Question

Problem: Using the alternative-parameter method, determine the parameters of the following distributions based on the given assessments.

Find the parameter value β for the exponential distribution given:

P_E (T ≤ 15 \ β) = 0.50.

Find the parameters μ and s for a normal distribution given:

P_N (Y ≤ 25\ μ, s) = 0.25 and P_N (Y ≤ 125\ μ, s) = 0.75

Find the Min, Most Likely, and Max for the triangular distribution given:

P_T (Y ≤ 15\Min, Most Likely, Max) = 0.15

P_T (Y ≤ 50\Min, Most Likely, Max) = 0.50, and

P_T (Y > 95\Min, Most Likely, Max) = 0.05

Find the parameters values a₁ and a₂ for the beta distribution given:

P_B (Q ≤ 0.3\a_1, a₂) ₌ 0.05 and P_B (Q ≤ 0.5\a_1, a₂) ₌ 0.25

Answer 1

1. pmf of exponential distribution with parameter β is: f(x) =(1/β)e^-x/βdx and

CDF = P(X ≤ t) = 1 - e^-t/β ……………………………………………..(1)

2. If X is Normal with parameters, µ and σ, P(X ≤ t) = P[Z ≤ {(t - µ)/σ}], where Z is the Standard Normal Distribution

Now to work out the solution,

Part (a)

Exponential distribution

Given, P(X ≤ 15) = 0.5, by (1), 1 - e^-15/β = 0.5. Transposing and taking natural log,

β = 21.64 ANSWER

Part (b)

Normal distributin

Given P(X ≤ 125) = 0.25, by (2), P[Z ≤ {(125 - µ)/σ)}] = 0.25

Extrapolating from Standard Normal Distribution Tables, {(125 - µ)/σ)} = - 0.6467 or

µ - 0.6467σ = 125 ………………………….(3)

Also given P(X ≤ 125) = 0.75, which obviously cannot be correct. So, the solution cannot be completed.

Just to give directions for completing,

In the second probability, the value cannot be 125. It must be a value greater than 125. Say that value is 250. Then, we will have P(X ≤ 250) = 0.75. Extrapolating from Standard Normal Distribution Tables, {(250 - µ)/σ)} = 0.6467 or

µ + 0.6467σ = 250 ………………………….(4)

(3) + (4): 2µ = 375 or µ = 187.5 ……………(5)

(5) in (4): σ = 62.5/0.6467 = 96.64 ANSWER 2