Using the alternative-parameter method, determine the parameters of the followin

Question

Using the alternative-parameter method, determine the parameters of the following distributions based on the given assessments. Refer to Step 5.5 if necessary. a. Find the parameter value beta for the exponential distribution given P_E(T lessthanorequalto 15 beta) = 0.50. b. Find the parameters p and cr for a normal distribution given P_N(Y lessthanorequalto 12.5 mu, sigma) = 0.25 and P_N (Y lessthanorequalto 236 mu, sigma) = 0.75. c. Find the Min, Most Likely, and Max for the triangular distribution given P_T(Y lessthanorequalto 15 Min, Most Likely, Max) = 0.15, P_T(Y lessthanorequalto 50| Min, Most Likely, Max) = 0.50, and P_T(Y > 95 Min. Most Likely, Max) = 0.05. d. Find the parameter values alpha_1 and alpha_2 for the beta distribution given P_B (Q lessthanorequalto 0.3| alha_1, alpha_2) = 0.05 and P_B (Q lessthanorequalto 0.5| alpha_1_1, alpha_2) = 0.25

Explanation / Answer

Solution

Back-up Theory

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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