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Chapter 5: Q22E (page 219)

An instructor has given a short quiz consisting of two parts. For a randomly selected student, let \({\rm{X = }}\) the number of points earned on the first part and \({\rm{Y = }}\) the number of points earned on the second part. Suppose that the joint pmf of \({\rm{X}}\) and \({\rm{Y}}\) is given in the accompanying table.

a. If the score recorded in the grade book is the total number of points earned on the two parts, what is the expected recorded score\({\rm{E(X + Y)}}\)?

b. If the maximum of the two scores is recorded, what is the expected recorded score?

Short Answer

Expert verified

a. The expected recorded score is \[{\rm{E(X + Y) = 14}}{\rm{.1}}\].

b. The expected recorded score is \[{\rm{E[max(X,Y)] = 9}}{\rm{.6}}\].

Step by step solution

01

Definition

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

02

Determining the expected recorded score \({\rm{E(X + Y)}}\)

We are given joint pmf of \({\rm{X}}\) and\({\rm{Y}}\).

(a):

Expected Value

(Mean value) of a random variable\({\rm{g(X,Y)}}\), where \({\rm{g( \times )}}\) is a function, denoted as \({\rm{E(g(X,Y))}}\) is given by \({\rm{E(g(X,Y)) = }}\left\{ {\begin{aligned}{*{20}{l}}{\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\rm{g}} } {\rm{(x,y) \times p(x,y)}}}&{{\rm{,X and Y discrete, }}}\\{\int_{{\rm{ - 楼}}}^{\rm{楼}} {\int_{{\rm{ - 楼}}}^{\rm{楼}} {\rm{g}} } {\rm{(x,y) \times f(x,y)dxdy}}}&{{\rm{,X and Y continuous}}{\rm{. }}}\end{aligned}} \right.\)

where \({\rm{p(x,y)}}\) is pmf and \({\rm{f}}\left( {{\rm{x, y}}} \right)\) pdf.

In our case, \({\rm{g(X,Y) = X + Y}}\), and the random variables are discrete, therefore the following is true

\(\begin{aligned}{\rm{E(X + Y) = }}\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\rm{g}} } {\rm{(x,y) \times p(x,y)}}\\{\rm{ = (0 + 0) \times 0}}{\rm{.02 + (0 + 5) \times 0}}{\rm{.06 + \ldots + (10 + 10) \times 0}}{\rm{.14 + (10 + 15) \times 0}}{\rm{.01}}\\{\rm{ = 14}}{\rm{.1}}\end{aligned}\)

03

Determining the expected recorded score

(b):

We are interested in expectation of random variable\({\rm{g(X,Y) = max(X,Y)}}\). Therefore,

\(\begin{aligned}{\rm{E(max(X,Y)) = }}\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\rm{g}} } {\rm{(x,y) \times p(x,y)}}\\{\rm{ = max(0,0) \times 0}}{\rm{.02 + max(0,5) \times 0}}{\rm{.06 + \ldots + max(10,10) \times 0}}{\rm{.14 + max(10,15) \times 0}}{\rm{.01}}\\{\rm{ = 0 \times 0}}{\rm{.02 + 5 \times 0}}{\rm{.06 + \ldots + \ldots 10 \times 0}}{\rm{.14 + 15 \times 0}}{\rm{.01}}\\{\rm{ = 9}}{\rm{.6}}\end{aligned}\)

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Most popular questions from this chapter

Consider a system consisting of three components as pictured. The system will continue to function as long as the first component functions and either component \({\rm{2}}\) or component \({\rm{3}}\)functions. Let \({{\rm{X}}_{{\rm{1,}}}}{{\rm{X}}_{\rm{2}}}\), and \({{\rm{X}}_{\rm{3}}}\) denote the lifetimes of components \({\rm{1}}\), \({\rm{2}}\), and \({\rm{3}}\), respectively. Suppose the \({{\rm{X}}_{\rm{i}}}\) 鈥檚 are independent of one another and each \({{\rm{X}}_{\rm{i}}}\) has an exponential distribution with parameter \({\rm{\lambda }}\).

a. Let \({\rm{Y}}\) denote the system lifetime. Obtain the cumulative distribution function of \({\rm{Y}}\)and differentiate to obtain the pdf. (Hint: \({{\rm{F}}_{\left( {\rm{Y}} \right)}}{\rm{P}}\left\{ {{\rm{Y}} \le {\rm{y}}} \right\}\); express the event \(\left\{ {{\rm{Y}} \le {\rm{y}}} \right\}\)in terms of unions and/or intersections of the three events \(\left\{ {{{\rm{X}}_{\rm{i}}} \le {\rm{y}}} \right\}\), \(\left\{ {{{\rm{X}}_{\rm{2}}} \le {\rm{y}}} \right\}\), and \(\left\{ {{{\rm{X}}_3} \le {\rm{y}}} \right\}\).)

b. Compute the expected system lifetime

Let \({\rm{X}}\) denote the number of Canon SLR cameras sold during a particular week by a certain store. The pmf of \({\rm{X}}\) is

Sixty percent of all customers who purchase these cameras also buy an extended warranty. Let \({\rm{Y}}\) denote the number of purchasers during this week who buy an extended warranty.

a. What is\({\rm{P(X = 4,Y = 2)}}\)? (Hint: This probability equals\({\rm{P(Y = 2}}\mid {\rm{X = 4) \times P(X = 4)}}\); now think of the four purchases as four trials of a binomial experiment, with success on a trial corresponding to buying an extended warranty.)

b. Calculate\({\rm{P(X = Y)}}\).

c. Determine the joint pmf of \({\rm{X}}\) and \({\rm{Y}}\)then the marginal pmf of\({\rm{Y}}\).

The difference between the number of customers in line at the express checkout and the number in line at the super-express checkout is\({{\rm{X}}_{\rm{1}}}{\rm{ - }}{{\rm{X}}_{\rm{2}}}\). Calculate the expected difference.

Two different professors have just submitted final exams for duplication. Let \({\rm{X}}\) denote the number of typographical errors on the first professor鈥檚 exam and \({\rm{Y}}\) denote the number of such errors on the second exam. Suppose \({\rm{X}}\) has a Poisson distribution with parameter \({{\rm{\mu }}_{\rm{1}}}\), \({\rm{Y}}\) has a Poisson distribution with parameter \({{\rm{\mu }}_{\rm{2}}}\), and \({\rm{X}}\) and \({\rm{Y}}\) are independent.

a. What is the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\)?

b. What is the probability that at most one error is made on both exams combined?

c. Obtain a general expression for the probability that the total number of errors in the two exams is m (where \({\rm{m}}\) is a nonnegative integer). (Hint: \({\rm{A = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:x + y = m}}} \right\}{\rm{ = }}\left\{ {\left( {{\rm{m,0}}} \right)\left( {{\rm{m - 1,1}}} \right){\rm{,}}.....{\rm{(1,m - 1),(0,m)}}} \right\}\)Now sum the joint pmf over \({\rm{(x,y)}} \in {\rm{A}}\)and use the binomial theorem, which says that

\({\rm{P(X + Y = m)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\sum\limits_{{\rm{k = 0}}}^{\rm{m}} {\left( {\begin{array}{*{20}{c}}{\rm{m}}\\{\rm{k}}\end{array}} \right){{\rm{a}}^{\rm{k}}}{{\rm{b}}^{{\rm{m - k}}}}{\rm{ = }}\left( {{\rm{a + b}}} \right)} ^{\rm{m}}}\)

Show that when \({\rm{X}}\) and \({\rm{Y}}\) are independent, \({\rm{Cov(X,Y) = Corr(X,Y) = 0}}\).
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