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.

(a)

Since\({\rm{f(x)}}\)is a legitimate pdf, hence it must satisfy following condition –

\(\int_{{\rm{ - }}\infty }^\infty {{\rm{f(x)}} \cdot {\rm{dx = 1}}} \)

For the given pdf –

\(\begin{aligned}{\rm{k}}\int_{{\rm{ - }}\infty }^\infty {\rm{f}} {\rm{(x)}} \cdot dx &= \int_{\rm{0}}^{\rm{2}} {\rm{k}} {{\rm{x}}^{\rm{2}}} \cdot {\rm{dx = 1}}\\\left( {\frac{{{\rm{k}}{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{\rm{0}}^{\rm{2}} &= 1\\\frac{{{\rm{k(2}}{{\rm{)}}^{\rm{3}}}}}{{\rm{3}}}- 0 &= 1\\ k &= \frac{{\rm{3}}}{{\rm{8}}}\end{aligned}\)

Hence the pdf can finally be written as –

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{{\rm{3}}{{\rm{x}}^{\rm{2}}}}}{{\rm{8}}}}&{{\rm{1}} \le {\rm{X}} \le {\rm{2}}}\\{{\rm{0}}\;\;\;}&{{\rm{otherwise}}}\end{array}} \right.\)

Conditions satisfied by a pdf: For a pdf to be a legitimate pdf, it must satisfy the following two conditions –

1. \({\rm{f(x)}} \ge {\rm{0}}\)for all\({\rm{x}}\)
2. \(\int_{{\rm{ - }}\infty }^\infty {{\rm{f(x)}} \cdot {\rm{dx = 1}}} \)

Therefore, the value is obtained as \({\rm{k = }}\frac{{\rm{3}}}{{\rm{8}}}\).

(b)

The probabilitythat lecture ends within \({\rm{1}}\) minute of the end of the hour is denoted as \({\rm{P(X < 1)}}\).

\(\begin{array}{c}{\rm{P(X < 1) = }}\int_{\rm{0}}^{\rm{1}} {\frac{{{\rm{3}}{{\rm{x}}^{\rm{2}}}}}{{\rm{8}}}} \cdot {\rm{dx}}\\{\rm{ = }}\left( {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{8}}}} \right)_{\rm{0}}^{\rm{1}}\\{\rm{ = }}\left( {\frac{{\rm{1}}}{{\rm{8}}}{\rm{ - 0}}} \right)\\{\rm{ = 0}}{\rm{.125}}\end{array}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.125}}\).

(c)

The probability that lecture continues beyond the hour for between \({\rm{60}}\) and \({\rm{90}}\) seconds is denoted as \({\rm{P(X < 1}}{\rm{.5)}}\).

\(\begin{aligned}{\rm{P(1 < X < 1}}.5) &= \int_{\rm{1}}^{{\rm{1}}{\rm{.5}}} {\frac{{{\rm{3}}{{\rm{x}}^{\rm{2}}}}}{{\rm{8}}}} \cdot {\rm{dx}}\\ &= \left( {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{8}}}} \right)_{\rm{1}}^{{\rm{1}}{\rm{.5}}}\\ &= \left( {\frac{{{{{\rm{(1}}{\rm{.5)}}}^{\rm{3}}}}}{{\rm{8}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{8}}}} \right)\\ &= 0 {\rm{.2969}}\end{aligned}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.2969}}\).

(d)

The probability that lecture continues at least \({\rm{90}}\) seconds beyond the end of the hour is denoted as \({\rm{P(X}} \ge {\rm{1}}{\rm{.5)}}\).

\(\begin{aligned}{\rm{P(X}} \ge {\rm{1}} .5) &= \int_{1.5}^2 {\frac{{{\rm{3}}{{\rm{x}}^{\rm{2}}}}}{{\rm{8}}}} \cdot {\rm{dx}}\\ &= \left( {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{8}}}} \right)_{1.5}^2\\ &= \left( {\frac{{{{(2)}^3}}}{{\rm{8}}}{\rm{ - }}\frac{{{{(1.5)}^3}}}{{\rm{8}}}} \right)\\ & = 0{\rm{.5781}}\end{aligned}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.5781}}\).