91Ó°ÊÓ

Probability is the likelihood that an event will occur and is calculated by dividing the number of favourable outcomes by the total number of possible outcomes. The simplest example is a coin flip. When you flip a coin there are only two possible outcomes, the result is either heads or tails.

(a)

\(X\)Has a gamma distribution that is generalised.

\(\alpha = 12\)

\(\beta = 7\)

\(\gamma = 40\)

The (regular) gamma distribution has been defined as \(X\). \(x\)is substituted by \(x - \gamma \) in \(Y.\)

\(Y = X - \gamma \)

Alternatively, by solving the \(X\)equation:

\(X = Y + \gamma \)

A (regular) gamma distribution's mean and variance are given by:

\(E(Y) = \alpha \beta \)

\(V(Y) = \alpha {\beta ^2}\)

Fill in the known values and evaluate:

\(E(Y) = \alpha \beta = 12 \times 7 = 84\)

\(V(Y) = \alpha {\beta ^2} = 12 \times {7^2} = 588\)

The mean and variance for the linear combination \(W = aX + b\)are as follows:

\({\rm{E(W) = a E(X) + b}}\)

\(V(W) = {a^2}V(X)\)

We can then calculate the mean and variance of \(X\)using these properties:

\(\mu = E(X) = E(Y + \gamma ) = E(Y) + \gamma = 84 + 40 = 124\)

\({\sigma ^2} = V(X) = V(Y + \gamma ) = V(Y) = 588\)

The standard deviation is the square root of the variance:

\(\sigma = \sqrt {{\sigma ^2}} = \sqrt {588} = 14\sqrt 3 \approx 24.2487\)

Therefore, the mean value and standard deviation of X is \(\mu = E(X) = 124,\,\,\sigma = 14\sqrt 3 \approx 24.2487\)

(b)

Given: \(X\)has a gamma distribution that is generalised.

\(\alpha = 12\)

\(\beta = 7\)

\(\gamma = 40\)

\({\rm{P(100 < X < 150)}}\)

The (regular) gamma distribution has been defined as \(X.\)\(x\) is substituted by \(x - \gamma \)in \(Y\).

\(Y = X - \gamma \)

or solving the equation to \(X\):

\(X = Y + \gamma \)

Rewrite the given probability in terms of \(Y\):

\(P(100 < X < 150) = P(100 < Y + \gamma < 150)\)

\( = P(100 - \gamma < Y < 150 - \gamma )\)

\( = P(100 - 40 < Y < 150 - 40)\)

\( = P(60 < Y < 110)\)

\( = P(Y < 110) - P(Y \le 60)\)

\( = P(Y \le 110) - P(Y \le 60)\)

Property gamma distribution:

\(P(Y \le x) = F(x;\alpha ,\beta ) = F\left( {\frac{x}{\beta },\alpha } \right)\)

The incomplete gamma function is represented by \(F\).

For \(x = 110\) and \(x = 60\), use this property:

\(P(Y \le 110) = F\)\((\frac{{110}}{\beta },\)\(\alpha )\)

\( = F\left( {\frac{{110}}{7},12} \right)\)

\(P(Y \le 60) = F\)\((\frac{{60}}{\beta },\alpha )\)

\( = F\left( {\frac{{60}}{7},12} \right)\)

Formula incomplete gamma function:

\(F(x;\alpha ) = \int_0^x {\frac{1}{{\Gamma (\alpha )}}} {y^{\alpha - 1}}{e^{ - y}}dy\)

Use technology to evaluate the integrals (or use technology to find the value of an incomplete gamma function, or assess the integral 11 times):

\(F\left( {\frac{{110}}{7},12} \right) = \int_0^{110/7} {\frac{1}{{\Gamma (12)}}} {y^{12 - 1}}{e^{ - y}}dy \approx 0.8582\)

\(F\left( {\frac{{60}}{7},12} \right) = \int_0^{60/7} {\frac{1}{{\Gamma (12)}}} {y^{12 - 1}}{e^{ - y}}dy \approx 0.1575\)

The probability then becomes:

\(P(100 < X < 150) = P(Y \le 110) - P(Y \le 60) = F\left( {\frac{{110}}{7},12} \right) - F\left( {\frac{{60}}{7},12} \right)\)

\({\rm{ = 0}}{\rm{.8582 - 0}}{\rm{.1575}}\)

\({\rm{ = 0}}{\rm{.7007}}\)

\({\rm{ = 70}}{\rm{.07 \% }}\)

Therefore, the probability that flood volume is between 100 and 150 is \(70.07\% \).

(c)

Given: \(X\)has a gamma distribution that is generalised.

\(\alpha = 12\)

\(\beta = 7\)

\(\gamma = 40\)

\(P(X > \mu + \sigma )\)

Result part (a):

\(\mu = E(X) = 124\)

\(\sigma = 14\sqrt 3 \approx 24.2487\)

The (regular) gamma distribution has been defined as \(X\). \(x\) is substituted by \(x - \gamma \) in \(Y\).

\(X = Y + \gamma \)

Complement rule:

\(P({\mathop{\rm not}\nolimits} {\rm{A}}) = 1 - P(A)\)

Rewrite the given probability in terms of \(Y\) (in the last step we use the complement rule):

\(P(X > \mu + \sigma ) = P(Y + \gamma > \mu + \sigma ) = P(Y > \mu + \sigma - \gamma )\)

\( = P(Y > 124 + 14\sqrt 3 - 40)\)

\( = P(Y > 84 + 14\sqrt 3 )\)

\( = 1 - P(Y \le 84 + 14\sqrt 3 )\)

Gamma distribution of a property:

\(P(Y \le x) = F(x;\alpha ,\beta ) = F\left( {\frac{x}{\beta },\alpha } \right)\)

The incomplete gamma function is represented by \(F\).

For \(x = 84 + 14\sqrt 3 :\), use this property:

\(P(Y \le 84 + 14\sqrt 3 ) = F\left( {\frac{{84 + 14\sqrt 3 }}{\beta },\alpha } \right) = F\left( {\frac{{84 + 14\sqrt 3 }}{7},12} \right) = F(12 + 2\sqrt 3 ,12)\)

Formula incomplete gamma function:

\(F(x;\alpha ) = \int_0^x {\frac{1}{{\Gamma (\alpha )}}} {y^{\alpha - 1}}{e^{ - y}}dy\)

Utilize technology to evaluate the integrals (or use technology to discover the value of the incomplete gamma function, or assess the integral 11 times):

\(F(12 + 2\sqrt 3 ,12) = \int_0^{12 + 2\sqrt 3 } {\frac{1}{{\Gamma (12)}}} {y^{12 - 1}}{e^{ - y}}dy \approx 0.8441\)

The probability then becomes:

\(P(X > \mu + \sigma ) = 1 - P(Y \le 84 + 14\sqrt 3 )\)

\( = 1 - F(12 + 2\sqrt 3 ,12)\)

\({\rm{ = 1 - 0}}{\rm{.8441}}\)

\({\rm{ = 0}}{\rm{.1559}}\)

\({\rm{ = 15}}{\rm{.59 \% }}\)

Therefore, the probability that flood volume exceeds its mean value by more than one standard deviation is \(15.59\% \)

(d)

Given: \(X\) has a gamma distribution that is generalised.

\(\alpha = 12\)

\(\beta = 7\)

\(\gamma = 40\)

The (regular) gamma distribution has been defined as \(X\). \(x\) is substituted by \(x - \gamma \) in \(Y\).

\(Y = X - \gamma \)

or solving the equation to \(X\) :

\(X = Y + \gamma \)

The 95th percentile \(P\) has the property that \(95\% \)of all data values are below it:

\(P(X \le P) = 95\% = 0.95\)

In terms of \(Y\), rewrite the probability:

\(P(X \le P) = P(Y + \gamma \le P) = P(Y \le P - \gamma ) = 0.95\)

Gamma distribution of a property:

\(P(Y \le x) = F(x;\alpha ,\beta ) = F\left( {\frac{x}{\beta },\alpha } \right)\)

The incomplete gamma function is represented by \(F\).

For \(x = P - Y\), use this property:

\(P(Y \le P - \gamma ) = F\left( {\frac{{P - \gamma }}{\beta },\alpha } \right) = F\left( {\frac{{P - \gamma }}{7},12} \right)\)

Incomplete gamma function formula:

\(F(x;\alpha ) = \int_0^x {\frac{1}{{\Gamma (\alpha )}}} {y^{\alpha - 1}}{e^{ - y}}dy\)

Substitute 12 for \(\alpha \).

\(F(x,12) = \int_0^x {\frac{1}{{\Gamma (12)}}} {y^{12 - 1}}{e^{ - y}}dy = 0.95\)

Using technology, find the value \(x\) for which the integral is \(0.95\):

\(x \approx 18.2075\)

The likelihood then becomes:

\(P(X \le P) = P(Y + \gamma \le P) = P(Y \le P - \gamma ) = F\left( {\frac{{P - \gamma }}{7},\alpha } \right)\)

If \(\frac{{P - \gamma }}{7}\) is identical to the found value of \(x\), this probability is \(0.95\):

\(\frac{{P - \gamma }}{7} = 18.2075\)

Multiply each side by 7:

\(P - \gamma = 127.4525\)

Replace \(\gamma \)by 40:

\(P - 40 = 127.4525\)

To either side of the equation, add 40:

\(P = 167.4525\)

Thus, the 95th percentile is about \(167.4525.\)