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Statistics / Applied Statistics and Probability for Engineers 6 / Chapter 4.12 / Problem 191E

Applied Statistics and Probability for Engineers | 6th Edition | ISBN: 9781118539712 | Authors: Douglas C. Montgomery, George C. Runger

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Chapter 4, Problem 191E

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QUESTION:

Suppose that the construction of a solar power station is initiated. The project’s completion time has not been set due to uncertainties in financial resources. The completion time for the

first phase is modeled with a beta distribution and the minimum, most likely (mode), and maximum completion times for the first phase are 1.0, 1.25, and 2.0 years, respectively. Also, the meantime is assumed to equal \(\mu=1+4(1.25)+2) / 6=1.333\). Determine the following in parts (a) and (b):

(a) Parameters α and β of the beta distribution.

(b) Standard deviation of the distribution.

(c) Sketch the probability density function.

Equation transcription:

$\mu{}=1+4(1.25)+2)/6=1.333$

Text transcription:

mu=1+4(1.25)+2) / 6=1.333

Questions & Answers

QUESTION:

Suppose that the construction of a solar power station is initiated. The project’s completion time has not been set due to uncertainties in financial resources. The completion time for the

first phase is modeled with a beta distribution and the minimum, most likely (mode), and maximum completion times for the first phase are 1.0, 1.25, and 2.0 years, respectively. Also, the meantime is assumed to equal \(\mu=1+4(1.25)+2) / 6=1.333\). Determine the following in parts (a) and (b):

(a) Parameters α and β of the beta distribution.

(b) Standard deviation of the distribution.

(c) Sketch the probability density function.

Equation transcription:

$\mu{}=1+4(1.25)+2)/6=1.333$

Text transcription:

mu=1+4(1.25)+2) / 6=1.333

ANSWER:

Answer

Step 1 of 3

(a)

The complete time for the first phase is modeled with a beta distribution and the minimum, most likely (mode), maximum completion times for the first phase are $1.0,\ 1.25,\ and\ 2.0\ years,$ respectively.

Also the mean time assumed to equal,

$\mu{}\ =\ \frac{1\ +4(1.25)+2}{6}\ =1.333.$

We are asked to determine the parameters $\alpha{}\ and\ \beta{}$ for the beta distribution.

We have given $a\ =1,\ m\ =1.25\ and\ b\ =2.0\ years.$

We have the formulas for mean and mode when above situation is given,

$mean\ =\mu{}\ =a\ +(b-a)\frac{\alpha{}}{\alpha{}+\beta{}}$

$mode\ =m\ =a\ +(b-a)\frac{\alpha{}-1}{\alpha{}+\beta{}-2}$

These equations can be solved for parameters $\alpha{}\ and\ \beta{}$ to obtain

$\alpha{}\ =\frac{(\mu{}-a)(2m-a-b)}{(m-\mu{})(b-a)}$

$\beta{}\ =\frac{\alpha{}(b-\mu{})}{(\mu{}-a)}$

Therefore,

$\alpha{}\ =\frac{(1.333-1)(2\times{}1.25-1-2)}{(1.25-1.33)(2-1)}\ =2.0625$

$\beta{}\ =\frac{2.0625(2-1.333)}{(1.333-1)}\ =4.1311$

Hence the value of parameters $\alpha{}\ and\ \beta{}$ is $2.0625\ and\ 4.1311$ respectively.

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