If we know the mean (**μ**) and standard deviation (**σ**) than we can use either the cumulative normal distribution function, accessible thru many scientific calculators with stat functions, or the standard normal distribution and its CDF, accessible thru a Z table.

Method 1. (from TI-8x calculator) n = 5000, **μ** = 86, **σ** = 10 limits [86 , 96]

Determine percentage under curve between 86 and 96

P = nmcdf(86,96,86,10) = .3413

Multiply by 5000 to get answer: 1706.72, which is rounded down to meet limit requirements.

**Answer : 1706**

Method 2. using Z table.

determine Z score for the upper bound by standard formula (x-μ)/σ or (96-86)/10 = 1

The area to the left of this Z score represents the total probability of having a score less than 96. Using the Z tables we find the percentage is = .8413

We can use a simple trick to then determine how much the probability is for between 86 and 96 by remembering that the area to the left (or right) of the mean (μ) is always equal to .5 Since the left bound of the area of interest is the same as the mean, then the area to the left that we must subtract from the area above (.8413) is .5

Thus to find the percentage between a z of 0 (i.e. the mean) and a z of 1 (the calculated z) we calculate P = .8413 - .5 = .3413

To get the total number of test scores fully between the mean and the upper bound we calculate .3413 x 5000 and round down.

**Answer: 1706**