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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

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls involve fax messages, and consider a sample of 25 incoming calls. What is the probability that

a. At most 6 of the calls involve a fax message?

b. Exactly 6 of the calls involve a fax message?

c. At least 6 of the calls involve a fax message?

d. More than 6 of the calls involve a fax message?

A manufacturer of integrated circuit chips wishes to control the quality of its product by rejecting any batch in which the proportion of defective chips is too high. To this end, out of each batch (10,000 chips), 25 will be selected and tested. If at least 5 of these 25 are defective, the entire batch will be rejected.

a. What is the probability that a batch will be rejected if 5% of the chips in the batch are in fact defective?

b. Answer the question posed in (a) if the percentage of defective chips in the batch is \({\bf{10}}\% \).

c. Answer the question posed in (a) if the percentage of defective chips in the batch is \({\bf{20}}\% \).

d. What happens to the probabilities in (a)鈥�(c) if the critical rejection number is increased from 5 to \({\bf{6}}\)?

A mail-order computer business has six telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table.

X

0

1

2

3

4

5

6

p(x)

.10

.15

.20

.25

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.06

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Calculate the probability of each of the following events.

a. {at most three lines are in use}

b. {fewer than three lines are in use}

c. {at least three lines are in use}

d. {between two and five lines, inclusive, are in use}

e. {between two and four lines, inclusive, are not in use}

f. {at least four lines are not in use}

The number of requests for assistance received by a towing service is a Poisson process with rate \({\rm{\alpha = 4}}\) per hour. a. Compute the probability that exactly ten requests are received during a particular \({\rm{2}}\)-hour period. b. If the operators of the towing service take a \({\rm{30}}\)-min break for lunch, what is the probability that they do not miss any calls for assistance? c. How many calls would you expect during their break?

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature of the project) in applying for a building permit. Let Y = the number of forms required of the next applicant. The probability that y forms are required is known to be proportional to y鈥攖hat is,\(p\left( y \right) = ky\)for\(y = 1, \ldots ,5\).

a. What is the value of k? (Hint:\(\sum\limits_{y = 1}^5 {p\left( y \right)} = 1\))

b. What is the probability that at most three forms arerequired?

c. What is the probability that between two and fourforms (inclusive) are required?

d. Could \(p\left( y \right) = \frac{{{y^2}}}{{50}}\)for \(y = 1, \ldots ,5\)be the pmf of Y?
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