Supposing, for the sake of illustration, that the mean is 31.2 and the std. dev. is 1.9.
This probability can be calculated by finding z-scores and their corresponding areas under the std. normal curve. 34 in - 31.2 in The area under this curve to the left of z = -------------------- = 1.47 (for 34 in) 1.9 32 in - 31.2 in and that to the left of 32 in is z = ---------------------- = 0.421 1.9
Know how to use a table of z-scores to find these two areas? If not, let me know and I'll go over that with you.
My TI-83 calculator provided the following result:
normalcdf(32, 34, 31.2, 1.9) = 0.267 (answer to this sample problem)
So what you would do is divide 988 by 26 and divide 731 by 17 and those two numbers you would add which gives you the amount of people that attended the performance
