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MBA Math Sample Exercise

Statistics: Normal Distribution 

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Suppose the daily customer volume at a call center has a normal distribution with mean 4,600 and standard deviation 950. What is the probability that the call center will get fewer than 3,400 calls in a day?

Click here to view a standard normal (z-score) table, if you know how to use it.


Solution Commentary


Manual Solution

"Standard" normal distributions have a mean of zero and a standard deviation of 1. Fortunately, nonstandard normal distributions can be easily converted into standard normal distributions. Solutions obtained for the standard distribution can be applied to the corresponding nonstandard problem. The motivation for this conversion from standard to nonstandard normal distribution is that there is no simple functional form for obtaining solutions to normal distribution problems. Rather, the manual solution process for standard normal distribution problems with the probability specified requires solving the problem using the normal distribution table of values.

The table entries consist of probabilities for the interval from the mean to positive z values. We use the first column to identify the row for the z value integer and tenth place. We then use the first row to identify the column for the z value hundredth place. The probability for the z value is at the intersection of the corresponding row and column. We interpolate for z-values with more than two decimal places.

We first convert the value of interest from the x scale and units of the original problem to the corresponding z-value, using the formula:

z = (x - m)/s, where x is value of interest, s is the standard deviation and m is the mean, all in the units of the original problem
z = (3,400 - 4,600)/ 950
z = -1.263

The clearest way to understand the transformations of the exercise into a form that can be answered using the standard normal distribution table of values is to create a set of pictures that illustrate the transformations. (Note that the pictures below show relationship to the mean and do not reflect the exact z-values of the exercise.)

P(x < 3,400) = P(z < -1.263)
= 0.5 -
P(x < 3,400) = 0.5 - P(0 < z < 1.263)

We compute the target probability for a z-value of 1.2632 in a series of steps:

1. Round the target z-value up to the nearest hundredth to find the 'upper' z-value.
The 'upper' z-value is 1.27
The corresponding 'upper' probability is 0.3980

2. Note that the 'lower' z-value is 1.26, which is the upper z-value minus 0.01.
The corresponding 'lower' probability is 0.3962

3. Because our probability solution is reported to the nearest percent, and the upper and lower probabilities round to the same percent, we can report the probability as 0.40 without further interpolation.

P(x < 3,400) = 0.5 - 0.40
= 0.10

Excel Solution

Peter Regan teaches decision science courses at Dartmouthís Tuck School and Dukeís Fuqua School. He also teaches pre-term quantitative skills courses at Tuck and Cornellís Johnson School. He created the MBA Math self-paced, online pre-MBA quantitative skills course covering finance, accounting, economics, statistics, and spreadsheets.

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