### CFA Practice Question

There are 985 practice questions for this topic.

### CFA Practice Question

For the density curve displayed below, which of the following is (are) true?

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.

The proportion is the area under the density curve between 1.5 and 2. The length is 0.5, but the height is only 0.5, so the overall area is 0.5 x 0.5 = 0.25.
Because the density curve is symmetrical, the mean and median must coincide.

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)