# Solved – How to find the upper bound of symmetric triangle distribution given median and lower bound

I have been told \$x\$ is drawn from a symmetric triangle distribution.
If the lower bound is 3 and the median is 9, how do I calculate the upper bound?

I'm not sure which value to use (I have to substitute the value I find into a formula). I was thinking if perhaps I had the upper bound, lower bound and median. I could work out the mean and use the mean value?

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Here's what you've been given: You know the lower bound is 3, and you know it's a triangle distribution so it increases from 3 towards the median at 9. At some point it starts to decrease again, either before, at, or after the median.

The median is the value of \$x\$ such that the probability of drawing a value less than the median is 50%. Pictorially this means that exactly half of the area under the triangle must lie between the lower bound and the median.

You're also told that the distribution is symmetric. What this means in the case of the triangle distribution is that you must be able to draw a vertical line through the highest point of the distribution such that the distribution to the left is a mirror image of the distribution to the right.

Hopefully this is enough information to let you find the answer for yourself.

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