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A manuscript is sent to a typing firm consisting of typists \(A, B\), and \(C .\) If it is typed by \(A\), then the number of errors made is a Poisson random variable with mean \(2.6\); if typed by \(B\), then the number of errors is a Poisson random variable with mean 3 ; and if typed by \(C\), then it is a Poisson random variable with mean \(3.4\). Let \(X\) denote the number of errors in the typed manuscript. Assume that each typist is equally likely to do the work. (a) Find \(E[X]\). (b) Find \(\operatorname{Var}(X)\).

Short Answer

Expert verified
(a) The expected value of errors in the typed manuscript is \(E[X] = 3\). (b) The variance of the number of errors in the typed manuscript is \(\operatorname{Var}(X) \approx 3.1067\).

Step by step solution

01

Identify given information

We are given that typists A, B, and C have Poisson distributions with means of 2.6, 3, and 3.4 respectively. Each typist is equally likely to have typed the manuscript.
02

Apply Law of Total Expectation

The Law of Total Expectation states that for a random variable, its expected value can be found by taking the expected values of each partition of the available information and then averaging them together. In our case, the partitions are the random variables for the number of errors typed by each typist (A, B, and C) and the available information is that each typist is equally likely to have done the work. So we can write the expected value of \(X\) as: \[E[X] = P(A)E[X|A] + P(B)E[X|B] + P(C)E[X|C]\]
03

Compute \(E[X]\)

Since each typist is equally likely to do the work, we know that \(P(A) = P(B) = P(C) = \frac{1}{3}\). We also know the means for each typist's Poisson distribution: \(E[X|A] = 2.6, E[X|B] = 3,\) and \(E[X|C] = 3.4\). Plug these values into the equation from Step 2 and compute \(E[X]\): \[E[X] = \frac{1}{3}(2.6) + \frac{1}{3}(3) + \frac{1}{3}(3.4) = \frac{9}{3} = 3\] (b) Find \(\operatorname{Var}(X)\):
04

Apply the Law of Total Variance

The Law of Total Variance states that the total variance of a random variable can be found by finding the expected value of the variance of each partition plus the variance of the expected values of each partition. In our case, the partitions are the random variables for the errors made by typists A, B, and C, and the available information is that each typist is equally likely to have done the work. So we can write the variance of \(X\) as: \[\operatorname{Var}(X) = E[\operatorname{Var}(X|\text{Typist})] + \operatorname{Var}(E[X|\text{Typist}])\]
05

Compute \(E[\operatorname{Var}(X|\text{Typist})]\)

Since the number of errors made by each typist is a Poisson random variable, the variance for each typist is equal to its mean. We can compute the expected value of the variance as: \[E[\operatorname{Var}(X|\text{Typist})] = \frac{1}{3}(2.6) + \frac{1}{3}(3) + \frac{1}{3}(3.4) = 3\]
06

Compute \(\operatorname{Var}(E[X|\text{Typist}])\)

To compute the variance of the expected values of each partition, we first find the squared difference between each partition's expected value and the overall expected value of \(X\), and then take the average: \[\operatorname{Var}(E[X|\text{Typist}]) = \frac{1}{3}((2.6-3)^2 + (3-3)^2 + (3.4-3)^2) = \frac{1}{3}(0.16 + 0 + 0.16) = 0.106666\]
07

Compute \(\operatorname{Var}(X)\)

Now, use the values from Steps 5 and 6 to find \(\operatorname{Var}(X)\): \[\operatorname{Var}(X) = E[\operatorname{Var}(X|\text{Typist})] + \operatorname{Var}(E[X|\text{Typist}]) = 3 + 0.106666 = 3.106666\] Thus, the variance of the number of errors in the manuscript is \(\approx 3.1067\).

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Key Concepts

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

Understanding the Law of Total Expectation
The Law of Total Expectation is a fundamental concept in probability theory that simplifies the process of finding the expected value of a random variable. It's especially useful when dealing with a mixture of different distributions, as in our exercise featuring typists with varying error rates.

Suppose you have a random variable that can be observed under several different conditions or scenarios. The Law of Total Expectation allows you to break down the expected value of this variable into the sum of expected values conditional on each scenario. In simpler terms, rather than attempting to find the expected value directly, you examine what the expected value would be under each condition and then take a weighted average of these, according to the likelihood of each condition.

For our manuscript example with typists A, B, and C, the expected number of errors (the expected value of the Poisson random variable) varies depending on who types the manuscript. Since each typist is equally likely to be the one typing, their contributions to the overall expected error count are averaged, with equal weights reflecting equal probabilities. This elegant property of the expected value ensures that no matter how the work is divided among the typists, you can always calculate the overall expected error count in a straightforward way.
Expected Value of a Poisson Random Variable
The expected value, often represented as E[X], is a measure of the central tendency of a random variable's probability distribution. It's essentially a weighted average of all possible values that a random variable can take on, with the probabilities acting as weights. When dealing with a Poisson random variable, the expected value is of particular interest because it represents both the mean and variance of the distribution.

For a Poisson distribution, which models the number of events happening in a fixed interval of time or space under certain conditions, the expected value is the average occurrence rate. In the context of our typists A, B, and C, each has a specific average error rate - these rates are 2.6, 3, and 3.4 errors, respectively. These averages become crucial in calculating the expected number of errors across all typists by using the Law of Total Expectation as shown in the provided step-by-step solution. Revealing a beautifully practical aspect of the Poisson distribution, the expected value directly informs us about the 'typical' error rate we can anticipate from any manuscript typed by these individuals.
Variance of a Random Variable
Variance gives us insight into the variability or spread of a random variable's probable outcomes. More technically, it is the expected value of the squared deviation of a random variable from its mean. When we say we're calculating the variance of a random variable, we're essentially quantifying how much the outcomes can differ from the expected value.

In the Poisson distribution scenario, since the expected value (mean) and variance are the same, knowing one statistic gives you the other - a unique property of the Poisson distribution. However, calculating the variance for a mixture of distributions, as in our manuscript problem where different typists may type the document, involves an additional consideration—the variance of the means of these distributions. The Law of Total Variance, which encompasses calculating both the expected variance within each group and the variance between the means of the groups, will help us determine this 'total' variance. The beauty of this approach is that it decomposes the overall variance into two understandable parts: one due to the variation within each typist's error rate and the other due to the difference in error rates between the typists.

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Most popular questions from this chapter

Let \(X_{1}, \ldots, X_{n}\) be independent random variables having a common distribution function that is specified up to an unknown parameter \(\theta\). Let \(T=T(\mathrm{X})\) be a function of the data \(\mathrm{X}=\left(X_{1}, \ldots, X_{n}\right) .\) If the conditional distribution of \(X_{1}, \ldots, X_{n}\) given \(T(\mathrm{X})\) does not depend on \(\theta\) then \(T(\mathrm{X})\) is said to be a sufficient statistic for \(\theta .\) In the following cases, show that \(T(\mathbf{X})=\sum_{i=1}^{n} X_{i}\) is a sufficient statistic for \(\theta\). (a) The \(X_{i}\) are normal with mean \(\theta\) and variance \(1 .\) (b) The density of \(X_{i}\) is \(f(x)=\theta e^{-\theta x}, x>0\). (c) The mass function of \(X_{i}\) is \(p(x)=\theta^{x}(1-\theta)^{1-x}, x=0,1,0<\theta<1\). (d) The \(X_{i}\) are Poisson random variables with mean \(\theta\).

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