Question

Suppose the proportion X of surface area in a randomly selected quadrat that is covered by a certain plant has a standard beta distribution with α = 4 and β = 3.(a) Compute E(X) and V(X). (Round your answers to four decimal places.)E(X) = Correct: Your answer is correct.V(X) = Correct: Your answer is correct.(b) Compute P(X ≤ 0.5). (Round your answer to four decimal places.)

Answer 1

Answer:

(a) The value of E (X) is 4/7.

    The value of V (X) is 3/98.

(b) The value of P (X ≤ 0.5) is 0.3438.

Step-by-step explanation:

The random variable X is defined as the proportion of surface area in a randomly selected quadrant that is covered by a certain plant.

The random variable X follows a standard beta distribution with parameters α = 4 and β = 3.

The probability density function of X is as follows:

$f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)} ; \hspace{.3in}0 \le x \le 1;\ \alpha, \beta > 0$

Here, B (α, β) is:

$B(\alpha,\beta)=\frac{(\alpha-1)!\cdot\ (\beta-1)!}{((\alpha+\beta)-1)!}$

            $=\frac{(4-1)!\cdot\ (3-1)!}{((4+3)-1)!}\\\\=\frac{6\times 2}{720}\\\\=\frac{1}{60}$

So, the pdf of X is:

$f(x) = \frac{x^{4-1}(1-x)^{3-1}}{1/60}=60\cdot\ [x^{3}(1-x)^{2}];\ 0\leq x\leq 1$

(a)

Compute the value of E (X) as follows:

$E (X)=\frac{\alpha }{\alpha +\beta }$

         $=\frac{4}{4+3}\\\\=\frac{4}{7}$

The value of E (X) is 4/7.

Compute the value of V (X) as follows:

$V (X)=\frac{\alpha\ \cdot\ \beta}{(\alpha+\beta)^{2}\ \cdot\ (\alpha+\beta+1)}$

         $=\frac{4\cdot\ 3}{(4+3)^{2}\cdot\ (4+3+1)}\\\\=\frac{12}{49\times 8}\\\\=\frac{3}{98}$

The value of V (X) is 3/98.

(b)

Compute the value of P (X ≤ 0.5) as follows:

$P(X\leq 0.50) = \int\limits^{0.50}_{0}{60\cdot\ [x^{3}(1-x)^{2}]} \, dx$

                    $=60\int\limits^{0.50}_{0}{[x^{3}(1+x^{2}-2x)]} \, dx \\\\=60\int\limits^{0.50}_{0}{[x^{3}+x^{5}-2x^{4}]} \, dx \\\\=60\times [\dfrac{x^4}{4}+\dfrac{x^6}{6}-\dfrac{2x^5}{5}]\limits^{0.50}_{0}\\\\=60\times [\dfrac{x^4\left(10x^2-24x+15\right)}{60}]\limits^{0.50}_{0}\\\\=[x^4\left(10x^2-24x+15\right)]\limits^{0.50}_{0}\\\\=0.34375\\\\\approx 0.3438$

Thus, the value of P (X ≤ 0.5) is 0.3438.