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The anticipated value is an extension of the weighted average in probability theory. The average score of a large number of randomly determined outcomes of a random variable is known as the expected value.

It is given that Y is the distance from the left end at which the break occurs. It is also given that Y has pdf \({\rm{f}}({\rm{y}})\), written as:

\(f(y) = \left\{ {\begin{array}{*{20}{l}}{\left( {\frac{1}{{24}}} \right) \cdot y \cdot \left( {1 - \frac{y}{{12}}} \right)}&{0 \le y \le 12}\\0&{{\rm{ otherwise }}}\end{array}} \right.\)

(a) We recall the definition of CDF as a continuous variable.

Definition: The cumulative distribution function F(y) for a continuous \(rv\;Y\)is defined for every number y by

\(F(y) = P(Y \le y) = \int_{ - \infty }^y f (x) \cdot dx\)

For any number y between \(0{\rm{ }}and{\rm{ }}12\)

\(\begin{aligned}F(y) & = \int_0^y {\left( {\frac{1}{{24}}} \right)} \cdot x \cdot \left( {1 - \frac{x}{{12}}} \right)\\ &= \frac{1}{{24}} \cdot \int_0^y {\left( {x - \frac{{{x^2}}}{{12}}} \right)} \cdot dx\\ & = \frac{1}{{24}} \cdot \left( {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{{36}}} \right)_0^y\\ & = \frac{1}{{24}} \cdot \left( {\frac{{{y^2}}}{2} - \frac{{{y^3}}}{{36}}} \right)\\F(y) & = \frac{{{y^2}}}{{48}} - \frac{{{y^3}}}{{864}}\end{aligned}\)

Thus F(y) can be given as:

\(F(y) = \left\{ {\begin{aligned}{{0}}&{y < 0}\\{\frac{{{y^2}}}{{48}} - \frac{{{y^3}}}{{864}}}&{0 \le y \le 12}\\1&{y > 12}\end{aligned}} \right.\)

(b) Computing\(P(X \le 4)\):

\(\begin{aligned}{l}P(X \le 4) & = F(4) = \frac{{{{(4)}^2}}}{{48}} - \frac{{{{(4)}^3}}}{{864}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & = \frac{1}{3} - \frac{2}{{27}} = \frac{9}{{27}} - \frac{2}{{27}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & = \frac{7}{{27}}\\P(X \le 4) &= 0.2592\end{aligned}\)

Computing\(P(6 < Y)\):

\(\begin{aligned}{l}P(6 < Y) & = 1 - F(6)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & = 1 - \left( {\frac{{{{(6)}^2}}}{{48}} - \frac{{{{(6)}^3}}}{{864}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & = 1 - \left( {\frac{3}{4} - \frac{1}{4}} \right) = 1 - (0.5)\\P(6 < Y) & = 0.5\end{aligned}\)

Computing\(P(4 \le Y \le 6)\):

\(\begin{aligned}{l}P(4 \le Y \le 6) & = F(6) - F(4)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & = 0.5 - 0.2592\\P(4 \le X \le 6) & = 0.2408\end{aligned}\)

Proposition: Let X be a continuous rv with p d f(x) and cdf F(x). Then for any number a,

\(\begin{aligned}{l}P(X \le a) & = F(a)\\P(a \le X) & = 1 - F(a)\end{aligned}\)

Proposition: Let\({\rm{X}}\)be a continuous rv with \(pdf({\rm{x}})\) and \(cdf\;{\rm{F}}({\rm{x}})\). Then for any two numbers a and b with a<b,

\(P(a \le X \le b) = F(b) - F(a)\)

(c) The mean value of the given distribution can be given as:

\(\begin{aligned}E(Y) & = \int_{ - \infty }^\infty y \cdot f(y) \cdot dy\\ & = \int_0^{12} y \cdot \left( {\frac{1}{{24}}} \right) \cdot y \cdot \left( {1 - \frac{y}{{12}}} \right) \cdot dx\\ & = \frac{1}{{24}} \cdot \int_0^{12} {\left( {{y^2} - \frac{{{y^3}}}{{12}}} \right)} \cdot dx\\ & = \frac{1}{{24}} \cdot \left( {\frac{{{{(y)}^3}}}{3} - \frac{{{{(y)}^4}}}{{48}}} \right)_0^{12}\\ & = \frac{1}{{24}} \cdot \left( {\frac{{{{(12)}^3}}}{3} - \frac{{{{(12)}^4}}}{{48}}} \right)\\E(Y) & = 6\end{aligned}\)

Definition: The expected or mean value of a continuous r v X with pdf f(x) is

\(\mu = E(X) = \int_{ - \infty }^\infty x \cdot f(x) \cdot dx\)

\(\begin{aligned}E\left( {{Y^2}} \right) & = \int_{ - \infty }^\infty {{y^2}} \cdot f(y) \cdot dy\\ & = \int_0^{12} {{y^2}} \cdot \left( {\frac{1}{{24}}} \right) \cdot y \cdot \left( {1 - \frac{y}{{12}}} \right) \cdot dx\\ & = \frac{1}{{24}} \cdot \int_0^{12} {\left( {{y^3} - \frac{{{y^4}}}{{12}}} \right)} \cdot dx\\ & = \frac{1}{{24}} \cdot \left( {\frac{{{{(y)}^4}}}{4} - \frac{{{{(y)}^5}}}{{60}}} \right)_0^{12}\\ & = \frac{1}{{24}} \cdot \left( {\frac{{{{(12)}^4}}}{4} - \frac{{{{(12)}^5}}}{{60}}} \right)\\E\left( {{Y^2}} \right) & = 43.2\end{aligned}\)

As we have already calculated E(X) in the last part: \(E(Y) = 6\), hence we use following proposition:

Proposition:\(V(Y) = E\left( {{Y^2}} \right) - E{(Y)^2}\)

Using this, we can write:

\(\begin{array}{l}V(Y) = 43.2 - {(6)^2}\\V(Y) = 7.2\end{array}\)

Definition: If \({\rm{X}}\)is a continuous \({\rm{rv}}\)with pdf f(x) and h(X) is any function of \({\rm{X}}\), then

\(E(h(x)) = \int_{ - \infty }^\infty h (x) \cdot f(x) \cdot dx\)

(d) Since the expected break point is calculated as\(6\)inches in last part, hence the probability that the break point occurs more than 2 inches from the expected break point can be represented as \(P(Y < 4\,or\,Y > 8)\).

This event is the complement of the event that the break point occurs within 2 inches from the expected break point. Hence

\(\begin{array}{l}P(Y < 4{\rm{ or }}Y > 8) & = 1 - P(4 \le Y \le 8)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & = 1 - (F(8) - F(4))\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & = 1 - \left( {\left( {\frac{{{{(8)}^2}}}{{48}} - \frac{{{{(8)}^3}}}{{864}}} \right) - \left( {\frac{{{{(4)}^2}}}{{48}} - \frac{{{{(4)}^3}}}{{864}}} \right)} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & = 1 - \left( {\frac{{(20}}{{27}} - \frac{7}{{27}}} \right) & = \frac{{14}}{{27}}\\P(Y < 4{\rm{ or }}Y > 8) & = 0.5185\end{array}\)

(e) Let X be the random variable denoting the length of the shorter segment when the break occurs. Then we can write X in terms of Y as:

\(x = \left\{ {\begin{array}{*{20}{l}}y&{0 \le y \le 6}\\{12 - y}&{6 < y \le 12}\end{array}} \right.\)

Hence expected value of X can be written as:

\(\begin{aligned}E(X) & = \int_0^6 y \cdot f(y) \cdot dy + \int_6^{12} {(12 - y)} \cdot f(y) \cdot dy\\ & = \int_0^6 y \cdot \left( {\frac{1}{{24}}} \right) \cdot y \cdot \left( {1 - \frac{y}{{12}}} \right) \cdot dx + \int_6^{12} {(12 - y)} \cdot \left( {\frac{1}{{24}}} \right) \cdot y \cdot \left( {1 - \frac{y}{{12}}} \right) \cdot dx\\ & = \frac{1}{{24}} \cdot \int_0^6 {\left( {{y^2} - \frac{{{y^3}}}{{12}}} \right)} \cdot dx + \frac{1}{2} \cdot \int_6^{12} {\left( {y - \frac{{{y^2}}}{{12}}} \right)} \cdot dx - \frac{1}{{24}} \cdot \int_6^{12} {\left( {{y^2} - \frac{{{y^3}}}{{12}}} \right)} \cdot dx\\ & = \frac{1}{{24}} \cdot \left( {\frac{{{y^3}}}{3} - \frac{{{y^4}}}{{48}}} \right)_0^6 + \frac{1}{2} \cdot \left( {\frac{{{y^2}}}{2} - \frac{{{y^3}}}{{36}}} \right)_6^{12} - \frac{1}{{24}} \cdot \left( {\frac{{{y^3}}}{3} - \frac{{{y^4}}}{{48}}} \right)_6^{12}\\ & = \frac{1}{{24}} \cdot {\left( {\frac{{{6^3}}}{3} - \frac{{{6^4}}}{{48}}} \right)_1} + \frac{1}{2} \cdot \left( {\left( {\frac{{{{12}^2}}}{2} - \frac{{{{12}^3}}}{{36}}} \right) - \left( {\frac{{{6^2}}}{2} - \frac{{{6^3}}}{{36}}} \right)} \right) - \frac{1}{{24}} \cdot \left( {\left( {\frac{{{{12}^3}}}{3} - \frac{{{{12}^4}}}{{48}}} \right) - \left( {\frac{{{6^3}}}{3} - \frac{{{6^4}}}{{48}}} \right)} \right)\\E(Y) & = \frac{{15}}{8} + 6 - \frac{{33}}{8} & = 3{\rm{ inches }}\end{aligned}\)