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1. I know how interpolate data, but I am not sure how to do this question because the class median = 26/2 = 13 so 13.5th value. This lies within the first class interval, 0 - 1 which is actually 0 - 1.5
Not sure how to go about this because it seems like I will have 18 at both ends of the line (the line that's drawn on to interpolate the datt...)

Zacken
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2. slowdive
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3. (Original post by Wolfram Alpha)
slowdive
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Hey sorry, I didn't do S1, I did M1 instead as I took Physics AS.
4. (Original post by slowdive)
Hey sorry, I didn't do S1, I did M1 instead as I took Physics AS.
Ahh that's alright then, thanks anyway
By the way do you have any tips for M1? I am also taking physics and we have started vectors (the complicated ones where there's T1 T2 T3 and you have to resolve vertically and horizontally and all that complicated stuff) and I am struggling...
Thankss
5. (Original post by Wolfram Alpha)
Ahh that's alright then, thanks anyway
By the way do you have any tips for M1? I am also taking physics and we have started vectors (the complicated ones where there's T1 T2 T3 and you have to resolve vertically and horizontally and all that complicated stuff) and I am struggling...
Thankss
I take a methodical approach to M1, the questions are very repetitive once you practice them. For the vectors, I always wrote a table with T1, T2 etc against vertical / horizontal. This may be obvious, but in M1 diagrams go a looong way
6. A couple of points on this one:

1) We're dealing with grouped data so there's no rounding when determining which member of the population represents the median, i.e. we go with 26/2 = 13.

2) The first class in the table takes the cumulative frequency from zero to 18. These are the end-points for your interpolation.
7. (Original post by old_engineer)
A couple of points on this one:

1) We're dealing with grouped data so there's no rounding when determining which member of the population represents the median, i.e. we go with 26/2 = 13.

2) The first class in the table takes the cumulative frequency from zero to 18. These are the end-points for your interpolation.
Hello old-engineer,
Thanks for your response, I just want to clarify something. You say there is no rounding as it is grouped data, but I thought what determined whether there is rounding or not is if the data is continuous or discrete. Or is my understanding incorrect? Also, my median value lies from 0-18 as you mentioned, but since it's discrete data (I think), my class width needs to be +- 0.5. But that means I would have -0.5 which makes no sense. So would my class interval just be 0-1.5?
8. (Original post by Wolfram Alpha)
...I thought what determined whether there is rounding or not is if the data is continuous or discrete....
No, GROUPED data is treated as continuous even if the original data was discrete. If you're using the Edexcel S1 textbook, Section 2.7 refers.

(Original post by Wolfram Alpha)
...So would my class interval just be 0-1.5?...
Yes that's right. There is no gap to the left of the first class, so you don't need to extend the lower limit of the class downwards,
9. (Original post by old_engineer)
No, GROUPED data is treated as continuous even if the original data was discrete. If you're using the Edexcel S1 textbook, Section 2.7 refers.

Yes that's right. There is no gap to the left of the first class, so you don't need to extend the lower limit of the class downwards,
Ah yes...I think I understand now. That means NO data can ever be discrete if the data is presented in a table like the one in the question, correct?
10. (Original post by Wolfram Alpha)
Ah yes...I think I understand now. That means NO data can ever be discrete if the data is presented in a table like the one in the question, correct?
The original data may be discrete in nature (e.g. collar size, or whatever), but if it is presented in a grouped table it is always handled as continuous data.

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