Peter Regan teaching MBA Math at Tuck 

MBA Math Sample Exercise
Statistics: Normal Distribution
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Exercise
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 (zscore) table, if you know how to use it.

Solution
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 zvalues
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 zvalue, 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
zvalues 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 zvalue of 1.2632 in a series of steps:
1. Round the target zvalue up to the nearest hundredth to find the 'upper'
zvalue.
The 'upper' zvalue is 1.27
The corresponding 'upper' probability is 0.3980
2. Note that the 'lower' zvalue is 1.26, which is the upper zvalue 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 Dartmouth’s Tuck School and
Duke’s Fuqua School. He also teaches preterm quantitative skills courses at
Tuck and Cornell’s Johnson School. He created the MBA Math selfpaced, online
preMBA quantitative skills course covering finance, accounting, economics,
statistics, and spreadsheets.

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