Problem : Using the alternative-parameter method, determine the parameters of th

Question

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

Find the parameter value for the exponential distribution given:

PE (T 15 ) = 0.50.

Find the parameters and s for a normal distribution given:

PN (Y 125 , s) = 0.25 and PN (Y 12 , s) = 0.75

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

PT (Y 15Min, Most Likely, Max) = 0.15

PT (Y 50Min, Most Likely, Max) = 0.50, and

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

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

PB (Q 0.31, a2) = 0.05 and PB (Q 0.51, a2) = 0.25

Explanation / Answer

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

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

2. If X is Normal with parameters, µ and , P(X t) = P[Z {(t - µ)/}], 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/ = 0.5. Transposing and taking natural log,

= 21.64 ANSWER

Part (b)

Normal distributin

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

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

µ - 0.6467 = 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)/)} = 0.6467 or

µ + 0.6467 = 250 ………………………….(4)

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

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

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