is it possible to have a mean-minimum-maximum distribution?

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Oliver H asked Jun 29 2017 at 4:43 PM Cliff King commented Jun 29 2017 at 7:12 PM

is it possible to have a mean-minimum-maximum distribution?

Hi,

I have some data for a process, who has different process times each process.

so the mean process time is 5 hours, but the process has at least 2 hours and 7 hours at the most. so the distribution is not symmetrical.

a normal distribution is symmetrical, so i can't use it. a uniform distribution is from 2 to 7, but the mean value can be anthing in between.

does anyone have a solution for this? thank you in advance!

Software Version:

FlexSim 17.0.2

distributions mean minimum maxim

· 1

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Answer 1 · 2017-06-29T16:45:51Z

Matthew Gillespie answered Jun 29 2017 at 4:45 PM Cliff King commented Jun 29 2017 at 7:12 PM

Yes, you can use the triangular distribution. You just need to specify the min, max, and mode.

triangular.png (5.2 KiB)

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Oliver H commented · Jun 29 2017 at 4:50 PM

@►Matthew Gillespie

Thank you for your fast answer. so "mode" in that case has the same meaning as "mean", am i right?

thanks anyways, you helped a lot.

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Cliff King Oliver H commented · Jun 29 2017 at 7:12 PM

Because (min + max + mode)/3 = avg, you can back calculate the mode as mode = (3 * avg) - min - max

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Matthew Gillespie ♦♦ Oliver H commented · Jun 29 2017 at 5:09 PM

(Sorry if this is isn't what you were asking) Mode and mean are both "averages", but they are a little bit different.

Mean is the traditional average everyone thinks of :

mean = a1 + a2 + ... an / n

Mode is the most repeated value:

1, 2, 3, 1, 2, 1	//mode = 1, mean = 1.67

That said, the triangular distribution is already fairly inaccurate so it probably doesn't matter too much which one you use. The mode value really just needs to represent the most likely value, so however you want to represent that.

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