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S1 Linear Interpolation - Solomon L Q7

Hi, I have performed linear interpolation on this question from a Edexcel solomon paper and am not sure why my method is different to the mark schemes. I was wondering if anyone could take a quick look at it and tell me why my method is wrong.

Thanks alot :smile:

Original post by ben789

You have made an error in your calculation for the median.

Reply 2

ben789

Original post by brianeverit

You have made an error in your calculation for the median.

Ah well spotted, seems I made a mistake in my upper quartile aswell. :colondollar:

Do you know why the mark scheme says 30.25, 60.5 and 90.75 instead of 30, 60 and 90?

Thanks :smile:

Reply 3

brianeverit

Original post by ben789

Ah well spotted, seems I made a mistake in my upper quartile aswell. :colondollar:

Do you know why the mark scheme says 30.25, 60.5 and 90.75 instead of 30, 60 and 90?

Thanks :smile:

With an even nuimber of items, the median is half way between the middle two, so with 120 items trhe median value is 60.5. Similar argumenbts for the upper and lower quartiles.

Reply 4

gdunne42

I think current edexcel text books recommend that all grouped data (discrete or continuous) should be treated as continuous And n/4, n/2, 3n/4 should be used directly to interpolate the quartiles.

The even number of data elements idea should only be applied to data that is not grouped. The process they recommend is:
Find n/2. If n/2 is an integer (as in this data 60) the median is the mid point of that term and the one above (mid point of of 69 and 61st term). If n/2 is not an integer round up to the next term and use that one. Same for other quantiles.

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(edited 8 years ago)

Reply 5

AJA1994

Original post by ben789

Ah well spotted, seems I made a mistake in my upper quartile aswell. :colondollar:

Do you know why the mark scheme says 30.25, 60.5 and 90.75 instead of 30, 60 and 90?

Thanks :smile:

Hi I am having the same issue as you here. I am following the text books approach and for continuous data I am did, as you did, 120/2 for median...

Reply 6

Bulbasaur10

Original post by AJA1994

Hi I am having the same issue as you here. I am following the text books approach and for continuous data I am did, as you did, 120/2 for median...

Yeah I'm having the same problem too... I think our method is actually what you are supposed to do. Also I thought if it was discrete anyway you would go halfway between the value found and the next one up e.g. 30.5 not 30.25...

Reply 7

Toasticide

quick thing. your method when done correctly will get you the marks (it's the one from the S1 textbook i think, so i cant see why it shouldnt)
but the mark schemes in the exam generally use this handy little formula that i was taught. it really helps!:
so after doing n/2 and identifying the group the middle value will be in you find the lower boundary (LB), the places into that group that the middle value lies (PG), the group frequency (GF) and the class width (including boundaries) (CW). the formula is:

median= LB+(PG/GF)xCW

if u can memorise that formula interpolation becomes a straight breeze. the only mistakes i make now are finding the wrong quartile or doing LB- instead of LB+