91Ó°ÊÓ

The appendix of a paper contains research-related information that isn't necessary to include in the main body.

a)
The cdf of a binomial random variable X with parameters n and p is the Cumulative Density Function.

\(\begin{aligned}B(x;n,p) = P(X \le x)\\ &= \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n.\end{aligned}\)

Theorem:

\(b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{\rm{ }}\end{array}} \right.\)

(a):

Using the codf and theorem given above, for \(n = 15,x = 4\) (at most 4), and \(p = 0.7\) ( 7 in 10 auto accidents invole a single vehicle), the following is true

\(\begin{aligned}B(4;15,0.7) = P(X \le 4)\\ &= \sum\limits_{y = 0}^4 b (y;15,0.7)\mathop = \limits^{(1)} 0.001\end{aligned}\)

(1): from the table of Appendix Table A.l.

Therefore,\(B\left( {4{\rm{ }};{\rm{ }}15,0.7} \right) = 0.001\).

b)

The cdf of a binomial random variable X with parameters n and p is a Cumulative Density Function.

\(\begin{aligned}B(x;n,p) = P(X \le x)\\ &= \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n\end{aligned}\)

Theorem:

\(b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{\rm{ }}\end{array}} \right.\)

(b):

Using the codf and theorem given above, for\(n = 15,x = 4\), and\(p = 0.7\)( 7 in 10 auto accidents involve a single vehicle), the following is true

\(\begin{aligned}b(4;15,0.7)\mathop = \limits^{(1)} B(4;15,0.7) - B{(3;15,0.7)^{(2)}}\\ &= 0.001 - 0.000\\ &= 0.001\end{aligned}\)

(1): We're expected to use Appendix Table A.l., where the only method to get exactly four that involve a single vehicle is to deduct at most four from at most three.

(2): from the Appendix Table A.l.

Therefore, \(b\,(4;15,\,0.7) = 0.001.\)

c)

The cdf of a binomial random variable X with parameters n and p is a Cumulative Density Function.

\(\begin{array}{l}B(x;n,p) = P(X \le x)\\ = \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n\end{array}\)

Theorem:

\(b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{\rm{ }}\end{array}} \right.\)

(c):

Using the cdf and theorem given above, for \(n = 15,x = 6\), and \(p = 0.3\) ( 3 in 10 auto accidents involve a multiple vehicles), the following is true

\(\begin{aligned}b(6;15,0.3)\mathop = \limits^{(1)} B(6;15,0.3) - B{(6;15,0.3)^{(2)}}\\ &= 0.869 - 0.722\\ &= 0.147\end{aligned}\)

(1) : We're expected to use Appendix Table A.l., where the only method to get exactly four that involve a single vehicle is to deduct at most four from at most three.

(2): from the Appendix Table A.l.

Therefore, \(\;b\,\left( {6{\rm{ }};{\rm{ }}15,0.3} \right) = 0.147\).

d)

Cumulative Density Function cdf of binomial random variable X with parameters n and p is

\(\begin{array}{c}B(x;n,p) = P(X \le x)\\ = \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n.\end{array}\)

(d):

For\(n = 15\), and\(p = 0.7\)( 7 in 10 auto accidents involve a single vehicle), the following is true

\(\begin{aligned}P(2 \le X \le 4)\mathop = \limits^{(1)} B(4;15,0.7) - B{(1;15,0.7)^{(2)}}\\ &= 0.001 - 0.000\\ &= 0.001\end{aligned}\)

(1) : we are supposed to use Appendix Table A.l. where the only way to do is to subtract at most four with at most 1 to get 2 , 3 or 4 single vehicle accidents;

(2) : from the Appendix Table A.l.

Therefore, \(P(2 \le X \le 4) = 0.001\).

e)

Cumulative Density Function cdf of binomial random variable X with parameters n and p is

\(\begin{array}{c}B(x;n,p) = P(X \le x)\\ = \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n\end{array}\)

(e):

For\(n = 15\), and\(p = 0.7\)( 7 in 10 auto accidents involve a single vehicle), the following is true

\(\begin{aligned}P(2 \le X)\mathop = \limits^{(1)} 1 - P(X \le 1) = 1 - B{(1;15,0.7)^{(2)}}\\ &= 1 - 0.000\\ &= 1\end{aligned}\)

(1): we are supposed to use Appendix Table A.l, this is one of the reasons we use complement (for any event A, \(P\left( {{A^\prime }} \right) = 1 - P(A)\));

(2) : from the Appendix Table A.l.

Therefore, P(2 \leq X)=1.

f)

Because four of them are single, the others must be many. The answer must be \(0.001\), which is the same as b).