Question 964030
probably not, based on the following test.


n = 9000
p (3 or 4) = 1/6 + 1/6 = 2/6 = 1/3
(1-p) = 2/3


mean = n * p = 9000 * 1/3 = 3000.


standard error equal sqrt (n * p * (1-p) = sqrt ( 9000 * 1/3 * 2/3) = sqrt(2000) = somewhere between 44.72 and 44.73, the exact value not being important for this problem.


a raw score of 3240 will generate a z-score of (x-m) / se which would be equal to:


z = (3240 - 3000) / 44.72 or 44.73.


therefore z = 5.367 or 5.366.


both are more than 3 standard deviations from the mean.


the probability of getting a z-score greater than 3 standard deviations from the mean would be equal to .00135.


this is less than .05 and also less than .01 which are the probabilities normally used as criteria.


the .05 is normally used at the .95 confidence level for a one tailed test.


the .01 is normally used as the .99 confidence level for a one tailed test.


bottom line is the probability of getting such a score out of 9000 tosses just based on random differences between different samples is very remote and therefore not considered feasible.


this indicates that the dice are probably biased in some way.