Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2007-02-20 18:28:47

nancy2007
Member
Registered: 2007-02-20
Posts: 1

Need help on percentage.

Hi,
How do I calculate the top 20% highest grade for the following students:

A got 99
B got 93
C got 91
D got 85
E got 60
F got 50

Thank you very much for your help.
Nancy

Offline

#2 2007-02-21 20:39:23

Jai Ganesh
Administrator
Registered: 2005-06-28
Posts: 45,956

Re: Need help on percentage.

Nancy,
The question doesn't seem clear enough, I am unable to understand.
Is the score given in percentage?
What are you expected to calculate?


It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

#3 2007-03-05 08:21:11

mykel
Member
Registered: 2007-02-11
Posts: 4

Re: Need help on percentage.

That would be quite difficult w/o the use of variables (assigning each grade a variable and then sorting numerically).  I can think of a solution by writing a computer program, but I'm not sure how I'd do it mathematically in a cut and dried method.  Mind you, I'm just starting Trig. smile

Offline

#4 2007-03-05 08:23:37

lightning
Real Member
Registered: 2007-02-26
Posts: 2,060

Re: Need help on percentage.

did you know that 111,111,111x111,111,111=12345678987654321


Zappzter - New IM app! Unsure of which room to join? "ZNU" is made to help new users. c:

Offline

#5 2007-03-05 08:39:20

mykel
Member
Registered: 2007-02-11
Posts: 4

Re: Need help on percentage.

roflol

Offline

#6 2007-03-05 08:40:57

lightning
Real Member
Registered: 2007-02-26
Posts: 2,060

Re: Need help on percentage.

what it's true and wise i think up


Zappzter - New IM app! Unsure of which room to join? "ZNU" is made to help new users. c:

Offline

#7 2007-03-05 08:50:47

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Need help on percentage.

True and wise it may be, but on-topic it is not.
Sorry to be a party-pooper, but next time you have an amazing fact like this to share, please put it in a new topic, preferably also on a different board.

"This is Cool" perhaps, or maybe Dark Discussions.


Why did the vector cross the road?
It wanted to be normal.

Offline

#8 2007-03-05 08:52:05

lightning
Real Member
Registered: 2007-02-26
Posts: 2,060

Re: Need help on percentage.

sorry what is this


Zappzter - New IM app! Unsure of which room to join? "ZNU" is made to help new users. c:

Offline

#9 2007-03-05 18:40:03

bluestar9cn
Member
Registered: 2007-03-04
Posts: 4

Re: Need help on percentage.

I'm not very sure about your problem.
but, I tried, maybe this is what your want.

100*(100-20)%=80

you have 4 students above 80.

Offline

#10 2007-03-06 04:07:00

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Need help on percentage.

nancy2007 wrote:

How do I calculate the top 20% highest grade for the following students:

A got 99
B got 93
C got 91
D got 85
E got 60
F got 50

For six people, the top 20% is the top [0.2×6] = [1.2] = 1 people (or rather person). ([x] denotes the greatest integer less than or equal to x.)

So only 1 person in the any group of 6 people can be in the top 20%; in this case, it’s A.

Offline

#11 2007-03-06 06:29:12

mykel
Member
Registered: 2007-02-11
Posts: 4

Re: Need help on percentage.

JaneFairfax wrote:

For six people, the top 20% is the top [0.2×6] = [1.2] = 1 people (or rather person). ([x] denotes the greatest integer less than or equal to x.)

So only 1 person in the any group of 6 people can be in the top 20%; in this case, it’s A.

Perhaps something like this?

((99+93+91+85+60+50)/6 )  *1.2

(1/6) <-for the six grades

(99+93+91+85+60+50) <- add the grades

* 1.2 <- to get the top 20%

Offline

#12 2007-03-06 07:09:38

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Need help on percentage.

I'm not sure that that would work.
If you wanted to find the top 10% instead, then by your method you'd multiply the mean by 1.1 instead of 1.2.

But that would lower the result, and that doesn't make sense.


Why did the vector cross the road?
It wanted to be normal.

Offline

#13 2007-03-06 07:21:47

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Need help on percentage.

A statistic like top 20% or top 10% would be much more relevant when there’s a much larger group (say, 100 or more). 6 is really too small a group.

Offline

Board footer

Powered by FluxBB