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Chapter 3: Q70E (page 130)

An instructor who taught two sections of engineering statistics last term, the first with\({\rm{20}}\)students and the second with\({\rm{30}}\), decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first\({\rm{15}}\)graded projects. a. What is the probability that exactly\({\rm{10}}\)of these are from the second section? b. What is the probability that at least\({\rm{10}}\)of these are from the second section? c. What is the probability that at least\({\rm{10}}\)of these are from the same section? d. What are the mean value and standard deviation of the number among these\({\rm{15}}\)that are from the second section? e. What are the mean value and standard deviation of the number of projects not among these first\({\rm{15}}\)that are from the second section?

Short Answer

Expert verified

(a) The probability is obtained as:\({\rm{0}}{\rm{.2070 = 20}}{\rm{.70\% }}\).

(b) The probability is obtained as: \({\rm{0}}{\rm{.3799 = 37}}{\rm{.99 \% }}\).

(c) The probability is obtained as: \({\rm{0}}{\rm{.3939 = 39}}{\rm{.39\% }}\).

(d) The mean value is obtained as: \({\rm{9}}\) and standard deviation is obtained as: \( \approx {\rm{1}}{\rm{.603}}\).

(e) The mean value is obtained as: \({\rm{6}}\) and standard deviation is obtained as: \( \approx {\rm{1}}{\rm{.603}}\).

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

Step 2: Evaluating the probability

(a) It is given that:

The value of N is the population size which is \({\rm{12}}\).

The value of n is the number of draws which is \({\rm{15}}\).

The value of r is the number of observed successes which is \({\rm{30}}\).

Then, definition of hypergeometric distribution is:

\({\rm{P(y) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{r}}\\{\rm{y}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{N - r}}}\\{{\rm{n - y}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{\rm{N}}\\{\rm{n}}\end{array}} \right)}}\)

Now, putting the value as \({\rm{y = 10}}\) and then evaluating it as:

\(\begin{array}{c}{\rm{P(10) = }}\frac{{\left( {\begin{array}{*{20}{c}}{{\rm{30}}}\\{{\rm{10}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{50 - 30}}}\\{{\rm{15 - 10}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{50}}}\\{{\rm{15}}}\end{array}} \right)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{c}}{{\rm{30}}}\\{{\rm{10}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{20}}}\\{\rm{5}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{50}}}\\{{\rm{15}}}\end{array}} \right)}}\\ \approx {\rm{0}}{\rm{.2070}}\\{\rm{ = 20}}{\rm{.70\% }}\end{array}\)

Therefore, the value is: \({\rm{0}}{\rm{.2070 = 20}}{\rm{.70\% }}\).

03

Step 3: Evaluating the probability

(b) Using the hypergeometric distribution, add the relevant probabilities:

\(\begin{array}{c}{\rm{P(X}} \ge {\rm{10) = P(10) + P(11) + }}.......{\rm{ + P(15)}}\\ \approx {\rm{0}}{\rm{.3799 }}\\{\rm{ = 37}}{\rm{.99 \% }}\end{array}\)

Therefore, the value is: \({\rm{0}}{\rm{.3799 = 37}}{\rm{.99 \% }}\).

04

Step 4: Evaluating the probability

(c) We also need to account for the likelihood that at least \({\rm{10}}\) come from the first segment (r \({\rm{ = 20}}\) rather than r \({\rm{ = 30}}\)).

\(\begin{array}{c}{\rm{P(10) = }}\frac{{\left( {\begin{array}{*{20}{c}}{{\rm{20}}}\\{{\rm{10}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{50 - 20}}}\\{{\rm{15 - 10}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{50}}}\\{{\rm{15}}}\end{array}} \right)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{c}}{{\rm{20}}}\\{{\rm{10}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{30}}}\\{\rm{5}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{50}}}\\{{\rm{15}}}\end{array}} \right)}}\\ \approx {\rm{0}}{\rm{.0117 }}\\{\rm{ = 1}}{\rm{.17\% }}\end{array}\)

Using the hypergeometric distribution, add the relevant probabilities:

\(\begin{array}{c}{\rm{P(X}} \ge {\rm{10) = P(10) + P(11) + }}.......{\rm{ + P(15)}}\\ \approx {\rm{0}}{\rm{.0140 }}\\{\rm{ = 1}}{\rm{.40 \% }}\end{array}\)

Part b's probability should be added to this one.

\(\begin{array}{c}{\rm{0}}{\rm{.3799 + 0}}{\rm{.0140 = 0}}{\rm{.3939}}\\{\rm{ = 39}}{\rm{.39\% }}\end{array}\)

Therefore, the value is: \({\rm{0}}{\rm{.3939 = 39}}{\rm{.39\% }}\).

05

Step 5: Evaluating the mean value and standard deviation

(d) The hypergeometric distribution's mean is:

\(\begin{array}{c}{\rm{\mu = }}\frac{{{\rm{nr}}}}{{\rm{N}}}\\{\rm{ = }}\frac{{{\rm{15(30)}}}}{{{\rm{50}}}}\\{\rm{ = }}\frac{{{\rm{450}}}}{{{\rm{50}}}}\\{\rm{ = 9}}\end{array}\)

The hypergeometric distribution's standard deviation is:

\(\begin{array}{c}{\rm{\sigma = }}\sqrt {\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}{\rm{ \times \mu \times (1 - }}\frac{{\rm{r}}}{{\rm{N}}}{\rm{)}}} \\{\rm{ = }}\sqrt {\frac{{{\rm{50 - 15}}}}{{{\rm{50 - 1}}}}{\rm{ \times 9 \times (1 - }}\frac{{{\rm{30}}}}{{{\rm{50}}}}{\rm{)}}} \\ \approx {\rm{1}}{\rm{.603}}\end{array}\)

Therefore, the mean value is \({\rm{9}}\) and standard deviation is: \( \approx {\rm{1}}{\rm{.603}}\).

06

Step 6: Evaluating the mean value and standard deviation

(e) The number of projects NOT among those from the second part is distributed using r \({\rm{ = 20}}\) (first section) rather than r \({\rm{ = 30}}\) (second section).

The hypergeometric distribution's mean is:

\(\begin{array}{c}{\rm{\mu = }}\frac{{{\rm{nr}}}}{{\rm{N}}}\\{\rm{ = }}\frac{{{\rm{15(20)}}}}{{{\rm{50}}}}\\{\rm{ = }}\frac{{{\rm{300}}}}{{{\rm{50}}}}\\{\rm{ = 6}}\end{array}\)

The hypergeometric distribution's standard deviation is:

\(\begin{array}{c}{\rm{\sigma = }}\sqrt {\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}{\rm{ \times \mu \times (1 - }}\frac{{\rm{r}}}{{\rm{N}}}{\rm{)}}} \\{\rm{ = }}\sqrt {\frac{{{\rm{50 - 15}}}}{{{\rm{50 - 1}}}}{\rm{ \times 6 \times (1 - }}\frac{{{\rm{20}}}}{{{\rm{50}}}}{\rm{)}}} \\ \approx {\rm{1}}{\rm{.6036}}\end{array}\)

Therefore, the mean value is \({\rm{6}}\) and standard deviation is: \( \approx {\rm{1}}{\rm{.603}}\).

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Most popular questions from this chapter

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.

The article 'should You Report That Fender Bender?" (Consumer Reports, Sept. 2013: 15) reported that 7 in 10 auto accidents involve a single vehicle (the article recommended always reporting to the insurance company an accident involving multiple vehicles). Suppose 15 accidents are randomly selected. Use Appendix Table A.l to answer each of the following questions.

a. What is the probability that at most 4 involve a single vehicle?

b. What is the probability that exactly 4 involve a single vehicle?

c. What is the probability that exactly 6 involve multiple vehicles?

d. What is the probability that between 2 and 4 , inclusive, involve a single vehicle?

e. What is the probability that at least 2 involve a single vehicle?

f. What is the probability that exactly 4 involve a single vehicle and the other 11 involve multiple vehicles?

Grasshoppers are distributed at random in a large field according to a Poisson process with parameter a \({\rm{\alpha = 2}}\) per square yard. How large should the radius \({\rm{R}}\) of a circular sampling region be taken so that the probability of finding at least one in the region equals \({\rm{.99}}\)?

Two fair six-sided dice are tossed independently. Let M = the maximum of the two tosses (so M(1,5) =5, M(3,3) = 3, etc.).

a. What is the pmf of M? (Hint: First determine p(1), then p(2), and so on.)

b. Determine the cdf of M and graph it.

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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