Explanation:
we assume a normal distribution.
the meanbox = (2+4+6+8)/4 = 20/4 = 5 (as we could see already just by looking at it).
the SDbox = 2.24 (as given), as it is the sum of the squares of the differences of each data point to the mean value, and this square is divided by the number of items in the "population". from this result we take the square root.
so, SDbox = sqrt(((2-5)²+(4-5)²+(6-5)²+(8-5)²)/4) =
= sqrt((9+1+1+9)/4) = sqrt(20/4) = sqrt(5) =
= 2.236067977... ≈ 2.24
I recalculated SDbox to check that the given number is correct (you would not believe how often given numbers are wrong). in our case, yes, it is correct.
so, now for the real "magic" :
the expected value EV of the sum of 100 draws is
100×meanbox = 100×5 = 500
the expected standard deviation ES of the 100 draws is
sqrt(100)×SDbox = 10×2.24 = 22.4
z for the normal distribution table is defined as
z = (x - mean)/standard deviation
applied to the sum of drawings we have
z = (x - EV)/ES = (550 - 500)/22.4 = 50/22.4 =
= 2.232142857...
actually, if we used the not rounded number, it should be again 2.236067977... ≈ 2.24
in the distribution table we find the area of left of 2.23 :
0.98713
left of 2.24
0.98745
the true middle between them would be 0.98729.
rounding a little bit up is then 0.9873
so, 0.9873 is the probability that the sum of the 100 drawings is less than (or equal to) 550.
the complimentary probabilty is then the probabilty that the sum is larger than (or equal to) 550 :
1 - 0.9873 = 0.0127 = 1.27% (as this is just the probabilty multiplied by 100).
if we used the rounded given number and therefore 2.23 for z, we would get
1 - 0.98713 = 0.01287 ≈ 1.29%
I guess, I would use the last number, as that is what your teacher was driving you to.
so, the percent chance for the 100 drawings to create a sum of over 550 is 1.29%.