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Chapter 3: Q68E (page 130)

Eighteen individuals are scheduled to take a driving test at a particular DMV office on a certain day, eight of whom will be taking the test for the first time. Suppose that six of these individuals are randomly assigned to a particular examiner, and let X be the number among the six who are taking the test for the first time. a. What kind of a distribution does X have (name and values of all parameters)? b. Compute P(X\({\rm{ = 2}}\)), P(X\( \le {\rm{ 2}}\)), and P(X\( \ge {\rm{ 2}}\)). c. Calculate the mean value and standard deviation of X.

Short Answer

Expert verified

(a) The value of X has hyper geometric distribution \({\rm{h(x;6,8,18)}}\).

(b) The values are obtained as: \(\begin{array}{c}{\rm{P(X = 2) = 0}}{\rm{.3167}}\\{\rm{P(X}} \le {\rm{2) = 0}}{\rm{.4366}}\\{\rm{P(X}} \ge {\rm{2) = 0}}{\rm{.8801}}\end{array}\).

(c) The mean value is \(\begin{array}{c}{\rm{E(X) = 2}}{\rm{.6667}}\\{\rm{V(X) = 1}}{\rm{.0458}}\end{array}\) and the standard deviation is: \({{\rm{\sigma }}_{\rm{X}}}{\rm{ = 1}}{\rm{.0226}}\).

Step by step solution

01

Define Discrete random variables

A discrete random variable is one that can only take on a finite number of different values.

02

What is the kind of distribution?

(a) This is an example of the Hyper geometric distribution in action. Consider the following scenario:

We have a total of N persons (in our example\({\rm{18}}\),) of which\({\rm{8}}\)are first-time test takers (we'll consider this a success). Out of those\({\rm{18}}\), we need to choose\({\rm{6}}\)people, and each subgroup is equally likely.

We also want to know how probable it is that we choose a specific amount of first-timers.

Taking everything into account, we can determine that this is a hyper geometric distribution, with the following properties:

\({\rm{h(x;n,M,N) = h(x;6,8,18)}}\)

Therefore, it is a hyper geometric distribution: \({\rm{h(x;6,8,18)}}\).

03

Computing the values

(b) Assume that the population has M successes (S) and N-M failures (F). If X is a random variable.

The value of X is denoted as number of successes in a random sample size n.

It has a probability mass function after that.

\(\begin{array}{c}{\rm{h(x;n,M,N) = P(X = x)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{M}}\\{\rm{x}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{N - M}}}\\{{\rm{n - x}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{\rm{N}}\\{\rm{n}}\end{array}} \right)}}\end{array}\)

for all x integers where

\({\rm{max\{ 0,n - N - M\} }} \le {\rm{x}} \le {\rm{min\{ n,M\} }}\)

Hyper geometric distribution is the name given to the probability distribution.

The parameters are \({\rm{N = 18}}\), \({\rm{M = 8}}\), and \({\rm{n = 6}}\), as shown in (a). The following statement is correct:

\(\begin{array}{c}{\rm{P(X = 2) = h(2;6,8,18)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{8}}\\{\rm{2}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{18 - 8}}}\\{{\rm{6 - 2}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{18}}}\\{\rm{6}}\end{array}} \right)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{8}}\\{\rm{2}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{10}}}\\{\rm{4}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{18}}}\\{\rm{6}}\end{array}} \right)}}\\{\rm{ = }}\frac{{{\rm{28 \times 210}}}}{{{\rm{18564}}}}\\{\rm{ = 0}}{\rm{.3167}}\end{array}\)

It then holds:

\(\begin{array}{c}{\rm{P(X}} \le {\rm{2) }}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P(X = 0) + P(X = 1) + P(X = 2)}}\\{\rm{ = h(0;6,8,18) + h(1;6,8,18) + h(2;6,8,18)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{8}}\\{\rm{0}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{18 - 8}}}\\{{\rm{6 - 0}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{18}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ + }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{8}}\\{\rm{1}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{18 - 8}}}\\{{\rm{6 - 1}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{18}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ + }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{8}}\\{\rm{2}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{18 - 8}}}\\{{\rm{6 - 2}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{18}}}\\{\rm{6}}\end{array}} \right)}}\\{\rm{ = }}\frac{{{\rm{1 \times 17280}}}}{{{\rm{18564}}}}{\rm{ + }}\frac{{{\rm{8 \times 252}}}}{{{\rm{18564}}}}{\rm{ + }}\frac{{{\rm{28 \times 17280}}}}{{{\rm{18564}}}}\\{\rm{ = 0}}{\rm{.0113 + 0}}{\rm{.1086 + 0}}{\rm{.3167}}\\{\rm{ = 0}}{\rm{.4366}}\end{array}\)

(1): see hyper geometric distribution pmf; it only accepts non-negative integer values.

It then again holds:

\(\begin{array}{c}{\rm{P(X}} \ge {\rm{2) }}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{1 - P(X < 2)}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{1 - (P(X = 0) + P(X = 1))}}\\{\rm{ = 1 - h(0;6,8,18) - h(1;6,8,18)}}\\{\rm{ = 1 - }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{8}}\\{\rm{0}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{18 - 8}}}\\{{\rm{6 - 0}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{18}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ - }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{8}}\\{\rm{1}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{18 - 8}}}\\{{\rm{6 - 1}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{18}}}\\{\rm{6}}\end{array}} \right)}}\\{\rm{ = }}\frac{{{\rm{1 \times 17280}}}}{{{\rm{18564}}}}{\rm{ + }}\frac{{{\rm{8 \times 252}}}}{{{\rm{18564}}}}\\{\rm{ = 0}}{\rm{.0113 + 0}}{\rm{.1086}}\\{\rm{ = 0}}{\rm{.8801}}\end{array}\)

(1):event \({\rm{(X}} \ge {\rm{2)}}\) counterpart is event \({\rm{X < 2}}\);

(2):X can only take non-negative integer values; \({\rm{0}}\) and \({\rm{1}}\) are the only two that are less than \({\rm{2}}\).

Therefore, the values are: \(\begin{array}{c}{\rm{P(X = 2) = 0}}{\rm{.3167}}\\{\rm{P(X}} \le {\rm{2) = 0}}{\rm{.4366}}\\{\rm{P(X}} \ge {\rm{2) = 0}}{\rm{.8801}}\end{array}\).

04

Evaluating the mean value and standard deviation

(c) For random variable X with hyper geometric distribution and pmf h(x; n, M, N), the following is true.

\(\begin{array}{c}{\rm{E(X) = n \times }}\frac{{\rm{M}}}{{\rm{N}}}\\{\rm{V(X) = }}\left( {\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}} \right){\rm{ \times n \times }}\frac{{\rm{M}}}{{\rm{N}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{\rm{M}}}{{\rm{N}}}} \right)\end{array}\)

X has a hyper geometric distribution with the parameters \({\rm{N = 18,M = 8 and n = 6}}\) (see in part a). As a result, the following is correct:

\(\begin{array}{c}{\rm{E(X) = 6 \times }}\frac{{\rm{8}}}{{{\rm{18}}}}\\{\rm{ = 2}}{\rm{.6667}}\end{array}\)

Also, based on the above premise, the following is true:

\(\begin{array}{c}{\rm{V(X) = }}\left( {\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}} \right){\rm{ \times n \times }}\frac{{\rm{M}}}{{\rm{N}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{\rm{M}}}{{\rm{N}}}} \right)\\{\rm{ = }}\left( {\frac{{{\rm{18 - 6}}}}{{{\rm{18 - 1}}}}} \right){\rm{ \times 6 \times }}\frac{{\rm{8}}}{{{\rm{18}}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{\rm{8}}}{{{\rm{18}}}}} \right)\\{\rm{ = 1}}{\rm{.0458}}\end{array}\)

The standard deviation of the data is:

\(\begin{array}{c}{{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{\rm{V(X)}}} \\{\rm{ = }}\sqrt {{\rm{1}}{\rm{.0458}}} \\{\rm{ = 1}}{\rm{.0226}}\end{array}\)

Therefore, the values are: \(\begin{array}{c}{\rm{E(X) = 2}}{\rm{.6667}}\\{\rm{V(X) = 1}}{\rm{.0458}}\\{{\rm{\sigma }}_{\rm{X}}}{\rm{ = 1}}{\rm{.0226}}\end{array}\).

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Most popular questions from this chapter

Suppose that trees are distributed in a forest according to a two-dimensional Poisson process with parameter\({\rm{\alpha }}\), the expected number of trees per acre, equal to\({\rm{80}}\). a. What is the probability that in a certain quarter-acre plot, there will be at most\({\rm{16}}\)trees? b. If the forest covers\({\rm{85,000}}\)acres, what is the expected number of trees in the forest? c. Suppose you select a point in the forest and construct a circle of radius\({\rm{.1}}\)mile. Let X = the number of trees within that circular region. What is the pmf of X? (Hint:\({\rm{1}}\)sq mile\({\rm{ = 640}}\)acres.)

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If the sample space S is an infinite set, does this necessarily imply that any rv X defined from S will have an infinite set of possible values? If yes, say why. If no, give an example.

Consider a disease whose presence can be identified by carrying out a blood test. Let \({\rm{p}}\) denote the probability that a randomly selected individual has the disease. Suppose \({\rm{n}}\) individuals are independently selected for testing. One way to proceed is to carry out a separate test on each of the \({\rm{n}}\) blood samples. A potentially more economical approach, group testing, was introduced during World War \({\rm{II}}\) to identify syphilitic men among army inductees. First, take a part of each blood sample, combine these specimens, and carry out a single test. If no one has the disease, the result will be negative, and only the one test is required. If at least one individual is diseased, the test on the combined sample will yield a positive result, in which case the n individual tests are then carried out. If \({\rm{p = }}{\rm{.1}}\) and \({\rm{n = 3}}\), what is the expected number of tests using this procedure? What is the expected number when \({\rm{n = 5}}\)? (The article 鈥淩andom Multiple-Access Communication and Group Testing鈥� (IEEE Trans. on Commun., \({\rm{1984: 769 - 774}}\)) applied these ideas to a communication system in which the dichotomy was active/ idle user rather than diseased/non-diseased.)

A test for the presence of a certain disease has probability\({\rm{.20}}\)of giving a false-positive reading (indicating that an individual has the disease when this is not the case) and probability\({\rm{.10}}\)of giving a false-negative result. Suppose that ten individuals are tested, five of whom have the disease and five of whom do not. Let\({\rm{X = }}\)the number of positive readings that result.

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