Posted by Lenny on Friday, October 3, 2008 at 5:20am.
Are you sure that second formula isn't Ó(x-ì)²F? Unless it is, I can't see a way of calculating a standard deviation from the statistics supplied.
On the assumption that the second statistic is Ó(x-ì)²F, the variance ought to be (1/150) times this figure, which is 196348/150 = 1309, so the standard deviation would be 36.2, with a mean of 10200/150=204. So any student with a score of between 204-(2x36.2) and 204+(2x36.2) would be within two standard deviations of the mean.
Having said all that, you asked for the actual number of students within that range - and it seems to me that you don't actually need either of the statistics given to answer that question. If the scores are Normally distributed, then 4.6% of the distribution ought to lie outside two standard deviations (i.e. either in the lower or the upper tail). Out of 150 students that's 7 students - so 143 ought to lie inside plus or minus two standard deviations. (I can't see any way to tell *exactly* how many students lie within that interval, even from the statistics supplied.)
(Sorry - the symbols Ó and ì ought to be a sigma sign and a mu respectively. You evidently can't just copy them from another posting - it doesn't work.)