When Area Under Probability Density Function Doesn't Equal 1?

Question

What do you do when a probability density function's area under the curve doesn't equal 1? The question given is f(x)=1-(x/30) from [0,30] and zero elsewhere. When I find the area, it is 15 units, which obviously doesn't equal 1. My teacher said to multiply by the reciprocal, but I don't understand that either.

Answer 1

You need to adjust your answer. Your teacher's point is that you can find the probabilty in the usual way, only at the end you multiply by 1/15 in order to scale the problem correctly.

Answer 2

These answers are all correct. I just wanted to add another explanation. The area under a PDF equals 1 by definition, in the same way that total probability in the usual sense always equals 1. The PDF allows us to graph total probabilities with the x-axis representing possible outcomes and the y-axis representing their relative frequency. As someone else remarked above, the 'units' listed on the graph are arbitrary and don't matter here; the total of all probabilities must equal 1 by definition. So your total of 15 must be multiplied by 1/15 to yield 1. The reciprocal is a way of 'standardizing' your results to the 0-1 range allowed by probability. Likewise, if you took an integral of a smaller subset of [0,30], say [0,15], you would need to multiply this result by 1/15 to yield the total probability of the subset, which would be less than 1, of course.

Answer 3

There is some problem here. How are you measuring the area under the curve. What do you mean by 15 units? The probability does not have a unit. So you will divide this area by the maximum area to get the probability. So what is the max? If it is 15 units then the probability is

15 (units - your area) divided by the maximum ( is it 15 units)

which will yield 1 and units get cancelled.

Strange . Good luck.

Sandy!

Answer 4

When the area is not 1, this is *not* a p.d.f.--not yet. But you can make it a p.d.f. by dividing by the area, as your teacher said.