Edexcel S1 median

Original post by krisshP

QP:
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202009/6683_01_que_20090520.pdf
MS:
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June09-ms/8371_9374_GCE_Maths_msc_20090717_UA021531.pdf

I need help on 4a. Here's what I did:
120/2=60
so Q2=60.5th observation

\frac{Q2-16}{18-16} = \frac{60.5-58}{87-58}

Q2=16.2 (3 S.F.)

What have I done wrong here? The MS says the correct answer is 17.1/17.2. I followed the method in the Edexcel S1 book can I still got it wrong, why?

THANKS

You do your interpolation differently to me, but the lower bound of the class with the median in is 17, so the median must be at least 17. :smile:

Reply 2

maggiehodgson

You have your median position as 60.5. OK. Which class is that in and what fraction of it is it? Once you've decided that you can find an estimate for the median value.

Is that clue enough?

Reply 3

avoxgirl

Original post by krisshP

QP:
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202009/6683_01_que_20090520.pdf
MS:
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June09-ms/8371_9374_GCE_Maths_msc_20090717_UA021531.pdf

I need help on 4a. Here's what I did:
120/2=60
so Q2=60.5th observation

\frac{Q2-16}{18-16} = \frac{60.5-58}{87-58}

Q2=16.2 (3 S.F.)

What have I done wrong here? The MS says the correct answer is 17.1/17.2. I followed the method in the Edexcel S1 book can I still got it wrong, why?

THANKS

You do your interpolating differently to the way I do it but here's how I'd go about it:

Q2 is the 60.5th observation, you then need to find which group that falls in, so cumulatively it falls in the 17-19th group and there are 58 values before it, meaning it is the 2.5th value in the 17-19 group.

Because the classes are all right next to each other you don't need to find the upper and lower bounds, so divide the class width by the frequency, to split the group into even parts (assuming all values are evenly spread). This is 2/29, so each value is an extra 2/29th into the group more than the last one.

Then you multiply it's position in the group by this (2.5*2/29) and add it to the lower boundary (in this case 17)

Also further note: for interpolating you're not supposed to bump up the median value, so you should leave it at n/2= 60. Some mark schemes accept both, but not all do.

(edited 10 years ago)

Reply 4

krisshP

OP

Original post by joostan

You do your interpolation differently to me, but the lower bound of the class with the median in is 17, so the median must be at least 17. :smile:

But the 58th observation is in the end of the 2nd class, so surely the 58th observation is 16.

(edited 10 years ago)

Reply 5

Original post by krisshP

But the 58th observation is in the end of the 2nd class, so surely the 58th observation is 16.

That doesn't change the fact that the 59th value must be 17. . .

Reply 6

krisshP

OP

Original post by joostan

that doesn't change the fact that the 59th value must be 17. . .

Q2-17 60.5-59
------=-------
18-17 87-59

Q2=17.053....
Q2=17.1

Do you think this would score the M1 mark since some values in the method are different from the MS method values, but the correct final answer is still obtained?

Reply 7

Original post by krisshP

Q2-17 60.5-59
------=-------
18-17 87-59

Q2=17.053....
Q2=17.1

Do you think this would score the M1 mark since some values in the method are different from the MS method values, but the correct final answer is still obtained?

I've not looked at the mark scheme, but my understanding is that provided your method is correct and your answer is correct, even if it differs from the mark scheme you get full credit.
Obviously there are many ways to do every question and the mark scheme can't account for all of them :smile:

Reply 8

krisshP

OP

Original post by joostan

I've not looked at the mark scheme, but my understanding is that provided your method is correct and your answer is correct, even if it differs from the mark scheme you get full credit.
Obviously there are many ways to do every question and the mark scheme can't account for all of them :smile:

Thanks

Can you please help me with one more question
http://www.examsolutions.net/a-level-maths-papers/worked-solution/worked-solution.php?paper_id=611&solution=3.4

Go to 3:19. He's substituted p=16, but I don't understand this. Shouldn't he substitute p=17 since the question mentions "ABOVE 16" which is >16, so is 17, 18, 19.... Since it talks about minimum thinning, surely p=17 should be substituted?

Thanks so much for the help!

(edited 10 years ago)

Reply 9