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?

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).

To find the chance that the sum of the draws will be bigger than 250, we need to calculate the number of ways this can happen and divide it by the total number of possible outcomes.

Let's break down the problem step by step:

Step 1: Determine the possible outcomes for each draw:
In this case, we have a box with numbers (1, 1, 2, 2, 2, 4). So, each draw can result in any of these numbers.

Step 2: Calculate the number of possible outcomes for all 100 draws:
Since each draw can result in any of the numbers in the box, there are 6 possible outcomes for each draw. Therefore, for 100 draws, there are a total of 6^100 possible outcomes.

Step 3: Calculate the number of outcomes where the sum is bigger than 250:
To determine the number of outcomes where the sum is bigger than 250, we need to consider all the combinations of draws that could result in a sum greater than 250. This involves calculating the sum of each possible combination and counting the number of combinations that meet the criteria.

This process can be quite complex, involving a significant number of calculations and exhaustive enumeration of all possible combinations. Therefore, it ideally requires computer programming or advanced mathematical techniques to obtain a precise answer.

Step 4: Determine the probability of the sum being bigger than 250:
To determine the probability, divide the number of outcomes where the sum is bigger than 250 by the total number of possible outcomes and express it as a percentage.

Given the complexity of the problem, let's assume that you have access to a programming language or a mathematical software package capable of performing these calculations efficiently.