Hello, i asked this question last year on the forum.....evidently it is very diffcult to answer. However there was formula on this site, which i found useful, and used that as rough measurement of DS standardized score, however that formula is not for age standardization but only taking the mean score of 100....so it is not accurate, but gave me a some sort of idea. I will post the formula if i find, but it should on this website.
If no age allowance were to be applied to the standardised scores, then the equation for converting raw scores to standardised scores with a mean (average) score of 100 is
S = 15(b — a)/sd + 100
where S is the pupil’s standardised score, b is the pupil’s raw score, a is the average raw score of all the pupils, and sd is the standard deviation of the raw scores.
I took SD as 15, average raw score of all pupils as 45 for an 80 questions paper, and 50 for 85 questions paper.
This is only rough estimate....by no means accurate.....
Oshosh - I got Data Protection details of my DD's raw scores from both CEM tests she took. The cohort average was presumed to be 55%. After wide spread reporting that the CEM exams were easier, people suggested at one point raw score cohort averages might be as high as 60%.
In facts they were between 51-53% from the stats we have knocked together, so Kids did not find them easier or harder. Still we have seen jumps of 3-4 standardised points needed to gain entry to the grammars this year. I played around with standard deviation and used 12.5 instead of 15 for modelling purposes.
In a massively competitive scenario ie the cohort is 6000, I would say approaching 70%, in a less competitive scenario say 1500 sitters, perhaps 65% on the test might do it.
Lot of people think only people getting 95% have a chance which is nonsense. At the margins Age can have a big effect. We saw instances where an older child lost 5 or so standardised marks for one question wrong if they were the very older in the cohort.
Hope that helps