Question

Question 1) The concentration, X, in ppm of chlorophyll in outfield grass be modeled as a gamma random variable with parameters α = 5 (α is called shape in R.) and β = 12 (β is called scale in R.). Thus, X has probability density:

f(x,α,β) = 1 / β^αΓ(α) x^α−1e^−x/β for 0 ≤ x ≤ ∞ and 0 otherwise with α = 5 and β = 12

Note that Γ(α) is the gamma function. In R, there is a built-in function gamma() which calculates this. Hint: Solve this with the built-in R functions for the gamma distribution (dgamma(),pgamma(), qgamma()) not f as defined above. Otherwise you will get numbers too large to use. For a) and b) the text (and notes) give formulas for the answers. You can calculate from these formulas.

a) What is the expected value of X?

b) What is the variance of X?

c) What is the standard deviation of X?

d) What is the probability that X is larger than its expected value?

e) What is the probability that X is > 80?

f) What is the probability that X is > 110?

g) What is the probability that X > 110 given that X > 80?

h) Calculate the median of X.

l) Enter any comments in the text box below.

Answer 1

a) E(X)=5*12=60

b)

c) SD(X)=square root of Var(x)=26.83282

d) P(X>E(X))=P(X>60)=1-P(X<=60)=1-0.5595067=0.4404933

e) P(X>80)=1-P(X<=80)=0.2056272

f)P(X>110)=1-P(X<=110)=0.04959444

g)P(X>110|X>80)=P(X>110)/P(X>80)=0.04959444/0.2056272=0.2411862

h) Repeating the calculation of P(X<=m) (by pgamma(m,shape=5,scale=12) in R) for different choices of m , we find that for  m= 56.051, the above becomes 0.5.

Thus median is 56.051.

i) Mean is more than the median