Weighted Percentile Problems

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Weighted Percentile Problems

by anuptvm » Wed Dec 29, 2010 4:55 am
Linda's grade was 90th percentile of 80 grades for her class. Of the 100 grades from another class, 19 were higher than Linda's, and the rest lower. If no other grade was the same as Linda's grade, then Linda's grade was what percentile of the grades of the two classes combined?

1. 72nd 2. 80th 3. 82nd 4. 85th 5. 92nd

OA is 4

Could someone explain this. I tried using the weighted average method but was wrong.
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by shovan85 » Wed Dec 29, 2010 5:10 am
anuptvm wrote:Linda's grade was 90th percentile of 80 grades for her class. Of the 100 grades from another class, 19 were higher than Linda's, and the rest lower. If no other grade was the same as Linda's grade, then Linda's grade was what percentile of the grades of the two classes combined?

1. 72nd 2. 80th 3. 82nd 4. 85th 5. 92nd

OA is 4

Could someone explain this. I tried using the weighted average method but was wrong.
Linda's grade was 90th percentile of 80 grades for her class.

Linda's grade is above 80*90/100 = 72 grades. Thus 8 grades are above Linda's grade.

Of the 100 grades from another class, 19 were higher than Linda's, and the rest lower.

OK

The two classes are then combined. Thus, total grades = 100+80 = 180

Total number of grades above Linda's grade = 19 + 8 = 27

Hence 180 - 27 = 153 grades are below Linda's grade

Linda's grade percentile = (153/180) * 100 = (17/20)*100 = 85
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by anuptvm » Wed Dec 29, 2010 5:17 am
shovan85 wrote: Linda's grade percentile = (153/180) * 100 = (17/20)*100 = 85
I messed up the calculations I guess :)

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by Night reader » Wed Dec 29, 2010 5:33 am
shovan85 wrote:
anuptvm wrote:Linda's grade was 90th percentile of 80 grades for her class. Of the 100 grades from another class, 19 were higher than Linda's, and the rest lower. If no other grade was the same as Linda's grade, then Linda's grade was what percentile of the grades of the two classes combined?

1. 72nd 2. 80th 3. 82nd 4. 85th 5. 92nd

OA is 4

Could someone explain this. I tried using the weighted average method but was wrong.
Linda's grade was 90th percentile of 80 grades for her class.

Linda's grade is above 80*90/100 = 72 grades. Thus 8 grades are above Linda's grade.

Of the 100 grades from another class, 19 were higher than Linda's, and the rest lower.

OK

The two classes are then combined. Thus, total grades = 100+80 = 180

Total number of grades above Linda's grade = 19 + 8 = 27

Hence 180 - 27 = 153 grades are below Linda's grade

Linda's grade percentile = (153/180) * 100 = (17/20)*100 = 85
shovan, Linda's grade was 85.5% exactly; also for GMAT purpose we may assign 85%

check => (90%+81%)/2 = 171%/2 = 85.5%;

Linda's grade is right between the upper and lower limits, i.e. 15% and 85%, so basically she is assigned 85,5%
But 85% is ok :)

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by shovan85 » Wed Dec 29, 2010 8:15 am
Night reader wrote: shovan, Linda's grade was 85.5% exactly; also for GMAT purpose we may assign 85%

check => (90%+81%)/2 = 171%/2 = 85.5%;

Linda's grade is right between the upper and lower limits, i.e. 15% and 85%, so basically she is assigned 85,5%
But 85% is ok :)
Hi!! I doubt this...

Keep the first line of data same and modify the second data in the same ratio
Let us say that total number of students in the second class is 500 (instead of 100) and keeping the ratio 95 (instead of 19) grades are above that of Linda.

According to you it will still be 85.5%ile (not percent)

However, actually it will be (500+80) - (95+8) = 580 - 103 = 477

Thus, percentile will be (477/580)*100 = 82.24 %ile

Though rounding off 85.5 to 85 is giving the correct answer but I believe the method what you are suggesting is dangerous. ;)

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by anshumishra » Wed Dec 29, 2010 8:23 am
Night reader wrote:
shovan85 wrote:
anuptvm wrote:Linda's grade was 90th percentile of 80 grades for her class. Of the 100 grades from another class, 19 were higher than Linda's, and the rest lower. If no other grade was the same as Linda's grade, then Linda's grade was what percentile of the grades of the two classes combined?

1. 72nd 2. 80th 3. 82nd 4. 85th 5. 92nd

OA is 4

Could someone explain this. I tried using the weighted average method but was wrong.
Linda's grade was 90th percentile of 80 grades for her class.

Linda's grade is above 80*90/100 = 72 grades. Thus 8 grades are above Linda's grade.

Of the 100 grades from another class, 19 were higher than Linda's, and the rest lower.

OK

The two classes are then combined. Thus, total grades = 100+80 = 180

Total number of grades above Linda's grade = 19 + 8 = 27

Hence 180 - 27 = 153 grades are below Linda's grade

Linda's grade percentile = (153/180) * 100 = (17/20)*100 = 85
shovan, Linda's grade was 85.5% exactly; also for GMAT purpose we may assign 85%

check => (90%+81%)/2 = 171%/2 = 85.5%;

Linda's grade is right between the upper and lower limits, i.e. 15% and 85%, so basically she is assigned 85,5%
But 85% is ok :)
Night reader,

That is not right. If you change these two values 90 and 80 here : Linda's grade was 90th percentile of 80 grades for her class. , you can see the difference getting more significant (your method vs Shovan's method- which is right). So, although here it gives you the right solution, that is just lucky.
Thanks
Anshu

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by GMATGuruNY » Wed Dec 29, 2010 8:50 am
anuptvm wrote:Linda's grade was 90th percentile of 80 grades for her class. Of the 100 grades from another class, 19 were higher than Linda's, and the rest lower. If no other grade was the same as Linda's grade, then Linda's grade was what percentile of the grades of the two classes combined?

1. 72nd 2. 80th 3. 82nd 4. 85th 5. 92nd

OA is 4

Could someone explain this. I tried using the weighted average method but was wrong.
Percentile means the percentage below.

Out of 80 grades, if Linda's grade was the 90th percentile, then 90% of the grades were lower than Linda's grade. Thus, .9*80 = 72 grades in Linda's class were lower than Linda's grade. Subtracting these 72 grades and Linda's grade from the total, we get that 80-72-1 = 7 grades were higher.

Since 19 grades in the other class were higher than Linda's grade, 19+7 = 26 grades in the combined classes were higher than Linda's grade. Subtracting these 26 grades and Linda's grade from the total number of grades in the combined classes, we get that 180-26-1 = 153 grades were lower than Linda's grade.

Thus, Linda's percentile in the combined classes was 153/180*100 = 85%.

The correct answer is D.
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by Night reader » Wed Dec 29, 2010 9:33 am
@Anshu, shovan: my take-aways are mismatch of percentage distribution with the percentage values; on your part placing of 8 grades above the percentile rank for Linda wasn't accurate (the same mismatch)
it's good we clarified.

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by shovan85 » Wed Dec 29, 2010 9:52 am
Night reader wrote:@Anshu, shovan: my take-aways are mismatch of percentage distribution with the percentage values; on your part placing of 8 grades above the percentile rank for Linda wasn't accurate (the same mismatch)
it's good we clarified.
Yes !! I agree :)
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