predicting the most likely duration of the final PVC pipe's length.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose the distribution of the time \(X\) (in hours) spent by students at a certain university on a particular project is gamma with parameters \(\alpha=50\) and \(\beta=2\). Because \(\alpha\) is large, it can be shown that \(X\) has approximately a normal distribution. Use this fact to compute the probability that a randomly selected student spends at most \(125 \mathrm{~h}\) on the project.

Suppose that for a certain individual, calorie intake at breakfast is a random variable with expected value 500 and standard deviation 50 , calorie intake at lunch is random with expected value 900 and standard deviation 100 , and calorie intake at dinner is a random variable with expected value 2000 and standard deviation 180 . Assuming that intakes at different meals are independent of each other, what is the probability that average calorie intake per day over the next (365-day) year is at most 3500 ? [Hint: Let \(X_{i}, Y_{i}\), and \(Z_{i}\) denote the three calorie intakes on day \(i\). Then total intake is given by \(\Sigma\left(X_{i}+Y_{i}+Z_{i}\right)\).]

A company maintains three offices in a region, each staffed by two employees. Information concerning yearly salaries (1000's of dollars) is as follows: $$ \begin{array}{lcccccc} \text { Office } & 1 & 1 & 2 & 2 & 3 & 3 \\ \text { Employee } & 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Salary } & 29.7 & 33.6 & 30.2 & 33.6 & 25.8 & 29.7 \end{array} $$ a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary \(\bar{X}\). b. Suppose one of the three offices is randomly selected. Let \(X_{1}\) and \(X_{2}\) denote the salaries of the two employees. Determine the sampling distribution of \(X\). c. How does \(E(X)\) from parts (a) and (b) compare to the population mean salary \(\mu\) ?

The National Health Statistics Reports dated Oct. 22,2008 stated that for a sample size of 277 18year-old American males, the sample mean waist circumference was \(86.3 \mathrm{~cm}\). A somewhat complicated method was used to estimate various population percentiles, resulting in the following values: \(\begin{array}{lllllll}5 \text { th } & 10 \mathrm{th} & 25 \mathrm{th} & 50 \mathrm{th} & 75 \mathrm{th} & 90 \mathrm{th} & 95 \mathrm{th} \\ 69.6 & 70.9 & 75.2 & 81.3 & 95.4 & 107.1 & 116.4\end{array}\) a. Is it plausible that the waist size distribution is at least approximately normal? Explain your reasoning. If your answer is no, conjecture the shape of the population distribution. b. Suppose that the population mean waist size is \(85 \mathrm{~cm}\) and that the population standard deviation is \(15 \mathrm{~cm}\). How likely is it that a random sample of 277 individuals will result in a sample mean waist size of at least \(86.3 \mathrm{~cm}\) ? c. Referring back to (b), suppose now that the population mean waist size is \(82 \mathrm{~cm}\) (closer to the median than the mean). Now what is the (approximate) probability that the sample mean will be at least \(86.3\) ? In light of this calculation, do you think that 82 is a reasonable value for \(\mu\) ?

The lifetime of a type of battery is normally distributed with mean value \(10 \mathrm{~h}\) and standard deviation \(1 \mathrm{~h}\). There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only \(5 \%\) of all packages?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.