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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

The manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are\(15,\;30,\;20\)min, respectively, and the standard deviations are\(1,\;2,\;1.5\)min, respectively. What is the probability that it takes at most\(1\)hour of machining time to produce a randomly selected component?

Suppose the amount of liquid dispensed by a certain machine is uniformly distributed with lower limit \({\rm{A = 8oz}}\) and upper limit\({\rm{B = 10oz}}\). Describe how you would carry out simulation experiments to compare the sampling distribution of the (sample) fourth spread for sample sizes\({\rm{n = 5,10,20}}\), and\({\rm{30}}\).

There are two traffic lights on a commuter's route to and from work. Let \({{\rm{X}}_{\rm{1}}}\) be the number of lights at which the commuter must stop on his way to work, and \({{\rm{X}}_{\rm{2}}}\) be the number of lights at which he must stop when returning from work. Suppose these two variables are independent, each with pmf given in the accompanying table (so \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}\) is a random sample of size \({\rm{n = 2}}\)).

a. Determine the pmf of \({{\rm{T}}_{\rm{o}}}{\rm{ = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}\).

b. Calculate \({{\rm{\mu }}_{{{\rm{T}}_{\rm{o}}}}}\). How does it relate to \({\rm{\mu }}\), the population mean?

c. Calculate \({\rm{\sigma }}_{{{\rm{T}}_{\rm{o}}}}^{\rm{2}}\). How does it relate to \({{\rm{\sigma }}^{\rm{2}}}\), the population variance?

d. Let \({{\rm{X}}_{\rm{3}}}\) and \({{\rm{X}}_{\rm{4}}}\) be the number of lights at which a stop is required when driving to and from work on a second day assumed independent of the first day. With \({{\rm{T}}_{\rm{o}}}{\rm{ = }}\) the sum of all four \({{\rm{X}}_{\rm{i}}}\) 's, what now are the values of \({\rm{E}}\left( {{{\rm{T}}_{\rm{a}}}} \right)\) and \({\rm{V}}\left( {{{\rm{T}}_{\rm{a}}}} \right)\)?

e. Referring back to (d), what are the values of \({\rm{P}}\left( {{{\rm{T}}_{\rm{o}}}{\rm{ = 8}}} \right)\) and \(\text{P}\left( {{\text{T}}_{\text{e}}}\text{ }\!\!{}^\text{3}\!\!\text{ 7} \right)\) (Hint: Don't even think of listing all possible outcomes!)

A restaurant serves three fixed-price dinners costing \({\rm{12, 15and 20}}\). For a randomly selected couple dining at this restaurant, let \({\rm{X = }}\)the cost of the man鈥檚 dinner and \({\rm{Y = }}\)the cost of the woman鈥檚 dinner. The joint \({\rm{pmf's}}\) of \({\rm{X and Y }}\)is given in the following table:

a. Compute the marginal \({\rm{pmf's}}\)of \({\rm{X and Y }}\)

b. What is the probability that the man鈥檚 and the woman鈥檚 dinner cost at most \({\rm{15}}\)each?

c. Are \({\rm{X and Y }}\)independent? Justify your answer.

d. What is the expected total cost of the dinner for the two people?

e. Suppose that when a couple opens fortune cookies at the conclusion of the meal, they find the message 鈥淵ou will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen.鈥 How much would the restaurant expect to refund?

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a. Determine the sampling distribution of\({\rm{\bar X}}\), calculate\({\rm{E(\bar X)}}\), and compare to\({\rm{\mu }}\).

b. Determine the sampling distribution of the sample variance\({{\rm{S}}^{\rm{2}}}\), calculate\({\rm{E}}\left( {{{\rm{S}}^{\rm{2}}}} \right)\), and compare to\({{\rm{\sigma }}^{\rm{2}}}\).

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