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Chapter 5: Q8E (page 211)

A stockroom currently has \({\rm{30}}\) components of a certain type, of which \({\rm{8}}\) were provided by supplier \({\rm{1,10}}\)by supplier \({\rm{2}}\) , and \({\rm{12}}\) by supplier \({\rm{3}}\). Six of these are to be randomly selected for a particular assembly. Let \({\rm{X = }}\) the number of supplier l's components selected, \({\rm{Y = }}\) the number of supplier \({\rm{2}}\) 's components selected, and \({\rm{p(x,y)}}\) denote the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\).

a. What is \({\rm{p(3,2)}}\) ? (Hint: Each sample of size \({\rm{6}}\) is equally likely to be selected. Therefore, \({\rm{p(3,2) = }}\) (number of outcomes with \({\rm{X = 3}}\) and \({\rm{Y = 2)/(}}\) total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.)

b. Using the logic of part (a), obtain \({\rm{p(x,y}}\) ). (This can be thought of as a multivariate hypergeometric distribution-sampling without replacement from a finite population consisting of more than two categories.)

Answer

Short Answer

Expert verified

a) The value of \({\rm{p}}\left( {{\rm{3,2}}} \right)\) is \({\rm{p}}\left( {{\rm{3,2}}} \right){\rm{ = 0}}{\rm{.0509}}\).

b) The solution is \({\rm{p(x,y) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{\left( {\begin{aligned}{*{20}{c}}{\rm{8}}\\{\rm{x}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{10}}}\\{\rm{y}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{12}}}\\{{\rm{6 - x - y}}}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}{{\rm{30}}}\\{\rm{6}}\end{aligned}} \right)}}}&{{\rm{,0~x + y~6 and x,y^I }}{{\rm{N}}_{\rm{0}}}}\\{\rm{0}}&{{\rm{, otherwise}}{\rm{. }}}\end{array}} \right.\)

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

Given in question

There are total of \({\rm{30}}\) components

\({\rm{8}}\)provided by supplier \({\rm{1}}\),

\({\rm{10}}\)provided by supplier \({\rm{2}}\),

\({\rm{12}}\)provided by supplier \({\rm{3}}\).

Six of these are randomly selected.

03

Determining the value of \({\rm{p}}\left( {{\rm{3,2}}} \right)\)

a)

As explained in the hint, we have that

\({\rm{p(3,2) = }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{8}}\\{\rm{3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{10}}}\\{\rm{2}}\end{array}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}{{\rm{30}}}\\{\rm{6}}\end{aligned}} \right)}}\) Where \(\left( {\begin{aligned}{*{20}{l}}{\rm{8}}\\{\rm{3}}\end{aligned}} \right)\)

represents the number of ways to take \({\rm{3}}\) components from supplier\({\rm{1}}\),

\(\left( {\begin{aligned}{*{20}{c}}{{\rm{10}}}\\{\rm{2}}\end{aligned}} \right)\)

represents the number of ways to take \({\rm{2}}\) components from supplier \({\rm{2}}\),

\(\left( {\begin{aligned}{*{20}{c}}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)\)

represents the number of ways to take \({\rm{1}}\) component from the supplier \({\rm{3}}\). By the product rule for counting, the numerator would be

\(\left( {\begin{aligned}{*{20}{l}}{\rm{8}}\\{\rm{3}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{10}}}\\{\rm{2}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right){\rm{ = 56 \times 45 \times 12 = 30,240}}\)

The denominator is total number of outcomes, which is the number of ways to take \({\rm{6}}\) components out of\({\rm{30}}\), hence

\(\left( {\begin{aligned}{*{20}{c}}{{\rm{30}}}\\{\rm{6}}\end{aligned}} \right){\rm{ = 593,775}}\)

Now, the requested probability is

\({\rm{p(3,2) = }}\frac{{{\rm{30,240}}}}{{{\rm{593,775}}}}{\rm{ = 0}}{\rm{.0509}}\)

04

Obtaining the value of \({\rm{p}}\left( {{\rm{x,y}}} \right)\)

(b):

The bivariate (multivariate) hypergeometric distribution, with the same logic as above, is given with

\({\rm{p(x,y) = }}\left\{ {\begin{aligned}{*{20}{l}}{\frac{{\left( {\begin{aligned}{*{20}{c}}{\rm{8}}\\{\rm{x}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{10}}}\\{\rm{y}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{12}}}\\{{\rm{6 - x - y}}}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}{{\rm{30}}}\\{\rm{6}}\end{aligned}} \right)}}}&{{\rm{,0~x + y~6 and x,y^I }}{{\rm{N}}_{\rm{0}}}}\\{\rm{0}}&{{\rm{, otherwise}}{\rm{. }}}\end{aligned}} \right.\)

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Most popular questions from this chapter

Suppose the proportion of rural voters in a certain state who favor a particular gubernatorial candidate is\(.{\bf{45}}\)and the proportion of suburban and urban voters favouring the candidate is\(.{\bf{60}}\). If a sample of\({\bf{200}}\)rural voters and\({\bf{300}}\)urban and suburban voters is obtained, what is the approximate probability that at least\(\;{\bf{250}}\)of these voters favour this candidate?

A more accurate approximation to \({\rm{E}}\left( {{\rm{h}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}} \right)} \right)\) in Exercise 95 is

\(h\left( {{\mu _1}, \ldots ,{\mu _n}} \right) + \frac{1}{2}\sigma _1^2\left( {\frac{{{\partial ^2}h}}{{\partial x_1^2}}} \right) + \cdots + \frac{1}{2}\sigma _n^2\left( {\frac{{{\partial ^2}h}}{{\partial x_n^2}}} \right)\)

Compute this for \({\rm{Y = h}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{,}}{{\rm{X}}_{\rm{4}}}} \right)\)given in Exercise 93 , and compare it to the leading term \({\rm{h}}\left( {{{\rm{\mu }}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{\mu }}_{\rm{n}}}} \right)\).

Compute the correlation coefficient \({\rm{\rho }}\) for \({\rm{X}}\) and \({\rm{Y}}\)(the covariance has already been computed).

Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable鈥擷 for the right tire and Y for the left tire, with joint pdf fsx, yd 5 5 Ksx2 1 y2 d 20 # x # 30, 20 # y # 30 0 otherwise

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{K(}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{)}}}&{,{\rm{20}} \le {\rm{x}} \le {\rm{30,20}} \le {\rm{y}} \le {\rm{30}}}\\{\rm{0}}&{,{\rm{ otherwise }}}\end{array}} \right.\)

a. What is the value of K?

b. What is the probability that both tires are underfilled?

c. What is the probability that the difference in air pressure between the two tires is at most 2 psi?

d. Determine the (marginal) distribution of air pressure in the right tire alone.

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Suppose that when the pH of a certain chemical compound is\(5.00\), the pH measured by a randomly selected beginning chemistry student is a random variable with a mean of\(5.00\)and a standard deviation .2. A large batch of the compound is subdivided and a sample is given to each student in a morning lab and each student in an afternoon lab. Let\(X = \)the average pH as determined by the morning students and\(Y = \)the average pH as determined by the afternoon students.

a. If pH is a normal variable and there are\(25\)students in each lab, compute\(P\left( { - .1 \le X - Y \le - .1} \right)\)

b. If there are\(36\)students in each lab, but pH determinations are not assumed normal, calculate (approximately)\(P\left( { - .1 \le X - Y \le - .1} \right)\).
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