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Chapter 3: Problem 37

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 number of errors \(E[X]\) is 3. (b) The variance of the number of errors \(\operatorname{Var}(X)\) is 3.

Step by step solution

01

Determine the probabilities for each typist

Since each typist is equally likely to do the work, we can calculate the probability assigned to each typist. There are three typists (A, B, and C) and each typist has an equal chance of doing the work, hence the probability for each typist is \(\frac{1}{3}\).
02

Calculate the expected value of X

To calculate the expected value of X, we use the Law of Total Expectations, also known as the Tower Property: \(E[X] = E[E[X|Y]]\), where Y is the source of the random variable (the typist) and X is the random variable (number of errors). In this case, Y takes on three outcomes: A, B, and C. So, \(E[X] = E[X|A] P(A) + E[X|B] P(B) + E[X|C] P(C)\) Plug in the given values: \(E[X] = (2.6) \left(\frac{1}{3}\right) + (3) \left(\frac{1}{3}\right) + (3.4)\left(\frac{1}{3}\right) \) Now, evaluate the expression: \(E[X] = \frac{2.6+3+3.4}{3} = \frac{9}{3} = 3\) The expected number of errors, \(E[X]\), is 3.
03

Calculate the variance of X

To calculate the variance of X, we use the Law of Total Variance: \(\operatorname{Var}(X) = E[\operatorname{Var}(X|Y)] + \operatorname{Var}(E[X|Y])\), where Y is the source of random variable (the typist) and X is the random variable (number of errors). Calculate the expected variance term: \(E[\operatorname{Var}(X|Y)] = \operatorname{Var}(X|A) P(A) + \operatorname{Var}(X|B) P(B) + \operatorname{Var}(X|C) P(C)\) Since the variance of a Poisson distribution is equal to its mean: \(E[\operatorname{Var}(X|Y)] = (2.6) \left(\frac{1}{3}\right) + (3) \left(\frac{1}{3}\right) + (3.4) \left(\frac{1}{3}\right) = 3\). Calculate the variance of the expected value term: \(\operatorname{Var}(E[X|Y]) = \operatorname{Var}(2.6|\textbf{A})P(A) + \operatorname{Var}(3|\textbf{B}) P(B) + \operatorname{Var}(3.4|\textbf{C})P(C) \). In this case, \(E[X|Y]=\mu_Y\), where \(\mu_Y\) is the mean of each Poisson distribution for each typist (fixed values). Hence, the conditional variance is not a function of Y, and thus, it is strictly equal to zero: \(\operatorname{Var}(E[X|Y]) = 0\). Now, plug the values into the Law of Total Variance: \(\operatorname{Var}(X) = E[\operatorname{Var}(X|Y)] + \operatorname{Var}(E[X|Y]) = 3 + 0 = 3\). The variance of X, \(\operatorname{Var}(X)\), is 3. #Summary# (a) The expected number of errors \(E[X]\) is 3. (b) The variance of the number of errors \(\operatorname{Var}(X)\) is 3.

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Key Concepts

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

Law of Total Expectation
The Law of Total Expectation, or Tower Property, allows us to compute the expected value of a random variable by conditioning on another random variable. When applied to practical problems, it simplifies calculations by breaking them into more manageable components.

For our typing firm exercise, it is employed to calculate the expected number of typing errors, denoted as \(E[X]\), by considering the possible selection of typists A, B, or C. Each typist, being a different condition, results in a different expected number of errors. Hence, the overall expectation is the average of these expected values, weighted by the probability of each typist being chosen. The Law of Total Expectation combines all the individual expectations into one average expectation which, in our case, happens to be 3 errors per manuscript.
Law of Total Variance
The Law of Total Variance, another useful concept in statistics, helps in understanding how the variability of a random variable is spread out. It's the sum of the expected variance given a conditional random variable and the variance of the expected values of the conditional random variable.

For the situation with our typists, we need to determine \(\operatorname{Var}(X)\), the variance in the number of errors. Since the number of errors follows a Poisson distribution, whose variance equals its mean, the Law of Total Variance allows us to say that if one typist was chosen at random, our expectation of the variance would simply be the average of the means. Additionally, since each typist's mean number of errors is fixed, the variance of the expected values of the conditional random variable is zero. Consequently, the total variance also calculates to 3, identical to the expected number of errors.
Expected Value
The expected value, or mean, of a random variable gives us a measure of the 'center' or the average outcome we expect if we were to repeat an experiment many times. In a Poisson distribution, which describes the number of events occurring within a fixed interval, the expected value also represents the average rate at which the events happen.

In our Poisson random variable scenario for typists in a manuscript typing firm, the expected value of errors for each typist defines the typical performance of that typist. When a document is typed, assuming each typist has an equal likelihood of typing it, the expected value of the total errors is the average of the expected errors by each typist. This is crucial for the firm to predict how many errors could occur on average and can also be essential for setting up quality checks.
Typing Errors Probability
When dealing with Poisson random variables, the probability of a specific number of events (in this case, typing errors) happening is determined by the Poisson distribution formula. For a manuscript typed by any of the typists (A, B, or C), the likelihood of a certain number of errors can be computed using their respective means.

In practice, the probability of encountering a specific number of errors can inform us about the reliability of each typist and the quality control processes that may be needed. It's also an application of the Poisson distribution in the real world, letting the typing firm project and mitigate any potential risks associated with typing errors.

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Most popular questions from this chapter

Suppose that we continually roll a die until the sum of all throws exceeds 100 . What is the most likely value of this total when you stop?

Give another proof of the result of Example \(3.17\) by computing the moment generating function of \(\sum_{i=1}^{N} X_{i}\) and then differentiating to obtain its moments. Hint: Let $$ \begin{aligned} \phi(t) &=E\left[\exp \left(t \sum_{i=1}^{N} X_{i}\right)\right] \\ &=E\left[E\left[\exp \left(t \sum_{i=1}^{N} X_{i}\right) \mid N\right]\right] \end{aligned} $$ Now, $$ E\left[\exp \left(t \sum_{i=1}^{N} X_{i}\right) \mid N=n\right]=E\left[\exp \left(t \sum_{i=1}^{n} X_{i}\right)\right]=\left(\phi_{X}(t)\right)^{n} $$ since \(N\) is independent of the \(X\) s where \(\phi_{X}(t)=E\left[e^{t X}\right]\) is the moment generating function for the \(X \mathrm{~s}\). Therefore, $$ \phi(t)=E\left[\left(\phi_{X}(t)\right)^{N}\right] $$ Differentiation yields $$ \begin{aligned} \phi^{\prime}(t) &=E\left[N\left(\phi_{X}(t)\right)^{N-1} \phi_{X}^{\prime}(t)\right] \\ \phi^{\prime \prime}(t) &=E\left[N(N-1)\left(\phi_{X}(t)\right)^{N-2}\left(\phi_{X}^{\prime}(t)\right)^{2}+N\left(\phi_{X}(t)\right)^{N-1} \phi_{X}^{\prime \prime}(t)\right] \end{aligned} $$ Evaluate at \(t=0\) to get the desired result.

An unbiased die is successively rolled. Let \(X\) and \(Y\) denote, respectively, the number of rolls necessary to obtain a six and a five. Find (a) \(E[X]\), (b) \(E[X \mid Y=1]\), (c) \(E[X \mid Y=5]\).

Consider a gambler who on each bet either wins 1 with probability \(18 / 38\) or loses 1 with probability \(20 / 38 .\) (These are the probabilities if the bet is that a roulette wheel will land on a specified color.) The gambler will quit either when he or she is winning a total of 5 or after 100 plays. What is the probability he or she plays exactly 15 times?

An urn contains three white, six red, and five black balls. Six of these balls are randomly selected from the urn. Let \(X\) and \(Y\) denote respectively the number of white and black balls selected. Compute the conditional probability mass function of \(X\) given that \(Y=3\). Also compute \(E[X \mid Y=1]\).
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