CFA Practice Question
There are 985 practice questions for this topic.
CFA Practice Question
In the following distribution:
0 to 9 5
10 to 19 12
20 to 29 4
30 to 39 7
40 to 49 10
0 to 9 5
10 to 19 12
20 to 29 4
30 to 39 7
40 to 49 10
The 3rd quartile is ______.
Correct Answer: A
User Contributed Comments 19
|Gina||why is the 3rd quartile not 36.750 - where 75% of the data would be underneath?|
|CWilkinson||it's frequency; values in the range 0 to 9 occur 5 times; values in the range 10-19 occur 12 times; etc. So, apparently we should assume 3rd quartile is the higher scores not lower, and thus the 9.5th occurence from bottom, or a midpoint of higher ranges of 40-49, as determined by range/frequency or .9 per occurence in that range, and the 9.5th is .45 from lower bound so 40.45.|
|Tcpa||Total observations are 5+12+4+7+10=38
3rd quartile would be 38*75%=28.5, the middle point between 28 and 29.
28th number is the largest one in 30-39 range;
29th number is the smallest one in 40-49 range.
In extreme scenario, the distribution could has all 7 numbers in 30-39 range be 30, and the first number in 40-49 range be 40, then the 3rd quartile would be (30+40)/2=35.
In another extreme scenario, the distribution could has the last numbers in 30-39 range be 39, and all 10 numbers in 40-49 range be 49, then the 3rd quartile would be (39+49)/2=44.
The first example gives the smallest number possible for 28th and 29th numbers, and the second example gives the largest possible numbers, so the 3rd could range from 35 to 44.
If the distribution is not that extreme, then I would say 40.450 would be a better choice then 37.500.
|jameel||To find quartiles you have to calculate the cumulative frequencies. Now for the cumulative frequencies are as under:
Interval Freq Cum Freq
0-9 5 5
10-19 12 17
20-29 4 21
30-39 7 28
40-49 10 38
The third quartile also known as upper quartile is calculated as 38*0.75 = 28.5, Now if you look in the table above the upper quartile i.e. 28.5 falls in the last interval which is 40-49. The choice that comes closest to this interval is 40.450.
|Arron||Jameel, in this case, 45.000 is also possible?|
|Arron||Oh, I see, 45.000 is impossible becoz (39+49)/2=44 <45|
L = (n+1)*y/100
where n = 38, number of data
for 3 quartile, y = 75.
So L = 29.25
That means the 3rd quartile should be little larger than the 29th data point.
|itconcepts||what about - 75% of 38 observations is 28.5
the cumulative frequency to the last class is 28
so 0.5 remains
0.5 / 9 (derived from 40-40 interval)/10 (frequency)=.45
add that to lower of class = 40+.45 = 40.45?
|itconcepts||oops, I meant the 40-49 interval
|bobert||I am quite lost. I got the correct answer by reasoning the wrong ones out, but is there a much easier way to come down with the answer in a quick fashion? I did this:
Formula: L = (n+1)*(y/100), subbed everything in and got L = (38+1)*.75 = 29.25.
I then started working my way up by reducing L by the number of observations in each frequency (29.25-5, 24.25-12, 12.25-4, 8.25-7 = 1.25) which left me in the low 40-49 range where 40.45 seemed to be the best choice, and was apparently correct.
Then I tried to just simply weight it such that the upper class contains 26.3% of the results (10/38). This left me now with 1.25 data points into the 40-49 class which contained 26.3%.
I also considered that the midpoint between the 30-39 and 40-49 would be a more proper starting place (39.5) but I again failed to see the connection there.
It concepts method works, but I also do not know if that is the correct way to go about it considering that the formula used is not correct in that it should be 39 observations. (38 + 1)
|mordja||itconcepts is on the mark|
|hannovanwyk||GUYS it can easily be calculated. chances are good that this concept will be tested and the values might be much closer, then you will need to calculate correctly...
1. 0.75 * 38 = 28.5
2. cum. frequency to 39 = 28.
3. 0.5 falls in range 40-49
4. 10 observations fall between 40 and 49
5. thus (49-40)/10 = 0.9
6. wheight by multiplying 0.9 with 0.5 = 0.45
7. FINAL answer = 40 + 0.45 = 40.45
|JKiro||a good question! thank you hannovanwyk|
|vinooka||I guess hannovanwyk's explanation is not correct. If the quartile is at 28.5th position then it should lie between 39 and 40. I think it should be calculated this way -
Quartile Position - (38 + 1)*3/4 = 29.25
so it should be the average of 29th and 30th values.
The range of the last class is 9 and the frequency is 10, so the estimated interval would be 0.9
29th - 40
30th - 40.9
31st - 41.18
average of 29th and 30th values - 40.45 (this is based on analystnotes formula)
However, based on the textbook, the quartile should actually lie at 25% distance between 40 and 40.9 and so should be - 40.225!
|8thlegend||Ok first of all if the formula is Ly=(n+1)*y/100
I dont know why everybody is multiplying buy 75% or 3/4 because you are not doing the algebra right at all.
It's [(38+1)* 75]/100 = 29.25 frequency, which is where the 75% begins.
Meaning if you add all the frequency from range 0-9 to 30-39 it is 28 frequency.
Since the 75% has to start from 29.5 frequency = 40.45 is the number that will fit closest
|alester83||Ly= (n+1) (y/100) this gives you the observation location for in this case where the 3rd quartile is. Since when we use that formula we get 29.25, by looking at the cumulative frequencies we know what the 3rd quartile is in between(30-39) (40-49). When this happens you need to reference the linear interpolation formula|
|gill15||Odd. The 29.25 is correct. The range Vinook gives is wrong --- its 49-39 = 10 so each estimated interval would be 1....29th value = 40 and 30th = 41 etc...
I would get .75(40) + .25(41) = 40.25 using linear interpolation. Close to the 40.45.
|ravi480000||n = 5 + 12 + 4 + 7 + 10 = 38
Class Frequency Cumulative Frequency
0 to 9 5 5
10 to 19 12 17
20 to 29 4 21
30 to 39 7 28
40 to 49 10 38
Using definition of quartiles as partitions which divide an ordered data set
into four equal parts. Note that these paritions could be actual observations
which are part of the data set or could be set up as average of previous
and next observation.
Each quartile has 38 / 4 = 9.5, roughly 9 observations.
If you put 9 observations in each quartile, you use up 9 * 4 = 36 obs
we are left with 2 observations to fill up 3 partitions of a quartile.
One of the quartiles is going to be an average observation, which
could occur at either 1st, 2nd or 3rd quartile. Here are different ways to
form quartiles for the data set
1-9 \ 10-18 | 20-28 | 30-38 Q3 is 29th obs
1-9 | 11-19 \ 20-28 | 30-38 Q3 is 29th obs
1-9 | 11-19 | 21-29 \ 30-38 Q3 is avg(29th obs, 30th obs)
\ indicates quartile which is an average observation
29th obs could be anything from 40 to 49
if 29th obs is 45, then B is a right answer
if 28th obs = 30 and 29th obs = 45, Q3 = avg(30, 45) = 37.5
and C is a right answer
if 28th obs = 32 and 29th obs = 48.9, Q3 = avg(32, 48.9) = 40.45
and A is the right answer
Seems to me any of the answers could be the 3rd quartile and we need more
information about the sample to get a specific number.
|msk500||Thanks to all of you for the above comments, but I'm still confused. Having computed the location using L75 = (n+1)*75/100, ie 29.25, we know that the value we seek lies between 29 and 30. We then use linear interpolation: take the value that relates to the 29th item, and then add the difference in value between the 29th and 30th items. The question is, what are the values?|