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One hundred draws will be made at random with replacement from the box 1, 1, 2, 2, 2, 4... The chance that the sum will be bigger than 250 is what percent?

  • statistics -

    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.

  • statistics -

    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

  • statistics -

    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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