In paper 5,question number 2, candidates are asked to plot error bars for a given quantity(either on the Y axis or X axis but till now not both axes!) on the graph which CIE Examiners always already mark the scales(this saves candidates a lot of time and avoid them using awkward scales as in usually the case in paper 3 question 1) along both X and Y axis. But when it comes to plotting the error bars on the graph, some of them are too small to be plotted accurately or to be at least visible, for instance in paper 42 of Oct/Nov 2011 question 2, the error bars ranges from: +-0.07 to +-0.0046
Here are the values of lg(T/s) and its associated absolute uncertainties which I calculated:
1st row: 4.38 +- 0.07
2nd row: 4.62+-0.04
3rd row: 5.188+-0.023
4th row: 5.483+-0.011
5th row: 5.771+-0.011
6th row: 6.1523+-0.0046
(Is the last two values of T in the table (ie 590 and 1420) given to 3 and 4 sf respectively or is it 2 or 3 sf respectively? Because if it is 3 and 4 sf then corresponding vales of lg(T/s) will be to 3 or 4 dp and 4 or 5 dp as opposed to the markscheme where the last value is given as 6.152. Please clarify me on this)
Now according to the vertical scale given (here lg(T/s)), one small square is 0.02, so it is almost impossible for candidates to plot error bar(which is represented as small vertical lines if errors bars are plotted on the Y axis) of +- 0.005(6th row) (this will be a quarter of a small square,so far too small to be plotted nicely as opposed to an error bar of +-0.08 on the same scale). The same is true for error bars of +-0.011 or +-0.023 (3rd,4th and 5th rows) which will be too small on the given scales.
So what candidates have to do when they encounter this situation given that they have to plot all the error bars?
Please clarify on this.
thanks...
Here are the values of lg(T/s) and its associated absolute uncertainties which I calculated:
1st row: 4.38 +- 0.07
2nd row: 4.62+-0.04
3rd row: 5.188+-0.023
4th row: 5.483+-0.011
5th row: 5.771+-0.011
6th row: 6.1523+-0.0046
(Is the last two values of T in the table (ie 590 and 1420) given to 3 and 4 sf respectively or is it 2 or 3 sf respectively? Because if it is 3 and 4 sf then corresponding vales of lg(T/s) will be to 3 or 4 dp and 4 or 5 dp as opposed to the markscheme where the last value is given as 6.152. Please clarify me on this)
Now according to the vertical scale given (here lg(T/s)), one small square is 0.02, so it is almost impossible for candidates to plot error bar(which is represented as small vertical lines if errors bars are plotted on the Y axis) of +- 0.005(6th row) (this will be a quarter of a small square,so far too small to be plotted nicely as opposed to an error bar of +-0.08 on the same scale). The same is true for error bars of +-0.011 or +-0.023 (3rd,4th and 5th rows) which will be too small on the given scales.
So what candidates have to do when they encounter this situation given that they have to plot all the error bars?
Please clarify on this.
thanks...
