91Ó°ÊÓ

Probability denotes the possibility of something happening. It's a field of mathematics that studies the probability of a random event occurring.

a)

We are given \(X\sim {\mathop{\rm Bin}\nolimits} (25,0.05)\)

(Binomial Distribution).

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

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

Therefore, the following is true

\(\begin{array}{l}B(3;25,0.05) = P(X \le 3)\\ = \sum\limits_{y = 0}^x b (y;n,p)\mathop = \limits^{(1)} 0.9659\end{array}\)

(1) : the value can be found in Appendix Table A.l. (\(n = 25\), column " and row " . The value given here is correct up to fourth decimal.

The following holds,

\(\begin{aligned}P(X < 3)\mathop = \limits^{(1)} P(X \le 2)\\ = B(2;25,0.05)\mathop = \limits^{(2)} 0.8729\end{aligned}\)

(1): X can take only non negative values, which for\(\{ X < 3\} \)means that it can take values\(0,1,2\)or equally\(\{ X \le 2\} \);

(2) : The value can be found in Appendix Table A.l. (\(n = 25\), column " and row " 2 "). The value given here is correct up to fourth decimal.

b)

We are given \(X\sim {\mathop{\rm Bin}\nolimits} (25,0.05)\) (Binomial Distribution).

Cumulative Density Function cdf of binomial random variable\({\rm{X}}\)with parameters n and p is

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

The following is true

\(\begin{aligned}P(X \ge 4)\mathop = \limits^{(1)} 1 - P(X < 4)\mathop = \limits^{(2)} 1 - P(X \le 3)\\ = 1 - B(3;25,0.05)\\\mathop = \limits^{(3)} 1 - 0.9659\\ = 0.0341\end{aligned}\)

(1) : the complement of event \(\{ X \ge 4\} \)is event \(\{ X < 4\} \);

(2): X can take only non negative values, which indicates that events\(\{ X < 4\} \) and\(\{ X \le 3\} \)are the same'

(3) : the value can be found in Appendix Table A.l. ( $n=25$, column and row ". The value given here is correct up to fourth decimal.

Therefore, \(P(X \ge 4) = 0.0341\)

c)

We are given \(X\sim {\mathop{\rm Bin}\nolimits} (25,0.05\) ) (Binomial Distribution).

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

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

The following is true

\(\begin{aligned}P(1 \le X \le 3)\mathop = \limits^{(1)} P(X \le 3) - P(0 \le X) = B(3;25,0.05) - B(0;25,0.05)\\\mathop = \limits^{(2)} 0.9659 - 0.2779\\ = 0.688\end{aligned}\)

(1) : probability of event\(\{ 1 \le X \le 3\} \)gives us probability that\(X = 1,2,3\), which we can obtain by subtracting probability of event\(X = 0,1,2,3\) with probability that\(X = 0\);

(2) : the value can be found in Appendix Table A.1. (\(n = 25\), column

and row $ and " 0. The values given here are correct up to fourth decimal.

Therefore, \(P(1 \le X \le 3) = 0.688\).

d)

We are given \(X\sim {\mathop{\rm Bin}\nolimits} (25,0.05)\) (Binomial Distribution).

Proposition: For a binomial random variable\({\rm{X}}\) with parameters\({\rm{n}},{\rm{p}}\), and\(q = 1 - p\), the following is true

\(\begin{aligned}E(X) &= np\\V(X) &= np\\(1 - p) = npq\\{\sigma _X} &= \sqrt {npq} \end{aligned}\)

From the proposition, the following is true

\(\begin{aligned} E(X) &= np\\ &= 25 \cdot 0.05\\ &= 1.25.\end{aligned}\)

And

\(\begin{aligned}V(X) &= npq\\ &= 25 \cdot 0.05 \cdot (1 - 0.05)\\ &= 1.1875.\end{aligned}\) And the Standard Deviation

\(\begin{aligned} {\sigma _X} &= \sqrt {npq} \\ &= \sqrt {1.1875} \\ &= 1.0897\end{aligned}\)

\(\begin{aligned} E(X) = 1.25\\{\sigma _X} = 1.0897.\end{aligned}\)

Therefore, the answer is \(1.0897.\)

e)

We are given \(X\sim {\mathop{\rm Bin}\nolimits} (50,0.05)\) (Binomial Distribution).

\(({\rm{d}}):\)Theorem:

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

The probability that none has a food allergy is

\(\begin{array}{c}b(0;50,0.05) = \left( {\begin{array}{*{20}{c}}{50}\\0\end{array}} \right){0.05^0}{(1 - 0.05)^{50 - 0}}\\ = 0.0769\end{array}\)

Therefore, the answer is \(0.0769\).