# continuous uniform dist?

A continuous uniform distribution has the parameters a=4 and b = 10. The F(20) is: A. .25 B. .50 C. 1.00 D. 2.00 correct answer: C, F(x) = 1 for all x > b. Remember F(x) is the cumulative probability, P(x<20) here. -question is from Schweser, could not make sense of this using Z table or text from CFAI Thanks, John

Are you sure that’s part of 2009 curriculum because I have never seen F(x) anywhere in the CFAI text…

question is from '08 schweser notes, study session 3, so not sure.

Think of a coninuous uniform distribution as an x and y axis. starts at 4(a) on the x axis and goes to 10(b). It is a rectangle. The F() part means how much of the distribution space would you cover if you drew a vertical line at 20 and shaded in the area back to 0 or all the way to the Y axis. It would cover all of the rectangle and then some right? Well a probability distribution F() can have a value between 0 and 1. So it must cover all of it ==1.0. 2.0 is impossible so throw that out right away.

To summarize BizBanker’s reply: The range from your distribution is anything between 4 and 10 (keep in mind since its continuous, the probability of any specific number is equal to zero). Therefore, no matter how you take a sample from the distribution, it will ALWAYS be greater than or equal to twenty. Asking F(11) would yield the same answer of 1.00. Hope that built on the previous response.

so would F(7) be 0.5 and say F(3) = 0.0? I’m a little confused about this concept…

sa.86 Wrote: ------------------------------------------------------- > so would F(7) be 0.5 and say F(3) = 0.0? > > I’m a little confused about this concept… You’re correct. If your continuous uniform distribution runs from values of 4 to 10, F(3) would have zero probability. F(7) of the same continuous distribution would have a cumulative probability of .5. Think of the rectangle bizbanker described above and visualize the shading.

You couldn’t make any sense of this using the z-table because the z-table refers to a normal distribution, not a uniform distribution. z-table has nothing to do with this problem.