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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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