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Statistics / Applied Statistics and Probability for Engineers 6 / Chapter 4.9 / Problem 150E

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

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Solved: The total service time of a multistep

Chapter 4, Problem 150E

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QUESTION:

The total service time of a multistep manufacturing operation has a gamma distribution with mean 18 minutes and standard deviation 6.

(a) Determine the parameters \(\lambda\) and r of the distribution.

(b) Assume that each step has the same distribution for service time. What distribution for each step and how many steps produce this gamma distribution of total service time?

Questions & Answers

QUESTION:

The total service time of a multistep manufacturing operation has a gamma distribution with mean 18 minutes and standard deviation 6.

(a) Determine the parameters \(\lambda\) and r of the distribution.

(b) Assume that each step has the same distribution for service time. What distribution for each step and how many steps produce this gamma distribution of total service time?

ANSWER:

Solution

Step 1 of 2

Given that total service time has a gamma distribution with mean 18 and standard deviation 6

a) We have to find the parameters of gamma distribution( $\lambda{}\ and\ r$ ) for the given data

The mean of the gamma distribution is $\mu{}=r/\lambda{}$

And the standard deviation of gamma distribution is $\sigma{}=\sqrt{r/\lambda{}{\ }^{2}}$

The shape of the distribution can be obtained by dividing mean by standard deviation

Then $\frac{\mu{}}{\sigma{}}=\frac{r/\lambda{}}{\sqrt{r/\lambda{}{\ }^{2}}}$

= $\frac{r}{\sqrt{r}}$

= $\sqrt{r}$

Given that $\mu{}=18\ and\ \sigma{}=6$

Now $r=(\mu{}/\sigma{}){\ }^{2}$

=(18/6)2

=9

And $\lambda{}=r/\mu{}$

=9/18

=1/2

Hence the parameters $r=9\ and\ \lambda{}=1/2$

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