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Basic Question 1 of 6
For the density curve displayed below, which of the following is (are) true?
II. The proportion of outcomes between 0.2 and 0.5 is equal to 0.3.
III. The proportion of outcomes exceeding 1.5 is equal to 0.25.
I. The mean is larger than the median.
II. The proportion of outcomes between 0.2 and 0.5 is equal to 0.3.
III. The proportion of outcomes exceeding 1.5 is equal to 0.25.
User Contributed Comments 13
| User | Comment |
|---|---|
| Sambo | I'm lost on this one. |
| danlan | I: mean=median=1 II: proportion between 0.2 and 0.5 is 0.3/2=0.15 III exceeding 1.5=0.5/2=0.25 |
| Damy4real | this is confusing...whats the rationale for it?anyone? |
| akanimo | You have to look at the graph in terms of a "uniform distribution" I. We learnt earlier (session 2) that for a uniform distribution the mean, mode and median will all lie at the same point. So this must be false II. This is a probability density function (it appears) since it satisfies both a) 0 <= P(X=x) <= 1 b) Sum( P(X=x) ) = 1 ... this is given by the area "under" the curve ... i.e. the square block with height = 0.5 and width 2 ... i.e 0.5x2=1.0 ... so we can find the proportion in terms of "area" ... the area between X=0.2 and X=0.5 = (height) x (width btw .2 and .5) = 0.5 x (0.5 - 0.2) = 0.15 we can get the proportion of this by comparing to the TOTAL area ... which gives: =(area btw 0.2 to 0.5) / (total area) =0.15/1 =0.15 so this is false too III. Using the same logic as in (II) you can calculate the area above 1.5 = 0.5 x ( 2.0 - 1.5 ) = 0.25 proportion is this area / total area = 0.25 / 1 = 0.25 |
| aspazia | where is the curve on this graph and what is the space between the axis at (0,0)???? |
| Janey | the horizontal line can be split into quarters - 0.5, 1, 1.5 and 2. Therefore the outcome exceeding 1.5 is .5 which equals a quarter (.25) |
| steved333 | Ooh, careful, Janey. True, the outcome exceeding 1.5 is 1/4, but that just happens to be a nice coincidence. You still have to consider the height of the rectangle. If the height were .6 instead of .5, the result would have been .3 instead of .25. You have to take the probability into consideration, so don't fall into the easy way out trap! |
| steved333 | Okay, since the height is always going to be proportionate to the length so as to equal 1, maybe that will work... |
| MiciMori | Wonderful akanimo. |
| bkballa | what do they mean by, because the density curve is symmetric the mean and median must coincide. how is it symmetric? |
| ashish100 | akanimo explanation on point!! thanks!! |
| ashish100 | wait!! akanimo - Shouldn't numero II = (.5 - .2) * 2 where 2 is the width? So .6 instead of .15 someone let me know please. ashishinvests@gmail.com Thanks! :D |
| dbedford | Height = Probabilities. Width = Outcomes. So for II we see they are referring to OUTCOMES between .2 and .5. this would be like drawing a vertical line on the graph on the horizontal line at .2 and .5 and finding the space in between. Area = Length x Width = .5(length of the max of the vert of the box) x (.3 the width between .2 and .5 on the horizontal line) |
Your review questions and global ranking system were so helpful.
Lina
Learning Outcome Statements
define a probability distribution and compare and contrast discrete and continuous random variables and their probability functions;
calculate and interpret probabilities for a random variable given its cumulative distribution function;
CFA® 2024 Level I Curriculum, Volume 1, Module 4.