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Old 2009 August 4th, 10:29 AM   #1 (permalink)
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Percentile problem

In a distribution of 8500 parameters, if 26.7 is 56 percentile and 37.1 is 78 percentile, then what is the percentile of x (26.7<=x<= 37), that is closest in this range?
A. 1888
B. 4750
C. 6650
& so on....
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Old 2009 August 4th, 10:50 AM   #2 (permalink)
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78 - 56 = 22% lies in [26.7; 37] range. Which corresponds to 1870 parameters. The closest answer choice is A.
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Old 2009 August 4th, 02:25 PM   #3 (permalink)
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I think the answer is B.

if 26.7 is 56 percentile and 37.1 is 78 percentile, then what is the percentile of x (26.7<=x<= 37),

56 % means 4760 parameters , 78 % means 6630 parameters.
If percentile is btw (26.7<=x<= 37), it will be between 4760.

so the closest number i found is 4750.

QA please
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Old 2009 August 4th, 03:29 PM   #4 (permalink)
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i go with A (1888)

What is OA?
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Old 2009 August 4th, 03:40 PM   #5 (permalink)
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Yes, answer (A)

8500(78/100 - 56/100) = 85 x 22 = 1870

What is the OA?
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Old 2009 August 4th, 11:14 PM   #6 (permalink)
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Quote:
Originally Posted by VinayDeore View Post
I think the answer is B.

if 26.7 is 56 percentile and 37.1 is 78 percentile, then what is the percentile of x (26.7<=x<= 37),

56 % means 4760 parameters , 78 % means 6630 parameters.
If percentile is btw (26.7<=x<= 37), it will be between 4760.

so the closest number i found is 4750.

QA please
I believe your explanation but the question asks for the range (Highest number-smallest number).
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Old 2009 August 5th, 02:15 AM   #7 (permalink)
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Quote:
Originally Posted by dissa View Post
Yes, answer (A)

8500(78/100 - 56/100) = 85 x 22 = 1870

What is the OA?


I have no OA for this problem
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Old 2009 August 5th, 03:45 AM   #8 (permalink)
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i think the question is bit wrongly framed.
It should ask for number of parameters present within the percentile range (26.7<=x<= 37). Also the answer choice is not percentile.

so parameter lists range will be 4760 <=Y<=6630.

simply think of any exam in which percentile is measured. 99 percentile means the candidate has score greater than 99 students in 100.i had the same appraoch for above problem.

The difference does not make sense.

If that helps...
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