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The propagator is a function in quantum mechanics and quantum field theory that determines the probability amplitude for a particle to travel from one point to another in a given amount of time or with a particular amount of energy and momentum.

It is assumed that the proportion \({\rm{X}}\) of surface area covered by a particular plant in a randomly chosen quadrat has a typical beta distribution with parameters. \({\rm{\alpha = 5}}\) and \({\rm{\beta = 2}}\)

(a) For the given standard beta distribution, \({\rm{E(X)}}\) can be expressed as follows:

\(\begin{aligned} E(X) &= \frac{\alpha }{{\alpha + \beta }} \\ &= \frac{5}{{5 + 2}} \\ &= \frac{5}{7} \\ E(X) &= 0.7143 \\\end{aligned} \)

For the given standard beta distribution, the variance \({\rm{V(X)}}\) can be expressed as:

\(\begin{aligned} V(X) &= \frac{{\alpha \beta }}{{{{(\alpha + \beta )}^2}(\alpha + \beta + 1)}} \\ &= \frac{{5 \times 2}}{{{{(5 + 2)}^2}(5 + 2 + 1)}} \\ &= \frac{{10}}{{{{(7)}^2}(8)}} \\ V(X) &= 0.0255 \\\end{aligned} \)

(b) The pdf for the standard beta distribution is given as:

\({\rm{f(x,\alpha ,\beta ) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{\Gamma {\rm{(\alpha + \beta }})}}{{\Gamma {\rm{(\alpha ) \times }}\Gamma {\rm{(\beta }})}}{\rm{ \times }}{x^{{\rm{\alpha - 1}}}}{\rm{ \times (1 - x}}{{\rm{)}}^{{\rm{\beta - 1}}}}}&{{\rm{0}} \le {\rm{x}} \le 1}\\0&{{\rm{ otherwise }}}\end{array}} \right.\)

Substituting the parameters' values \({\rm{\alpha = 5}}\)and \({\rm{\beta = 1}}\), we get :

\({\rm{f(x,5,2) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{30 \times }}{{\rm{x}}^{\rm{4}}}{\rm{ \times (1 - x)}}}&{0 \le x \le 1}\\0&{{\rm{ otherwise }}}\end{array}} \right.\)

The \({\rm{P(X}} \le 0.2)\)can be written as :

\(\begin{aligned}P(X \leqslant 0.2) &= \int_0^{0.2} f (x) \times dx \hfill \\&= \int_0^{0.2} 3 0 \times {x^4} \times (1 - x) \times dx \hfill \\&= 30\int_0^{0.2} \times \left( {{x^4} - {x^5}} \right) \times dx \hfill \\&= 30\left[ {\frac{{{x^5}}}{5} - \frac{{{x^6}}}{6}} \right]_0^{0.2} \hfill \\&= 30\left[ {\frac{{{{(0.2)}^5}}}{5} - \frac{{{{(0.2)}^6}}}{6}} \right] \hfill \\P(X \leqslant 0.2) &= 0.0016 \hfill \\\end{aligned} \)

(c) The \({\rm{P(0}}{\rm{.2}} \le {\rm{X}} \le 0.4)\)can be written as:

\(\begin{aligned}P(0.2 \leqslant X \leqslant 0.4)&=\int_{0.2}^{0.4} f (x)\times dx\\&= \int_{0.2}^{0.4} 3 0\times {x^4}\times (1 - x) \times dx \\&= 30\int_{0.2}^{0.4}\times\left( {{x^4} - {x^5}} \right) \times dx \\\left.{= 30\left[ {\frac{{{x^5}}}{5} - \frac{{{x^6}}}{6}} \right]_{0.2}^{0.4} - \left( {\frac{{{{(0.2)}^5}}}{5} - \frac{{{{(0.2)}^6}}}{6}} \right)}\right]\\&= 30\left[ {\left( {\frac{{{{(0.4)}^5}}}{5} - \frac{{{{(0.4)}^6}}}{6}} \right)} \right]\\P(0.2 \leqslant X \leqslant 0.4) &= 0.0394 \\\end{aligned}\)

(d) Because \({\rm{X}}\) is the percentage of a plant's surface area that it covers.

As a result, the proportion of the sampling region not covered by the plant is predicted to be \({\rm{ }}\left( {{\rm{1 - X}}} \right)\).

Its predicted value is as follows:

\(\begin{aligned} E(1 - X) &= 1 - E(X) \\ &= 1 - 0.7143E(1 - X) \\ &= 0.2857 \\\end{aligned} \)