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 = 5 and β = 2. (a) Compute E(X) and V(X). (Round your answers to four decimal places.) E(X) = 0.7143 Correct: Your answer is correct. V(X) = 0.02551 Correct: Your answer is correct. (b) Compute P(X ≤ 0.3). (Round your answer to four decimal places.)

For this case we have defined the random variable X ="surface area in a randomly selected quadrant that is covered by a certain plant", and we know the distribution for X

[tex] X \sim Beta(\alpha =5, \beta =2)[/tex]

Part a

For this case we want to find the expected value and the variance for X.

The expected value is given by:

[tex] E(X)= \frac{\alpha}{\alpha +\beta}= \frac{5}{5+2}= \frac{5}{7}=0.7143[/tex]

And the variance is given by:

[tex] Var(X) = \frac{\alpha \beta}{(\alpha +\beta)^2 (\alpha +\beta +1)}[/tex]

And replacing we got:

[tex] Var(X) = \frac{5*2}{(5 +2)^2 (5 +2 +1)}=0.02551[/tex]

Part b

For this case we want the following probability:;

[tex] P(X \leq 0.3)[/tex]

For this case we can calculate the regularized incomplete beta function

[tex] I(0.3, 5,2)= \frac{B(0.3;5,2)}{B(5,2)}[/tex]

Where [tex]B(x;a,b)[/tex] is given by:

[tex] B(x; a;b) =\int_{0}^{0.3} t^{a-1} (1-t)^{b-1} dt[/tex]

Replacing we got:

[tex] B(0.3; 5;2) =\int_{0}^{0.3} t^4 (1-t) dt = \int_{0}^{0.3} t^4 -t^5 [/tex]

[tex] B(0.3; 5;2) = \frac{0.3^5}{5} -\frac{0.3^6}{6} =0.0003645[/tex]

We can calculate the probability with the following R code:

> pbeta(0.3,5,2,TRUE)

[1] 0.007885565

And as we can see we got 0.007885565