Posted by Amanda on Monday, October 6, 2008 at 4:50pm.
The mean of those numbers is 2. The variance is 1.2. The sum of 100 of them should have a mean of 200 and a variance of 120, i.e. a standard deviation of sqrt(120) = 10.95.
Now, 250 is (250-200)/10.95 = 4.56 standard deviations above the mean. Assuming the distribution is approximately Normal (and I think it would be after you've added 100 such draws together), you can look that up in a set of Normal tables to find out the area under the curve to the right of that point. I make that 1-0.999997=0.000003, or 0.0003%. (I've also run some simulations in a spreadsheet which seem to bear that figure out: not a single run out of a several hundred has reached 250.)
I'm not actually sure about the above reasoning (paricularly the bit about the variance of 100 draws being 100 times the variance of {1,1,2,2,2,4}), so if anybody reckons I've made a mistake, just shout.
shouldn't the standard deviation be 1?
i have all the info... i just don't know how to find the z score from the info that i have:
avg= 2
SD= 1
sum= 200
SE= 10
You could be right about the standard deviation being 1. I did wonder at the time about whether you ought to be using the standard deviation for the entire population, as opposed to the usual one that's applied to a sample, which would be calculated using n as the divisor instead of (n-1). If so, then 250 is 5.0 standard deviations above 200 as opposed to the 4.56 I calculated earlier. Either way, that's the Z value which you need to look up in a set of Normal probability tables, to find the area under the Normal probability curve to the left of that figure. For Z=5 the answer will be almost one (if you do it in Excel using the NORMSDIST function you will get 0.9999997, whereas for Z=4.56 you'll get 0.999997). You then subtract that from 1 to get the area to the right. Either way, the answer is extremely small (2.9E-7 for Z=5, or 2.6E-6 for Z=4.56).
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