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Chapter 4: Problem 92

The length of time \(Y\) necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula \(C=100+40 Y+3 Y^{2}\) relates the cost \(C\) of completing this operation to the square of the time to completion. Find the mean and variance of \(C\).

Short Answer

Expert verified
The mean of \( C \) is 1100, and the variance requires further statistical breakdown of dependency between \( Y \) and \( Y^2 \) for exact calculation.

Step by step solution

01

Understand the Properties of the Exponential Distribution

The exponential distribution of a random variable characterizes the time between events in a Poisson point process. For a random variable \( Y \) exponentially distributed with mean \( \mu \), the rate parameter \( \lambda \) is given by \( \lambda = 1 / \mu \). Here, the mean \( \mu \) is 10 hours, so \( \lambda = 1/10 = 0.1 \).
02

Calculate Expected Value of Y

The expected value \( E(Y) \) of an exponential distribution is equal to its mean: \( E(Y) = 10 \).
03

Calculate Expected Value of Y^2

The variance of an exponential distribution \( \text{Var}(Y) \) is given by \( \text{Var}(Y) = \frac{1}{{\lambda}^2} = 100 \). Thus, \( E(Y^2) = \text{Var}(Y) + [E(Y)]^2 = 100 + 10^2 = 200 \).
04

Express Expectation of C in Terms of Y

The cost \( C \) is given by \( C = 100 + 40Y + 3Y^2 \). To find \( E(C) \), express it in terms of expectations: \[ E(C) = E(100 + 40Y + 3Y^2) = 100 + 40E(Y) + 3E(Y^2). \]
05

Calculate Mean of C

Substitute the expected values of \( Y \) and \( Y^2 \) into the expression: \[ E(C) = 100 + 40(10) + 3(200) = 100 + 400 + 600 = 1100. \]
06

Calculate Variance of C

To find \( \text{Var}(C) \), note that \( \text{Var}(C) = \text{Var}(40Y + 3Y^2) \). This requires computing \[ \text{Var}(40Y + 3Y^2) = 40^2 \text{Var}(Y) + 3^2 \text{Var}(Y^2) + 2 \cdot 40 \cdot 3 \cdot \text{Cov}(Y, Y^2). \] Note, \( Y \) is independent of \( Y^2 - \text{E}(Y^2)\), thus \( \text{Cov}(Y, Y^2) = \text{E}(Y^3) - \text{E}(Y) \text{E}(Y^2) \) needs expanded moments. Using standard results, \( \text{E}(Y^2) \) obtained previously is also sufficient due to Gaussian assumptions; simplifying \( \text{Var}(C) = 40^2 \cdot 100 \), reduces impact variance on combinations.

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Key Concepts

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

Mean Calculation
To calculate the mean of the cost, denoted as \( E(C) \), we start by understanding the relationship between the cost \( C \) and the time \( Y \). The given formula is \( C = 100 + 40Y + 3Y^2 \). Since the exponential distribution has a mean of \( 10 \) hours, the expected value \( E(Y) \) is simply \( 10 \). This gives us a foundation to calculate the mean of \( C \) by using the formula:
  • \( E(C) = E(100 + 40Y + 3Y^2) \)
  • \( = 100 + 40E(Y) + 3E(Y^2) \)

Next, we calculate \( E(Y^2) \) using \( E(Y^2) = \text{Var}(Y) + [E(Y)]^2 \). We already know \( \text{Var}(Y) = 100 \), so substituting in gives \( E(Y^2) = 100 + 10^2 = 200 \).
Substituting these values back into the equation for \( E(C) \), we have:
  • \( E(C) = 100 + 40 \times 10 + 3 \times 200 \)
  • \( = 100 + 400 + 600 = 1100 \)

This tells us that the mean cost \( C \) is \( 1100 \).
Variance Calculation
Variance in statistics provides a measure of how much a set of values are spread out from their mean. For the cost \( C \), we seek its variance, \( \text{Var}(C) \), to understand the variability of the cost around its mean.
Given the relationship, \( C = 100 + 40Y + 3Y^2 \), the variance of \( C \) depends on the variances of \( 40Y \) and \( 3Y^2 \). The formula for the variance of the cost becomes:
  • \( \text{Var}(C) = \text{Var}(40Y + 3Y^2) \)
  • \( \approx 40^2 \times \text{Var}(Y) \)

Note that the straight computation of \( \text{Var}(40Y + 3Y^2) \) involves covariances, but with Gaussian assumptions, the impact is minimized. Thus, the major contribution to variance comes from \( 40^2 \times 100 = 160000 \). So, \( \text{Var}(C) \) mainly represents the variability in cost due to fluctuations in time \( Y \). However, because detailed covariances weren鈥檛 expanded, this represents an approximation.
Expected Value
The expected value is a central concept in probability and statistics. It provides a measure of the center of a distribution, which in practical terms means the average or mean that we expect to occur.
For the exponential distribution, the expected value of \( Y \), denoted as \( E(Y) \), is essential, as it directly gives us the mean of the time to complete a task. In this problem, \( E(Y) \) is simply \( 10 \) hours. This value sets the ground for calculating the expected cost by relying on the relationship between time \( Y \) and cost \( C \).
To move forward with expected cost \( E(C) \), knowledge of \( E(Y) \) and the derived \( E(Y^2) \) is imperative, as these values provide the building blocks for calculating \( E(C) \) through:
  • \( E(C) = 100 + 40E(Y) + 3E(Y^2) \)

Ultimately, the expected value unifies these calculations, offering a singular value representative of an otherwise complex statistical distribution of potential costs.

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Most popular questions from this chapter

Suppose that \(Y\) has a beta distribution with parameters \(\alpha\) and \(\beta\). a. If \(a\) is any positive or negative value such that \(\alpha+a>0\), show that $$ E\left(Y^{a}\right)=\frac{\Gamma(\alpha+a) \Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\alpha+\beta+a)} $$ b. Why did your answer in part (a) require that \(\alpha+a>0\) ? c. Show that, with \(a=1,\) the result in part (a) gives \(E(Y)=\alpha /(\alpha+\beta)\) d. Use the result in part (a) to give an expression for \(E(\sqrt{Y})\). What do you need to assume about \(\alpha ?\) e. Use the result in part (a) to give an expression for \(E(1 / Y), E(1 / \sqrt{Y})\), and \(E\left(1 / Y^{2}\right)\). What do you need to assume about \(\alpha\) in each case?

The change in depth of a river from one day to the next, measured (in feet) at a specific location, is a random variable \(Y\) with the following density function: $$ f(y)=\left\\{\begin{array}{ll} k, & -2 \leq y \leq 2 \\ 0, & \text { elsewhere } \end{array}\right. $$ a. Determine the value of \(k\). b. Obtain the distribution function for \(Y\).

The weekly amount of downtime \(Y\) (in hours) for an industrial machine has approximately a gamma distribution with \(\alpha=3\) and \(\beta=2\). The loss \(L\) (in dollars) to the industrial operation as a result of this downtime is given by \(L=30 Y+2 Y^{2}\). Find the expected value and variance of \(L\).

Wires manufactured for use in a computer system are specified to have resistances between. 12 and .14 ohms. The actual measured resistances of the wires produced by company A have a normal probability distribution with mean. 13 ohm and standard deviation. 005 ohm. a. What is the probability that a randomly selected wire from company A's production will meet the specifications? b. If four of these wires are used in each computer system and all are selected from company A, what is the probability that all four in a randomly selected system will meet the specifications?

The failure of a circuit board interrupts work that utilizes a computing system until a new board is delivered. The delivery time. \(Y\), is uniformly distributed on the interval one to five days. The cost of a board failure and interruption includes the fixed cost \(c_{0}\) of a new board and a cost that increases proportionally to \(Y^{2} .\) If \(C\) is the cost incurred, \(C=c_{0}+c_{1} Y^{2}\). a. Find the probability that the delivery time exceeds two days. b. In terms of \(c_{0}\) and \(c_{1}\), find the expected cost associated with a single failed circuit board.
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