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Chapter 5: Problem 10

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 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 \(34.5\) in. and 35 in.?

Short Answer

Expert verified
The probability that the total length is between 34.5 and 35 inches is approximately 0.15.

Step by step solution

01

Determine Total Length Distribution

The total length after insertion, denoted as \( L_T \), is the sum of the lengths of both pieces minus the overlap: \( L_T = L_1 + L_2 - O \), where \( L_1 \) is the length of the first piece, \( L_2 \) is the length of the second piece, and \( O \) is the overlap. Since \( L_1 \), \( L_2 \), and \( O \) are independent normals:\[ L_1 \sim N(20, 0.5^2), \ L_2 \sim N(15, 0.4^2), \ O \sim N(1, 0.1^2) \] then \( L_T \sim N(\mu_T, \sigma_T^2) \), where \( \mu_T = 20 + 15 - 1 \) and \( \sigma_T^2 = 0.5^2 + 0.4^2 + 0.1^2 \).
02

Calculate Mean and Variance

Calculate the mean and variance for the total length distribution. The mean \( \mu_T \) is \( 20 + 15 - 1 = 34 \). The variance \( \sigma_T^2 \) is \( 0.5^2 + 0.4^2 + 0.1^2 = 0.25 + 0.16 + 0.01 = 0.42 \). Hence, \( \sigma_T = \sqrt{0.42} \).
03

Standardize the Normal Distribution

To find the probability that \( L_T \) is between 34.5 and 35, standardize \( L_T \) to the standard normal distribution: \[ Z = \frac{L_T - \mu_T}{\sigma_T} \] where \( \mu_T = 34 \) and \( \sigma_T = \sqrt{0.42} \). Calculate the standardized values \( z_1 \) and \( z_2 \) for the intervals: \[ z_1 = \frac{34.5 - 34}{\sqrt{0.42}} \] \[ z_2 = \frac{35 - 34}{\sqrt{0.42}} \].
04

Compute Z-scores

Compute \( z_1 \) and \( z_2 \): \[ z_1 = \frac{0.5}{\sqrt{0.42}} \approx 0.772 \] \[ z_2 = \frac{1}{\sqrt{0.42}} \approx 1.545 \].
05

Find Probability from Z-table

Use the standard normal distribution Z-table to find the probability for \( z_1 \) and \( z_2 \). The probabilities correspond to the areas under the curve for each \( Z \) score: \( P(Z < 0.772) \) and \( P(Z < 1.545) \). Calculate this difference to find \( P(0.772 < Z < 1.545) \).
06

Calculate Final Probability

Look up \( P(Z < 0.772) \approx 0.78 \) and \( P(Z < 1.545) \approx 0.93 \). Thus, the probability that the total length is between 34.5 and 35 inches is \( P(0.772 < Z < 1.545) = P(Z < 1.545) - P(Z < 0.772) = 0.93 - 0.78 = 0.15 \).

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Key Concepts

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

Standard Deviation
Understanding standard deviation is crucial when working with normal distributions. It tells us how much the data spreads out or deviates from the mean. For example, if you have a set of pipe lengths that are normally distributed, with a mean length, the standard deviation indicates the variability of these lengths.

Let's break it down:
  • A small standard deviation means the lengths are close to the mean. For example, if the mean length is 20 inches with a standard deviation of 0.5 inches, most pipes will be between 19.5 and 20.5 inches.
  • A large standard deviation means a wider spread of lengths around the mean.
In our case, the standard deviations for the pipe lengths are 0.5 inches and 0.4 inches, while the overlap has a standard deviation of 0.1 inches. This tells us each length's variability and helps us determine the total length's variability after insertion.

To calculate the total standard deviation for the combined length of the pipes minus the overlap, you sum the variances (squared standard deviations) and then take the square root. This is because variances, rather than standard deviations, directly add up when summing independent normal random variables.
Probability Calculation
Probability calculation involves finding the likelihood of a particular event within a normal distribution range. To do this for our case, you need to understand the distribution of the total length after the two pipes are combined, accounting for the overlap.

First, determine the total mean (or expected value) and standard deviation for the sum of both pipes minus the overlap. Here, the total mean length is calculated as 34 inches, while the variance is calculated as 0.42.

Then, you standardize your data, converting it to the standard normal distribution using Z-scores. This step simplifies the probability calculation by referencing a standard normal distribution table, where you can look up probability values for any given Z-score. This table gives you the probability that a random value is less than or equal to a specific value (cumulative probability).

Once standardized, find the probability that the total pipe length is between two value limits (such as 34.5 and 35 inches in our example). By calculating the cumulative probabilities for both Z-scores and then finding their difference, you determine the actual probability of the length being within that range. The probability result shows how likely or unlikely a particular length range is when inserting the PVC pipes.
Z-score
The Z-score is a statistical measure that indicates how many standard deviations an element is from the mean. It standardizes the normal distribution, allowing for comparison across different sets of data or different parts of one set.

To calculate a Z-score:
  • Subtract the mean from the value you're interested in.
  • Divide the result by the standard deviation.

In formula terms:
\( Z = \frac{X - \mu}{\sigma} \)
where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

In our PVC pipe example, we use Z-scores to find out how extreme certain lengths are compared to typical lengths. By calculating Z-scores for the lengths of 34.5 and 35 inches, we transformed our problem around the total length into a standard form suitable for analysis using a standard normal distribution table.

These Z-scores provided were 0.772 and 1.545, which were then used to reference the Z-table to get cumulative probabilities, which ultimately helped calculate the probability that the total pipe length falls within a certain range. Using the Z-table is crucial here because it simplifies the complex probability calculations and provides a fast and reliable method of checking how likely certain outcomes are.

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Most popular questions from this chapter

a. Use the general formula for the variance of a linear combination to write an expression for \(V(a X+Y)\). Then let \(a=\sigma_{Y} / \sigma_{X}\), and show that \(\rho \geq-1\). [Hint: Variance is always \(\geq 0\), and \(\left.\operatorname{Cov}(X, Y)=\sigma_{X} \cdot \sigma_{Y} \cdot \rho .\right]\) b. By considering \(V(a X-Y)\), conclude that \(\rho \leq 1\). c. Use the fact that \(V(W)=0\) only if \(W\) is a constant to show that \(\rho=1\) only if \(Y=a X+b\).

A particular brand of dishwasher soap is sold in three sizes: \(25 \mathrm{oz}, 40 \mathrm{oz}\), and \(65 \mathrm{oz}\). Twenty percent of all purchasers select a \(25-0 z\) box, \(50 \%\) select a \(40-0 z\) box, and the remaining \(30 \%\) choose a \(65-\mathrm{oz}\) box. Let \(X_{1}\) and \(X_{2}\) denote the package sizes selected by two independently selected purchasers. a. Determine the sampling distribution of \(\bar{X}\), calculate \(E(\bar{X})\), and compare to \(\mu\). b. Determine the sampling distribution of the sample variance \(S^{2}\), calculate \(E\left(S^{2}\right)\), and compare to \(\sigma^{2}\).

Suppose that when the \(\mathrm{pH}\) of a certain chemical compound is \(5.00\), the \(\mathrm{pH}\) measured by a randomly selected beginning chemistry student is a random variable with mean \(5.00\) and standard deviation .2. A large batch of the compound is subdivided and a sample given to each student in a morning lab and each student in an afternoon lab. Let \(\bar{X}=\) the average \(\mathrm{pH}\) as determined by the morning students and \(\bar{Y}=\) the average \(\mathrm{pH}\) as determined by the afternoon students. a. If \(\mathrm{pH}\) is a normal variable and there are 25 students in each lab, compute \(P(-.1 \leq \bar{X}-\bar{Y} \leq .1)\). [Hint: \(\bar{X}-\bar{Y}\) is a linear combination of normal variables, so is normally distributed. Compute \(\mu_{\bar{x}-\bar{Y}}\) and \(\sigma_{\bar{x}-\bar{y}}\). b. If there are 36 students in each lab, but \(\mathrm{pH}\) determinations are not assumed normal, calculate (approximately) \(P(-.1 \leq \bar{X}-\bar{Y} \leq .1)\).

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let \(X\) denote the number of hoses being used on the self-service island at a particular time, and let \(Y\) denote the number of hoses on the full-service island in use at that time. The joint pmf of \(X\) and \(Y\) appears in the accompanying tabulation. \begin{tabular}{ll|ccc} \(p(x, y)\) & & 0 & 1 & 2 \\ \hline \multirow{x}{*}{\(x\)} & 0 & \(.10\) & \(.04\) & \(.02\) \\ & 1 & \(.08\) & \(.20\) & \(.06\) \\ & 2 & \(.06\) & \(.14\) & \(.30\) \end{tabular} a. What is \(P(X=1\) and \(Y=1)\) ? b. Compute \(P(X \leq 1\) and \(Y \leq 1)\). c. Give a word description of the event \(\\{X \neq 0\) and \(Y \neq 0\\}\), and compute the probability of this event. d. Compute the marginal pmf of \(X\) and of \(Y\). Using \(p_{X}(x)\), what is \(P(X \leq 1)\) ? e. Are \(X\) and \(Y\) independent rv's? Explain.

Let \(X\) and \(Y\) be independent standard normal random variables, and define a new rv by \(U=.6 X+.8 Y\). a. Determine \(\operatorname{Corr}(X, U)\). b. How would you alter \(U\) to obtain \(\operatorname{Corr}(X, U)=\rho\) for a specified value of \(\rho\) ?
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