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Chapter 5: Q81SE (page 244)

We have seen that if\(E\left( {{X_1}} \right) = E\left( {{X_2}} \right) = . . . = E\left( {{X_n}} \right) = \mu \), then\(E\left( {{X_1} + . . . + {X_n}} \right) = n\mu \). In some applications, the number of\({X_i} 's\)under consideration is not a fixed number n but instead is a rv N. For example, let N\(5\)be the number of components that are brought into a repair shop on a particular day, and let Xi denote the repair shop time for the\({i^{th}}\)component. Then the total repair time is\({X_1} + {X_2} + . . . + {X_n}\), the sum of a random number of random variables. When N is independent of the\({X_i} 's\), it can be shown that

\(E\left( {{X_1} + . . . + {X_N}} \right) = E\left( N \right) \times \mu \)

a. If the expected number of components brought in on a particular day is\({\bf{10}}\)and the expected repair time for a randomly submitted component is\({\bf{40}}\)min, what is the expected total repair time for components submitted on any particular day?

b. Suppose components of a certain type come in for repair according to a Poisson process with a rate of\(5\)per hour. The expected number of defects per component is\(3.5\). What is the expected value of the total number of defects on components submitted for repair during a four-hour period? Be sure to indicate how your answer follows from the general result just given.

Short Answer

Expert verified

\(\begin{array}{l}{\rm{a}}{\rm{. }}E\left( {{X_1} + {X_2} + \ldots + {X_N}} \right) = 400{\rm{; }}\\{\rm{b}}{\rm{. }}E(N) \cdot \mu = 70\end{array}\)

Step by step solution

01

Definition of Variance

In probability and statistics, variance is the expected value of a random variable's squared departure from its mean value. Informally, variance is a measurement of how far a group of numbers (random) deviates from their mean value. The square of standard deviation, another important tool, equals the value of variance.

02

Calculation for the determination of the expected time in part a.

(a).

The expected number of components that need repair is\(E(N)\), and the expected repair time is\(\mu = 40\), therefore, the expected total repair time is

\(\begin{aligned}E\left( {{X_1} + {X_2} + \ldots + {X_N}} \right) &= E(N) \cdot \mu \\ &= 10 \cdot 40\\ &= 400\end{aligned}\)

03

Calculation for the determination of the expected time in part b.

(b):

Proposition: Number of events during a time interval of length \({\rm{t}}\)can be modeled using Poisson random variable with parameter\(\mu = \alpha t\). This indicates that

\({P_k}(t) = {e^{ - \alpha t}} \cdot \frac{{{{(\alpha t)}^k}}}{{k!}}\)

It is also known as Poisson process (not formally defined).

The rate is\(\alpha = 5\)per hour, and time interval is \(t = 4\)hours. Therefore, the parameter of given Poisson random variable is

\({\mu _N} = \alpha \cdot t = 5 \cdot 4 = 20\)

04

Calculation for the determination of the expected time in part b.

Proposition: For a random variable X with Poisson Distribution with parameter\(\mu > 0\), the following is true

\(E(X) = V(X) = \mu \)

Hence, the expected number of components which need repair is

\(E(N) = {\mu _N} = 20\)

Given in the exercise, the expected number of defects per component is $\mu=3.5$. Finally, the expected total number of defects is

\(E(N) \cdot \mu = 20 \cdot 3.5 = 70\)

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Most popular questions from this chapter

a. Let \({{\rm{X}}_{\rm{1}}}\)have a chi-squared distribution with parameter \({{\rm{\nu }}_{\rm{1}}}\) (see Section 4.4), and let \({{\rm{X}}_{\rm{2}}}\)be independent of \({{\rm{X}}_{\rm{1}}}\)and have a chi-squared distribution with parameter\({{\rm{v}}_{\rm{2}}}\). Use the technique of to show that \({{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}\)has a chi-squared distribution with parameter\({{\rm{v}}_{\rm{1}}}{\rm{ + }}{{\rm{v}}_{\rm{2}}}\).

b. You were asked to show that if \({\rm{Z}}\)is a standard normal \({\rm{rv}}\), then \({{\rm{Z}}^{\rm{2}}}\)has a chi squared distribution with\({\rm{v = 1}}\). Let \({{\rm{Z}}_{\rm{1}}}{\rm{,}}{{\rm{Z}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{Z}}_{\rm{n}}}\)be \({\rm{n}}\)independent standard normal \({\rm{rv}}\) 's. What is the distribution of\({\rm{Z}}_{\rm{1}}^{\rm{2}}{\rm{ + }}...{\rm{ + Z}}_{\rm{n}}^{\rm{2}}\)? Justify your answer.

c. Let \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)be a random sample from a normal distribution with mean \({\rm{\mu }}\)and variance\({{\rm{\sigma }}^{\rm{2}}}\). What is the distribution of the sum \({\rm{Y = }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - \mu }}} \right){\rm{/\sigma }}} \right)}^{\rm{2}}}} {\rm{?}}\)Justify your answer?

A student has a class that is supposed to end at\({\rm{9:00 A}}{\rm{.M}}\). and another that is supposed to begin at \({\rm{9:10 A}}{\rm{.M}}\).

Suppose the actual ending time of the \({\rm{9:00 A}}{\rm{.M}}\). class is a normally distributed \({\rm{rv}}\) \({{\rm{X}}_{\rm{1}}}\)with mean \({\rm{9:02 A}}{\rm{.M}}\)and standard deviation \({\rm{1}}{\rm{.5\;min}}\)and that the starting time of the next class is also a normally distributed \({\rm{rv}}\) \({{\rm{X}}_{\rm{2}}}\)with mean \({\rm{9:10}}\)and standard deviation \({\rm{1\;min}}{\rm{.}}\)Suppose also that the time necessary to get from one classroom to the other is a normally distributed \({\rm{rv}}\) \({{\rm{X}}_{\rm{3}}}\)with mean \({\rm{6\;min}}\) and standard deviation\({\rm{1\;min}}\). What is the probability that the student makes it to the second class before the lecture starts? (Assume independence of\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}\), and\({{\rm{X}}_{\rm{3}}}\), which is reasonable if the student pays no attention to the finishing time of the first class.)

According to the article 鈥淩eliability Evaluation of Hard Disk Drive Failures Based on Counting Processes鈥� (Reliability Engr. and System Safety,\({\bf{2013}}:{\rm{ }}{\bf{110}}--{\bf{118}}\)), particles accumulating on a disk drive come from two sources, one external and the other internal. The article proposed a model in which the internal source contains a number of loose particles W having a Poisson distribution with mean value of m; when a loose particle releases, it immediately enters the drive, and the release times are independent and identically distributed with cumulative distribution function G(t). Let X denote the number of loose particles not yet released at a particular time t. Show that X has a Poisson distribution with parameter\(\mu \left( {1 - G\left( t \right)} \right)\). (Hint: Let Y denote the number of particles accumulated on the drive from the internal source by time t so that\(X + Y = W\). Obtain an expression for\(P\left( {X = x, Y = y} \right)\)and then sum over y.)

Let \({\rm{A}}\)denote the percentage of one constituent in a randomly selected rock specimen, and let \({\rm{B}}\)denote the percentage of a second constituent in that same specimen. Suppose \({\rm{D}}\)and \({\rm{E}}\)are measurement errors in determining the values of \({\rm{A}}\)and \({\rm{B}}\)so that measured values are \({\rm{X = A + D}}\)and\({\rm{Y = B + E}}\), respectively. Assume that measurement errors are independent of one another and of actual values.

a. Show that

\({\rm{Corr(X,Y) = Corr(A,B) \times }}\sqrt {{\rm{Corr}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right)} {\rm{ \times }}\sqrt {{\rm{Corr}}\left( {{{\rm{Y}}_{\rm{1}}}{\rm{,}}{{\rm{Y}}_{\rm{2}}}} \right)} \)

where \({{\rm{X}}_{\rm{1}}}\)and \({{\rm{X}}_{\rm{2}}}\)are replicate measurements on the value of\({\rm{A}}\), and \({{\rm{Y}}_{\rm{1}}}\)and \({{\rm{Y}}_{\rm{2}}}\)are defined analogously with respect to\({\rm{B}}\). What effect does the presence of measurement error have on the correlation?

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Two components of a minicomputer have the following joint pdf for their useful lifetimes \({\rm{X}}\)and \({\rm{Y}}\)

a. What is the probability that the lifetime \({\rm{X}}\) of the first component exceeds \({\rm{3}}\)?

b. What are the marginal pdf鈥檚 of \({\rm{X}}\)and \({\rm{Y}}\)? Are the two lifetimes independent? Explain.

c. What is the probability that the lifetime of at least one component exceeds\({\rm{3}}\)?
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