Seven, no?
The margin of error could be as much as 7 and as little as 3, so we should account for a margin of error at greatest of seven.
I'm not sure, maybe I don't have the best idea of what is meant by margin of error, but it makes sense to me.
Suppose someone tells you that the number of marbles in a jar is 100 with a margin of error of 5, and that 5 has a margin of error 2.
What is the true margin of error of the number of marbles in the jar expressed as a single number?
Hopefully this question should be hard enough to be a puzzle.
Seven, no?
The margin of error could be as much as 7 and as little as 3, so we should account for a margin of error at greatest of seven.
I'm not sure, maybe I don't have the best idea of what is meant by margin of error, but it makes sense to me.
Well, the margin of error on 100 could be 7 or 6 or 5 or 4 or 3
If all margins of error are equally likely, would it not make sense to take the average margin of error?
(7 + 6 + 5 + 4 + 3)/5 = 5
My guess is margin of error stays 5, but I don't like it.
If the largest margin of error could be 7, then shouldn't we consider that...just to be safe?
As eulcer observed?
I think I figured it out.
If the margin of error is 5 with a margin of error of 2, then, the number of marbles can be
Does anyone see where to go from here, how do I express the margin of error as a single number? Seems like you can have a margin of error but some parts of it are more likely of appearing.Code:93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 // error of 7 94 95 96 97 98 99 100 101 102 103 104 105 106 // error of 6 95 96 97 98 99 100 101 102 103 104 105 // error of 5 96 97 98 99 100 101 102 103 104 // error of 4 97 98 99 100 101 102 103 // error of 3
So there's 55 numbers.
93: 1/55
94: 2/55
95: 3/55
96: 4/55
97: 5/55
98: 5/55
99: 5/55
100: 5/55
101: 5/55
102: 5/55
103: 5/55
104: 4/55
105: 3/55
106: 2/55
107: 1/55
You should first decide what the term "margin of error" means, it is ambiguous. From what you post it seems you think is is the semi-range.
You seem to assume that you have uniform distributions for your first two margin of error values, but you will not have a uniform distribution for the final result.
CB