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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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