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Chapter 5: Problem 92

The permeability of a membrane used as a moisture barrier in a biological application depends on the thickness of three integrated layers. Layers \(1,2,\) and 3 are normally distributed with means of \(0.5,1,\) and 1.5 millimeters, respectively. The standard deviations of layer thickness are 0.1,0.2 , and \(0.3,\) respectively. Also, the correlation between layers 1 and 2 is \(0.7,\) between layers 2 and 3 is \(0.5,\) and between layers 1 and 3 is \(0.3 .\) (a) Determine the mean and variance of the total thickness of the three layers. (b) What is the probability that the total thickness is less than 1.5 millimeters?

Short Answer

Expert verified
(a) Mean: 3 mm, Variance: 0.246 mm^2. (b) Probability is nearly 0.

Step by step solution

01

Define Variables

Let's denote the thickness of layers 1, 2, and 3 as random variables \(X_1\), \(X_2\), and \(X_3\) respectively. We know \(X_1 \sim N(0.5, 0.1^2)\), \(X_2 \sim N(1, 0.2^2)\), and \(X_3 \sim N(1.5, 0.3^2)\).
02

Calculate the Mean of Total Thickness

The total thickness of the three layers is \(T = X_1 + X_2 + X_3\). The mean of the total thickness is the sum of the means of the individual layers: \[ E(T) = E(X_1) + E(X_2) + E(X_3) = 0.5 + 1 + 1.5 = 3 \text{ mm}. \]
03

Calculate the Variance of Total Thickness

The variance of the total thickness involves the sum of variances and the covariances. The formula for the variance is: \[ \text{Var}(T) = \text{Var}(X_1) + \text{Var}(X_2) + \text{Var}(X_3) + 2\text{Cov}(X_1, X_2) + 2\text{Cov}(X_2, X_3) + 2\text{Cov}(X_1, X_3) \] Using the correlation \(\rho_{ij}\) and the standard deviations \(\sigma_i\, \sigma_j\), we have: \[ \begin{align*} \text{Var}(T) = & \ 0.1^2 + 0.2^2 + 0.3^2 + 2(0.7)(0.1)(0.2) + 2(0.5)(0.2)(0.3) + 2(0.3)(0.1)(0.3) \ = & \ 0.01 + 0.04 + 0.09 + 0.028 + 0.06 + 0.018 \ = & \ 0.246 \text{ mm}^2. \end{align*} \]
04

Use Total Thickness Distribution

Since \(T\) is a sum of normally distributed variables, \(T\) itself is normally distributed: \(T \sim N(3, 0.246)\). We need this information to find probabilities concerning \(T\).
05

Calculate Probability for Thickness Less Than 1.5 mm

We need to find \(P(T < 1.5)\). Using the standard normal distribution, convert the problem to a standard score (\(z\)-score). \[ z = \frac{1.5 - 3}{\sqrt{0.246}} = \frac{-1.5}{0.496} \approx -3.024 \]. Consulting a standard normal distribution table, \(P(Z < -3.024)\) is nearly 0.

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Key Concepts

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

Mean and Variance
When dealing with probability distributions, two fundamental concepts are the mean and variance. They help us understand the central tendency and variability of a distribution. In this exercise, we determine these for a total thickness of three layers.

The **mean** is the arithmetic average of a set of values. Here, it represents the expected total thickness of the layers. Each layer has its own mean:
  • Layer 1: 0.5 mm
  • Layer 2: 1.0 mm
  • Layer 3: 1.5 mm
The mean of the total thickness is simply the sum: 0.5 + 1.0 + 1.5 = 3 mm.

The **variance** indicates how much the thickness values can vary, or spread out, from the mean. For the layers, the variance is calculated using the standard deviations and the correlations between each pair of layers:
  • Variance for layer 1: 0.1虏 = 0.01
  • Variance for layer 2: 0.2虏 = 0.04
  • Variance for layer 3: 0.3虏 = 0.09
The variance of the total thickness also factors in the correlations, which result in additional covariance terms. The overall variance of the combined thickness of the layers is 0.246 mm虏.
Normal Distribution
A normal distribution is a continuous probability distribution that is symmetrical around the mean. It looks like a bell shape and is often referred to as a "bell curve."

In the context of this problem, we are dealing with normally distributed layer thicknesses, meaning that the thickness of each layer follows a normal distribution pattern. This is indicated by the notation:
  • Layer 1: Normally distributed with mean 0.5 mm and variance 0.1虏.
  • Layer 2: Normally distributed with mean 1.0 mm and variance 0.2虏.
  • Layer 3: Normally distributed with mean 1.5 mm and variance 0.3虏.
One important property of normal distributions is that the sum of normally distributed variables is also normally distributed. Thus, when the total thickness of the layers is calculated, it continues to follow a normal distribution:
The total thickness is normally distributed with a mean of 3 mm and a variance of 0.246 mm虏, or in terms of standard deviation \( \sqrt{0.246} \approx 0.496 \).
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It indicates how much the values deviate, on average, from the mean of the distribution.

For each layer, we have a given standard deviation, which is the square root of its variance:
  • Layer 1: 0.1 mm
  • Layer 2: 0.2 mm
  • Layer 3: 0.3 mm
The standard deviation helps estimate the spread or dispersion of the thickness values for each layer.

For the total thickness, the standard deviation is \( \sqrt{\text{Variance of total thickness}} \), which was calculated as 0.496 mm. This tells us that most thickness values fall within this distance from the mean (3 mm) in a normal distribution setting. Understanding this spread is crucial when determining probabilities, as it shows the likelihood of the total thickness being under or over certain limits.

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Most popular questions from this chapter

An order of 15 printers contains four with a graphics-enhancement feature, five with extra memory, and six with both features. Four printers are selected at random, without replacement, from this set. Let the random variables \(X, Y,\) and \(Z\) denote the number of printers in the sample with graphics enhancement only, extra memory only, and both, respectively. (a) Describe the range of the joint probability distribution of \(X, Y,\) and \(Z\). (b) Is the probability distribution of \(X, Y,\) and \(Z\) multinomial? Why or why not? (c) Determine the conditional probability distribution of \(X\) given that \(Y=2\). Determine the following: (d) \(P(X=1, Y=2, Z=1)\) (e) \(P(X=1, Y=1)\) (f) \(E(X)\) and \(V(X)\) pe (g) \(P(X=1, Y=2 \mid Z=1)\) (h) \(P(X=2 \mid Y=2)\) to (i) Conditional probability distribution of \(X\) given that \(Y=0\) and \(Z=3\).

A manufacturing company employs two devices to inspect output for quality control purposes. The first device is able to accurately detect \(99.3 \%\) of the defective items it receives, whereas the second is able to do so in \(99.7 \%\) of the cases. Assume that four defective items are produced and sent out for inspection. Let \(X\) and \(Y\) denote the number of items that will be identified as defective by inspecting devices 1 and \(2,\) respectively. Assume the devices are independent. Determine: (a) \(f_{X Y}(x, y)\) (b) \(f_{X}(x)\) (c) \(E(X)\) (d) \(f_{Y \mid 2}(y)\) (e) \(E(Y \mid X=2)\) (f) \(V(Y \mid X=2)\) (g) Are \(X\) and \(Y\) independent?

The weights of adobe bricks used for construction are normally distributed with a mean of 3 pounds and a standard deviation of 0.25 pound. Assume that the weights of the bricks are independent and that a random sample of 20 bricks is selected. (a) What is the probability that all the bricks in the sample exceed 2.75 pounds? (b) What is the probability that the heaviest brick in the sample exceeds 3.75 pounds?

If \(X\) and \(Y\) have a bivariate normal distribution with \(\rho=0,\) show that \(X\) and \(Y\) are independent.

The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of \(1 / 8\) inch. The width of a door is normally distributed with a mean of \(23-7 / 8\) inches and a standard deviation of \(1 / 16\) inch. Assume independence. (a) Determine the mean and standard deviation of the difference between the width of the casing and the width of the door. (b) What is the probability that the width of the casing minus the width of the door exceeds \(1 / 4\) inch? (c) What is the probability that the door does not fit in the casing?
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