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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?
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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?
Yea I know, I got your pointMate 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.
So you support Dug's theory that class boundaries aren't required when calculating midpoints? I am considering 0. It's driving me mental!Yea I know, I got your point
But, we cant have -ve number in some cases (many cases) so consider 0!
Nope, I don't support that, I agree to your point in that case.So you support Dug's theory that class boundaries aren't required when calculating midpoints? I am considering 0. It's driving me mental!
For 0 it stays zero, keep it like a rule.!!
u know about these class boundaries and stuff our sir told us that we can either subtract and add respectively to the 1st and the second value or we cud just add 1 to the second one or subtract one from the first one... just have to make a difference with 1. and given this question i dont think there's much we can do about the negative value so its better to add 1 to the second one.Yeah I know, and that's what confuses me the most!
Youtube doesn't work here mateNope, I don't support that, I agree to your point in that case.
But I think this will satisfy you
http://www.examsolutions.net/maths-revision/statistics/representing-data/histograms/tutorial-1.php
Not 10.5, 9.5!u know about these class boundaries and stuff our sir told us that we can either subtract and add respectively to the 1st and the second value or we cud just add 1 to the second one or subtract one from the first one... just have to make a difference with 1. and given this question i dont think there's much we can do about the negative value so its better to add 1 to the second one.
btw the answers to these class widths are ambigous because they are different in almost every marking scheme. just make ur best judgement according to the question is pretty much all u can do.
also if u take 10.5 till 20.5 in this specific example under discussion u will still get a decimal value for the mean (15.5) ryt?
I've watched the video. It still doesn't make sense. If you take the class boundaries as 0 and 9.5, you'll get the midpoint as 4.75, whereas the mark scheme clearly states it as 4.5. That's what I don't understand!!!!!!!!!!!!!!!!!!!!!!!!!Nope, I don't support that, I agree to your point in that case.
But I think this will satisfy you
http://www.examsolutions.net/maths-revision/statistics/representing-data/histograms/tutorial-1.php
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