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How much do the middle 68% of customers purchase?

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2 Answers

This URL is a good presentation of the problem that you pose:
In the figure on this page, you can see the normal distribution.  The top of the curve represents the mean value of the data.  For the standard normal distribution, the mean (Greek letter mu, µ) is 0, and the standard deviation (Greek letter sigma, σ) is 1.  Along the horizontal axis is the Z value.  A Z = -1 value means the data value that is 1 standard deviation less than (below) the mean, while a Z = +1 (you can omit the positive sign) is the data value 1 standard deviation higher than (above) the mean.
Area under the curve.  The total area under the normal distribution curve is 100%.  It is a characteristic of the curve that the area between Z = -1 and Z = 1 represents approximately 68% of the curve (you use calculus to determine that).  The term "middle 68%" that is in your question essentially refers to the area between Z = -1 and Z = 1.  So what you are being asked to calculate are the values at the Z = -1 and Z = 1 values for your normal distribution.  The formula Z = (x - µ) / σ is used in one of the ways:  when you know the x value relative to the mean and standard deviation, it determines Z which is the position determined as the number of standard deviations units from the mean.  If less than zero, it is below the mean, if above, then above the mean.  In this case, we know Z, so we calculate x, because it represents the data value that is a certain number of standard deviation units from the mean.  You want to know it for both Z = -1 and Z = 1, because this range represents the middle 68%.
x =  µ + Zσ 
For Z = 1, x = $55 + $18(1) = $73
For Z = -1, x = $55 + $18(-1) = $55 - $18 = $37
Therefore the middle 68% pay in the range from $37 to $73
Hello Jamie, 
How are you?  I have answered your question below:
For a normal distribution, 68% of the values fall within one standard deviation of the mean.  This means that 68% of the values are between ($55 - $18) and ($55 + $18).  
Hope this helps,