Answer to Question #236620 in Statistics and Probability for john k

Question

Suppose the proportion X of surface area in a randomly selected quad-rate that is covered by a certain plant has a standard beta distribution with α = 5 and β = 2.

(a) Compute E(X) and Var(X).

(b) Compute P(X ⩽ 0.2).

(c) Compute P(0.2 ⩽ X ⩽ 0.4).

(d) What is the expected proportion of the sampling region not covered by the plant?

Answer 1

Given the proportion X of surface area in a randomly selected quadrat that is covered by a certain plant has a standard beta distribution with

The probability function of a standard beta distribution and its mean and variance is

$f(x) = \frac{1}{β(α,β)}x^{α-1}(1-x)^{β-1} \\ Mean = \frac{α}{α+β} \\ Variance = \frac{αβ}{(α +β)^2 ( α+ β+1) }$

(a)

Mean of the beta distribution is

$E(X) = \frac{α}{α+β} \\ = \frac{5}{5+2} \\ = 0.714286$

Variance of the beta distribution is

$V(X) = \frac{αβ}{(α +β)^2 ( α+ β+1) } \\ = \frac{10}{7^2 \times 8} \\ = 0.02551$

(b)

$P(X≤0.2) = \int^{0.2}_0 \frac{1}{β(α,β)}x^{α-1}(1-x)^{β-1} dx \\ = \int^{0.2}_0 \frac{1}{β(5,2)}x^{4}(1-x)^{β-1} dx \\ = \frac{Г(5+2)}{Г5Г2} [\frac{x^5}{5} -\frac{x^6}{6}]^{0.2}_0 \\ = \frac{6!}{4!1!} [\frac{0.2^5}{5} -\frac{0.2^6}{6}] \\ = 0.0016$

Therefore, the required probability is 0.0016.

(c)

$P(0.2≤X≤0.4) = \int^{0.4}_{0.2} \frac{1}{β(α,β)}x^{α-1}(1-x)^{β-1} dx \\ = \int^{0.4}_{0.2} \frac{1}{β(5,2)}x^{4}(1-x)^{β-1} dx \\ = \frac{Г(5+2)}{Г5Г2} [\frac{x^5}{5} -\frac{x^6}{6}]^{0.4}_{0.2} \\ = \frac{6!}{4!1!} [\frac{0.4^5}{5} -\frac{0.4^6}{6}-\frac{0.2^5}{5} + \frac{0.2^6}{6}] \\ = 0.03936$

Therefore, the required probability is 0.03936.

(d)

The expected proportion of the sampling region not covered by the plant is

$1 -E(X) = 1 - \frac{5}{7} = 0.285714$

The expected proportion of the sampling region not covered by the plant is 0.285714.