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#1 2016-05-03 12:57:25

Mathegocart2
Member
Registered: 2016-05-03
Posts: 8

Help with box- and whiskers?

rhXwZq3.jpg
Help with 55?
I get 17, 3/4(Q1(Quartile 1))*14 + 1/2(median)*13 = 10
1/4 is bc. where I draw the line is 25%

Last edited by Mathegocart2 (2016-05-03 13:00:40)

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#2 2016-05-03 13:11:28

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Help with box- and whiskers?

When blown up, it is not clear enough to read sad
146232414950819?6889138799

Last edited by Relentless (2016-05-03 13:14:58)

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#3 2016-05-03 13:43:50

Mathegocart2
Member
Registered: 2016-05-03
Posts: 8

Re: Help with box- and whiskers?

Relentless wrote:

When blown up, it is not clear enough to read sad
http://www150.lunapic.com/editor/working/146232414950819?6889138799

I get 3/4(14) because the first line represents 1/4, this is 10.5. Bc 1/2 is where Team B is, they get 6 games, so 16.5.
I don't know how you could get 16, but that was the closest one so

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#4 2016-05-03 14:37:47

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Help with box- and whiskers?

Hi smile

This question was meant to be tricky and make you think! Applying your method, you would be getting 10.5 and 6.5 for 17. But what you need to do is separate the games on either side of 55 and the games on exactly 55. If you think hard, you may reason that team B has 6 games above 55, 6 games below, and 1 at exactly 55. So you count 7. And similarly, that team A has 10 games above 55, 3 games below, and 1 at exactly 55. So you count 11. This gives you the answer of 18, which they wanted.

However, I think this is a faulty question. It is possible that there are more scores at exactly 55. I will try to be more precise about this later.

Last edited by Relentless (2016-05-03 15:59:18)

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#5 2016-05-03 19:13:02

Bob
Administrator
Registered: 2010-06-20
Posts: 10,143

Re: Help with box- and whiskers?

hi Melcalci

Welcome to the forum.

Had to look at the 'original' imgur image to clarify that scale.

The 'formulas' for determining the fixed points are lower quartile = (n+1)/4  median = (n+1)/2 and upper = (n+1)x3/4

For A the lower quartile is at 55 and n = 14.  From above the LQ is at 15/4 = 3 and 3/4 so the scores above this are 4th,5th,6th,7th,8th...14th = 11 scores in total.

For B the median is at 55 and n = 13.  median  is at (13+1)/2 = 7th score.  So there are 6 scores above this.

So the total number of scores above or equal to 55 is 11 + 6 + 1 (the median for B) = 18.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#6 2016-05-03 22:07:24

Mathegocart2
Member
Registered: 2016-05-03
Posts: 8

Re: Help with box- and whiskers?

bob bundy wrote:

hi Melcalci

Welcome to the forum.

Had to look at the 'original' imgur image to clarify that scale.

The 'formulas' for determining the fixed points are lower quartile = (n+1)/4  median = (n+1)/2 and upper = (n+1)x3/4

For A the lower quartile is at 55 and n = 14.  From above the LQ is at 15/4 =(we always ROUND DOWN??) 3 and 3/4 so the scores above this are 4th,5th,6th,7th,8th...14th = 11 scores in total.

For B the median is at 55 and n = 13.  median  is at (13+1)/2 = 7th score.  So there are 6 scores above this.
4
So the total number of scores above or equal to 55 is 11 + 6 + 1 (the median for B) = 18.

Bob

Can you explain why (x+1)/4 and not x/4?

Last edited by Mathegocart2 (2016-05-03 22:19:39)

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#7 2016-05-03 23:39:42

Bob
Administrator
Registered: 2010-06-20
Posts: 10,143

Re: Help with box- and whiskers?

hi Melcalci

The simple answer is that it is defined that way.  But think first about the median and you'll see the logic behind it.

The median is the middle number when n is odd and half way between the two middle numbers when n is even.  You can express this algebraically by saying it's the (n+1)/2  number in the list.  So, eg., when n = 13 we compute (13+1)/2 = 7 and this correctly identifies the middle number as half way along the list; 6 before it and 6 after it.

The quartiles must be half way between the start and the median and half way between the median and the end.  So using (n+1)/4 and 3(n+1)/4 seems to do the trick.

Here's an example where the list does divide neatly into 4 sets:

1,3,8,9,11,13,16,17,19,22,24,26.

There are 12 numbers in the list so the positions are LQ at 13/4 = 3.25, M at 13/2 = 6.5 and UQ at 39/4 = 9.75

Those markers split the list into (1,3,8), (9,11,13), (16,17,19), and (22,24,26)  That gives you equal quarters as you'd want.

If you use n/4 = 12/4 = 3 then 8 lies on the border and isn't in either the bottom quartile nor in the next.

Of course it gets harder to see the logic where n+1 doesn't neatly divide by 4 but you still get the best split available .  Eg. for A the LQ = 3.75, M = 7.5 and UQ = 11.25

That puts 3 numbers in the lower quartile, 4 numbers in the next quartile, 4 numbers in the next and 3 in the upper quartile.  The sets are not equal but they never were going to be with n = 14.  A split of 3-4-4-3 is reasonable and the LQ is half way between 0 and 7.5

Hope that helps,

Bob

ps.  I did a google search for 'formula for lower quartile'.  Both BBC bytesize and mathsteacher.com.au give the formula as I stated.  The Wikipedia page gives 3 methods for calculating the quartiles which do not all give the same results.

The maths national curriculum (UK) does not specify how you're meant to calculate a quartile position, nor do the exam boards AQA, Edexcel and OCR. If you are preparing for an exam you really need to check with the board concerned.  If you say what exam I'll try to research this further.


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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