IA Help: Are Trapezoidal Sums for Integrals Higher Level Math?

[gamerseth]

Basically what my title asks. Unfortunately my data won't work with a chi squared test, so I am using a trapezoidal summation (which basically is an approximation of an integral) for some of my data to calculate the average value. Would this approximation, since it represents calculus even though the mathematics is not calculus, work for my higher math process?

[kw0573]

Quote

Examples of further processes are differential calculus, mathematical modelling, optimization, analysis of exponential functions, statistical tests and distributions, compound probability. For this level [level 4 out of 5 in Criterion 3: Math processes] to be achieved, it is not required that the calculations of the further process be without error. At least one further process must be calculated, showing full working

^ From the markscheme. 
Trapezoidal summation, as you call it, should count as a further process if there is a purpose to figure out the area underneath the curve. However, the average for discrete values are simply calculated by adding up all values and dividing by the number of numbers. If you cannot demonstrate that you know the purpose of the further process then marks cannot be allocated. You can try to find another statistical test, model the data, or find some purpose for the area underneath. 

[gamerseth]

sWell basically what im doing is seeing how accurately the weather channel predicts. What i did was plotted their predicted minus actual over 50 days (if they predicted 55 degrees and the high was actually 57, -2 would show up on the graph.) I then divide the area under the graph by the 50 days to get the average amount they were off by, which ended up being 0 (since overand underpredictions cancel out, if you get what i mean)

[kw0573]

I see what you are saying. But you have to realize that you would also get 0 if you just find the arithmetic mean. It's quite pointless to find area then divide by 50, because all you did is instead of
(x1 + x2 + ... + x50) / 50
you did
(x1*1 + x2 *1 + ... + x50 * 1) / (1 + 1 + 1 ... [50 times]) which is the identical calculation. I don't think it's going to work. But you might want to check with your teacher.