You are Here: Home >< Maths

# Edexcel S1 median Watch

1. QP:
http://www.edexcel.com/migrationdocu...e_20090520.pdf
MS:
http://www.edexcel.com/migrationdocu...7_UA021531.pdf

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

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
2. (Original post by krisshP)
QP:
http://www.edexcel.com/migrationdocu...e_20090520.pdf
MS:
http://www.edexcel.com/migrationdocu...7_UA021531.pdf

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

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.
3. 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?
4. (Original post by krisshP)
QP:
http://www.edexcel.com/migrationdocu...e_20090520.pdf
MS:
http://www.edexcel.com/migrationdocu...7_UA021531.pdf

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

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.
5. (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.
But the 58th observation is in the end of the 2nd class, so surely the 58th observation is 16.
6. (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. . .
7. (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?
8. (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
9. (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
Thanks

http://www.examsolutions.net/a-level...1&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!
10. (Original post by krisshP)
Thanks

http://www.examsolutions.net/a-level...1&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 talk bout minimum thinning, survey p=17 should be substituted?

Thanks so much for the help!
I don't think p has to be an integer so:
16.000000000001 is enough. . .
That said, the distinction between and is questionable in this case.
Though considering he's rounded to 1dp it shouldn't make a lot of difference.

No problem
11. 6 An economics student is trying to model the daily movement points, in a stock market indicator. The student assumes that the value of X on one day is independent of the value on the next day. A fair die is rolled and if an odd number is uppermost then the indicator is moved down that number of points. If an even number is uppermost then the indicator is moved up that number of points.

(a) Write down the distribution of X as specified by the student's model.

12. (Original post by noona77)
6 An economics student is trying to model the daily movement points, in a stock market indicator. The student assumes that the value of X on one day is independent of the value on the next day. A fair die is rolled and if an odd number is uppermost then the indicator is moved down that number of points. If an even number is uppermost then the indicator is moved up that number of points.

(a) Write down the distribution of X as specified by the student's model.

something along the lines of:

x value: -5 -3 -1 2 4 6
P(X=x): 1/6 1/6 1/6 1/6 1/6 1/6

sorry it's hard to type out, but a table with the probability underneath each x value being 1/6

the die is fair, so the probability of each value being rolled is 1/6. Take x to be how high or low the indicator is moved.
13. (Original post by avoxgirl)
something along the lines of:

x value: -5 -3 -1 2 4 6
P(X=x): 1/6 1/6 1/6 1/6 1/6 1/6

sorry it's hard to type out, but a table with the probability underneath each x value being 1/6

the die is fair, so the probability of each value being rolled is 1/6. Take x to be how high or low the indicator is moved.
thanxx a lot but how did you find the x values

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: May 16, 2013
Today on TSR

### What is the latest you've left an assignment

And actually passed?

### Simply having a wonderful Christmas time...

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.