91Ó°ÊÓ

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

The Binomial Random Variable\({\rm{X}}\)is defined as

\({\rm{X = }}\)the number of customers who qualify for membership out of\({\rm{n}}\)

Where conditions one to four on page\({\rm{117}}\)are satisfied (binomial experiment).

The described random variable has Binomial Distribution with parameters

\({\rm{n}}\)and\({\rm{p}}\)where\({\rm{p = 0}}{\rm{.1}}\).

(a)

There are\({\rm{25}}\)randomly selected current customers, which means that –

\({\rm{X}} \sim {\rm{Bin(25,0}}{\rm{.1)}}\)

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

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

The theorem is –

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

The following is true –

\(\begin{aligned}{\rm{P(2}} \le {\rm{X}} \le 6) &= B(6;25,0) {\rm{.1 - B(1;25,0)}}{\rm{.1}}\\ &= 0 {\rm{.991 - 0}}{\rm{.271}}\\ &= 0 {\rm{.720}}\end{aligned}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.720}}\).

(b)

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

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

Since in this it is given –

\({\rm{X\sim Bin(100,0}}{\rm{.1)}}\)

The following holds –

\(\begin{aligned} E(X) = np &= 100 \cdot {\rm{0}}{\rm{.1 = 10}}\\ V(X) &= npq = 100 \cdot {\rm{0}}{\rm{.1}} \cdot {\rm{0}}{\rm{.9 = 9}}\\{{\rm{\sigma }}_{\rm{X}}} &= \sqrt {{\rm{V(X)}}} {\rm{ = }}\sqrt {\rm{9}} {\rm{ = 3}}\end{aligned}\)

Therefore, the value is obtained as \({\rm{E(X) = 10}}\).

(c)

Find the following probability –

\(\begin{gathered}P(X \geqslant 7{\text{ }}when{\text{ }}p = 0.1) &= 1 - B(6;25,0.1) \\ &= 1 - 0.991 = 0.009 \\ \end{gathered} \)

Therefore, the value is obtained as \({\rm{0}}{\rm{.009}}\).

(d)

Do not reject the premise when –

\({\rm{x}} \le {\rm{6}}\)

Which is complement of –

\({\rm{x}} \ge {\rm{7}}\)

Thus, find the probability of event –

\({\rm{\{ X}} \le {\rm{6 when p = 0}}{\rm{.2\} }}\)

Which is –

\({\rm{P(X}} \le {\rm{6 when p = 0}}{\rm{.2) = B(6;25,0}}{\rm{.2) = 0}}{\rm{.78}}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.78}}\).