Re: Re: Maths Questions- absolute errors
new drkitten said:
No number is 100% exact. If I say that a stick is 10.5 cm long,
I'm "really" saying that it's between 10.45 and 10.55 cm long. This idea is often formallized as the notion of "significant figures." The number 10.500, although ostensibly the same, is actually much more accurate -- it means my stick is between 10.4995 and 10.5005 cm long.
My understanding is that the last significant digit is considered to be a guess. So if you say that a stick is 10.5 cm long, that means that you're
sure that it's between 10cm and 11cm, and you're
guessing that it's between 10.45 and 10.55. Significant digits are a very rough method of presenting error; saying that a stick is 10.5 cm long means that the error is
at most .5cm, but it might be as small as .05cm.
So for the question in the OP, the absolute errors are <.05, <.005, respectively. Now, the rule on significant digits is that when numbers are multiplied, the number of significant figures in the product is equal to the minimum number of significant digits in the factors. In this case, the factors have 3 and 4 signficant figures, respectively. The smaller of those two numbers is 3, so the product has 3 significant figures; xy= 23.0 +- .5. (I'm a bit confused here, however, because it looks like something got dropped from part ii. Perhaps it had some Greek letters that didn't make it in the copy and paste?) For relative error, you divide the error by the actual value (which in this case can be anywhere from 22.5 to 23.5, but that's not significant to the answer). That's .5/23, or about 2.2% (errors usually aren't given in more than two significant figures; 23.0 +- .50123, for instance, would be rather silly).
jzs:
There are two problems with your calculations. First, you are using a normal/Gaussian error method for error propagation. While this is more precise than the significant digit method, the fact that no explicit error ranges are given suggests that the student is expected to use the significant digit method. Second, when you were calculating the error, you treated the error as standard deviation, converted it to variance, added the two variances, then converted it back to standard deviation. This is the correct method
for calculating the standard deviation of a sum. When you
multiply two numbers together, the rule is that you multiply the error of one number by the other number (taking absolute values, of course). In this case, the error would be (x error)*y+(y error)*x=.37*4.291+.291*537~=3.2.
69dodge
The formula you used yields an estimate of the relative error of xy, not an estimate of its absolute error. That's why 20 lies far outside the range you calculated, even though 20 is the right answer if x = 5 and y = 4.
Using the method I described, the range would be [19.8, 26.2]. It actually is possible to get a number that is out of the range; usually the range given is a range for which we have a 95% confidence that it contains the correct value. If both of the factors are at the extreme edges of their ranges, and in the same direction, then the product may actually be outside of its range. However, this is unlikely. Also note that the
factors may be outside their ranges as well; the 95% confidence applies to both factors and product.
