- Messages
- 227
- Reaction score
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Class Intervals | X
1 - 100 | 50.5
101 - 150 | 125.5
151 - 200 | 175.5
...
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Class Intervals | X
I had same doubt, but apparently the reason is, it is less than or equal to the specific value. When it is only less than then only 50 not 50.5
i guess! not sure.
Interesting. Dug's method make sense. Thanks for sharing, Dug! Any thoughts, PhyZac?Class Intervals | X
1 - 100 | 50.5
101 - 150 | 125.5
151 - 200 | 175.5
...
i thikn it will be like for the first one: 0=< x < 10http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_63.pdf
ms http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_63.pdf
Have a look at Q3 part iii. Could you tell me the class boundries for this one Dug
0 - 9http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_63.pdf
ms http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_63.pdf
Have a look at Q3 part iii. Could you tell me the class boundries for this one Dug
I think its something like 19.5-20.5 etc. No?
Yeah exactly. So the boundries are something like:0 - 9
10 - 19
20 - 34
But that means the class width is 10, and there is no way you can get a midpoint of 4.5, as it says in the markscheme.i thikn it will be like for the first one: 0=< x < 10
include the first value and not the last one.... there can be no decimals i think bcz there are marks
Class boundaries are not needed for calculating mean. We do that only in histograms and cmf curves!Yeah exactly. So the boundries are something like:
-0.5-9.5
9.5-19.5
19.5-34.5
Correct or not?
You need the class boundaries to calculate the midpoint, don't you?Class boundaries are not needed for calculating mean. We do that only in histograms and cmf curves!
Nope!You need the class boundaries to calculate the midpoint, don't you?
Dug I'm pretty sure you calculate the class width by using the class boundaries. Then using the class width you can calculate the midpoint! If not, could you provide me with some reference stating otherwise?Nope!
Class Interval | Mid-point | Frequency
0 - 9 | 4.5 | 25
10 - 19 | 14.5 | 43
20 - 34 | 27 | 91
35 - 49 | 42 | 75
50 - 69 | 59.5 | 26
70 - 99 | 84.5 | 40
Mean = (4.5)(25) + (14.5)(43) + (27)(91) + (42)(75) + (59.5)(26) + (84.5)(31) = 11270/300
I don't have a reference but you could try calculating the mean using your method. The answer will come out wrong!Dug I'm pretty sure you calculate the class width by using the class boundaries. Then using the class width you can calculate the midpoint! If not, could you provide me with some reference stating otherwise?
Hmm, I will have do understand it first. But seems very sensible.Interesting. Dug's method make sense. Thanks for sharing, Dug! Any thoughts, PhyZac?
but the marks cannot be in negative :OYeah exactly. So the boundries are something like:
-0.5-9.5
9.5-19.5
19.5-34.5
Correct or not?
Actually it would be correct. You can try yourselfI don't have a reference but you could try calculating the mean using your method. The answer will come out wrong!
Yeah I know, and that's what confuses me the most!but the marks cannot be in negative :O
For 0 it stays zero, keep it like a rule.!!Yeah I know, and that's what confuses me the most!
What about the mid-point of the first class 0.5 - 9.5?Actually it would be correct. You can try yourself
It's the midpoints that matter, Dug. So take the class 19.5-34.5. It's midpoint is 27. And so on. Do you have that stats book by Chambers and Cranshaw?
Mate look: if you have two classes such as 10-19 and 20-29, then their class boundaries should be 9.5-19.5 and 19.5-29.5. That's what I was taught. I think you would concur.For 0 it stays zero, keep it like a rule.!!
No. I was thinking of something along the lines of -0.5-9.5. You need class boundaries! The negative confuses me, yes. The only possible reason I can think of is negative markingWhat about the mid-point of the first class 0.5 - 9.5?
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