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MATH 2441
Probability and Statistics for Biological Sciences
The Normal Distribution
The so-called normal probability distribution (sometimes also called the Gaussian distribution after the
name of the mathematician who first studied it extensively), is a continuous probability distribution. It turns
out to be the most important probability distribution in statistics because of some very fundamental
mathematical results in probability theory.
We will use the symbol x to represent a generic normal random variable. The probability density function
contains two constants, denoted  and , for reasons that will become clear shortly. The formula for the
density function is
f ( x) 
1
2 
e ( x   )
2
/ 2 2
(ND-1)
The graph of this function has the familiar bell shape, centered on, and symmetric with respect to, the line
x = .
µ
x
The value of  determines the width of the bell. Small values of  give very tall narrow bells, whereas larger
values of  give shorter flatter bells:
µ
All three of these "bells" represent a normal distribution with the same value of , but different values of .
You can see from formula (ND-1) that the height of the bell at the peak, where x = , is inversely
proportional to . This means that the lowest broadest bell in the figure has a value of  about three times
as big as does the tallest narrowest bell. Although it may be difficult to gauge this precisely from the figure,
David W. Sabo (1999)
The Normal Distribution
Page 1 of 10
the area under each of the three bells is the same (and equal to 1 when appropriate physical units of
measurement are used).
As with all continuous random variables, probabilities for the normal random variable are calculated as areas
underneath this probability density curve, and so, mathematically, the computation of probabilities involves
the evaluation of a definite integral:
x b
Pr(a  x  b)   f ( x ) dx 
x a
1
2 
b
e
( x   )2 / 2 2
dx
(ND-2)
a
Here, a, b, , and  stand for specific numerical values that would be known before you could attempt to
compute the probability. Note the following general properties:
(i.)
(ii.)
(iii.)
the probability density curve is defined for x running from - and +, though when the
value of x gets quite a bit different from the value of , the negative exponent becomes
large enough that the value of the density function is negligible for practical purposes.
The total area under this curve for x running from - and + is 1, with most of this area
occurring in the near vicinity of x = .
the probability density function is symmetric about the vertical line at x = . This line
divides the bell into two mirror-image halves, each with an area of 0.5.
Using formulas (RV-7a) and (RV-7b) from the preceding document, we find that the mean
value for the normal distribution works out to be precisely the quantity denoted above as 
and the variance works out to be the quantity denoted above as 2, hence the use of
these symbols in the defining formula (ND-1). Thus, the formula for a normal distribution
probability density function (which is equivalent to the formula for the relative frequency
distribution of a normally-distributed population) requires specification of the mean and
variance of that distribution: the values of  and 2 as numerical parameters.
Unfortunately, even when the numbers for a, b, , and  are plugged into this formula, the definite integral is
too difficult to calculate exactly by hand (though approximation formulas of more than adequate accuracy
are known, and are simple enough to build into even relatively inexpensive hand-held electronic calculators).
As a result, statisticians have come to rely on printed tables of values for this integral.
We need to simplify the situation somewhat first though. We need the normal probability tables to be able to
handle any values of a, b, , and  that may arise. In principle at least, a, b, and  can have values
anywhere between - and +, and  can have values anywhere between 0 and +. Suppose we set up
pages of the tabulation so that the value of 'a' varies down the rows, and the value of 'b' vary across
columns. Thus one page of the table would give probabilities for each combination of values of a and b for a
particular pair of values of  and . (These would be pretty big pages because we'd need one row for each
value of 'a' between - and +, and one column for each value of 'b' between - and +). Now, for each
value of , we'd need a separate page for each possible value of  (between 0 and +), forming a rather
large book. Then we'd need one book like that for each possible value of  between - and +. So, at first
look, it seems like we might be facing having to set up a library of an infinite number of books, each with an
infinite number of pages, with each page having an infinite number of rows and columns in order to tabulate
any possible value of the definite integral above that may be required.
Fortunately, there are ways of reducing the required amount of information considerably  in fact, we'll now
describe how to get it down to a single standard size page, and you won't even need a magnifying glass!
Simplification #1: The Standard Normal Probability Distribution
A very simple substitution in (ND-2) removes both  and  from the integrand (reducing our requirement for
a library of an infinite number of volumes, each with an infinite number of pages which are infinitely large to
just one infinitely large page!  we'll whittle that page down next.)
One of the main methods you used in MATH 1441 for calculating integrals was the method of substitution.
We will apply it here, by making a substitution for x based on the formula:
Page 2 of 10
The Normal Distribution
David W. Sabo (1999)
z
x
x    z


(ND-3)
You should recognize z here as the so-called standard score associated with x. Substituting for x in
(ND-2), recognizing that dx = dz, and noting that:
x=a
z

a
and

x=b
z

b

,
we get
Pr(a  x  b) 
1
2 
b
e
( x   ) / 2
2
2
dx 
a
1
2
z
b 


z
a 
e z
2
/2
dz

b
a  
 Pr 
z

 
 
(ND-4)
When you compare the second integral in (ND-4) with the integral in (ND-2), you see that z is itself a
normally distributed random variable  it is the normally distributed random variable that has  = 0 and
 = 1. Condensing (ND-4) somewhat, we get:
 a  x
b  x
Pr(a  x  b)  Pr 
z

x
x




(ND-5)
We've put subscripts, x, on the  and  in this formula to emphasize that these stand for the mean and
standard deviation of the probability distribution of x. Formula (ND-5) can be used to get any probabilities
for any normally distributed random variable as long as we can calculate probabilities for that one special
normally distributed random variable: z. Because probabilities of all normally-distributed random variables
are ultimately related to probabilities of this variable z, it is called the standard normal random variable,
and its probability distribution is called the standard normal probability distribution. The symbol 'z' will
be reserved exclusively in this course to stand for the standard normal random variable. (Some more
mathematically rigorous books on probability and statistics use the upper case symbol, Z, to represent the
standard normal random variable, and the lower case symbol, z, to represent generic values of Z. The
distinction between the name of the variable and values of the variable will usually be clear enough in this
course that we don't have to complicate our notation to that extent.)
Notice that the standard normal random variable has properties analogous to those listed above for
normally-distributed random variables in
general:
(i.)
the probability density curve is
defined for all values of z
between - and +, though in
most areas of application, it is
considered to have values
which are zero for practical
purposes for values of z much
greater than +3 or much less
than -3 (this is important). The
area under the entire curve is exactly 1.
the vertical line z = 0 partitions the curve into two mirror-image halves, each having an area of
exactly 0.5.
z
0
(ii.)
David W. Sabo (1999)
The Normal Distribution
Page 3 of 10
Illustration:
Formula (ND-5) is so fundamental to nearly everything that follows in this course that it is absolutely
essential you understand its meaning. As an initial example, suppose we define the random variable:
x = mass in grams of an apple of a particular variety
and suppose we have independent information that
x is approximately normally distributed. (We say "approximately" because the normal distribution is
a mathematical idealization of a common occurrence in nature. However, real populations of
things in nature probably never exactly match this mathematical idealization, and even if they
did by chance, we don't really have any way of proving that something matches a
mathematical ideal exactly. In this course, we will often insert the word "approximately" as
done here to remind ourselves of this limitation of applied mathematics. However, we will
perform calculations using the exact normal probability distribution formulas.)
x has a mean value, , of 250 g
x has a standard deviation, , of 25 g.
Then, formula (ND-5) says:
275  250 
 225  250
Pr( 225  x  275 )  Pr 
z
  Pr( 1  z  1)
25
25


That is, the area under the probability density curve for x between x = 225 g and x = 275 g is the same as
the area under the standard normal probability density curve between z = -1 and z = +1. You can see this
from actual graphs if we adjust the horizontal scales appropriately.
x
200
250
300
z
-2
0
2

Simplification #2: Using the Properties of the Standard Normal Distribution
It still looks like we need a table that gives probabilities of the form
Pr(a  z  b)
for all values of 'a' between - and + and all values of 'b' between - and +. However, we can do three
things to reduce the number of entries actually necessary in a useful table.
First, as already noted, the values of the probability density function are negligible for z less than about -3 or
-4, and for z greater than about 3 or 4. Most reference books fix these two cutoffs at -3 and +3, respectively.
So you can see why this is adequate, the table provided in this course goes from z = -4.1 to z = +4.1. This
restriction gets rid of part of the infinity of rows and columns it looked like we needed.
Page 4 of 10
The Normal Distribution
David W. Sabo (1999)
Secondly, we can make use of the symmetry of the distribution to be able to express any probability,
Pr(a  z  b) in terms of probabilities of the form Pr(0  z  b). This procedure will be illustrated by several
examples below. Notice that this gets rid of one of the two constants that can take on different values in the
problem. In effect, this means our table only has to handle one value of 'a', namely a = 0. That means that
we can use the entire sheet to display probabilities corresponding to different values of 'b'.
Finally, in practical terms, we don't really need Pr(0  z  b) for every possible value of b between 0 and say
4. As far as working with tables is concerned, most people are comfortable with being able to get
probabilities for values of b rounded to two decimal places. Thus, in a 10-column wide table, we need only
30 or 40 rows  we're down to the one page table promised earlier.
The resulting table is shown on the next page. It is known as a Standard Normal Probability Table or a
z-table. Although there are other ways of preparing tables of standard normal probabilities, their principles
of organization are similar enough the table given here that if you master the use of the table we present,
you should be able to adapt to other variations with little difficulty.
Determination of Standard Normal Probabilities
We will illustrate the use of the z-table with a number of examples. The overall strategy in each case is as
follows:
(i.)
make a rough sketch of the probability/area to be calculated. This is useful in organizing
your thoughts.
(ii.)
express the required area in terms of sums or differences of areas of regions of the form
0zb
(iii.)
look up the areas of the components in the z-table, and compute the required overall area.
The body of the table contains the values of Pr(0  z  b). The first two figures of b are listed down the left
edge of the table labeling the rows. The second decimal place of b is given across the top of the table,
labeling the columns.
Example 1: Find Pr(0  z  1.57).
Solution: This probability has the standard form,
Pr(0  z  b), with b = 1.57, so we can read its value straight
from the table with no further computation. Locate the row
labeled 1.50 on the left, and move across the row to the
column labeled 0.07 at the top. The table entry there is
0.4418. Thus, we conclude that
z
0
1.57
Pr(0  z  1.57)  0.4418.

Example 2: Find Pr(z  2.43).
The region of interest here is shown in the sketch to the right.
It is the area under the probability density curve starting at
z = 2.43 on the right, and extending leftwards to -. You see
that this overall region can be regarded as the combination of
two parts:
z
0
2.43
-the part to the left of z = 0, which is half the bell, and so has an area of 0.5
-the part between z = 0 and z = 2.43, which is Pr(0  z  2.43), of the form available from the table.
Consulting the row labeled 2.40 and the column labeled 0.03 in the standard normal probability table, we
find that Pr(0  z  2.43)  0.4925. Thus, as a final result, we get
Pr(z  2.43).= 0.5 + Pr(0  z  2.43)  0.5 + 0.4925 = 0.9925.
David W. Sabo (1999)
The Normal Distribution

Page 5 of 10
(MATH 2441/99)
Table of Standard Normal Distribution Probabilities (0 < z < b):
b
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
2.10
2.20
2.30
2.40
2.50
2.60
2.70
2.80
2.90
3.00
3.10
3.20
3.30
3.40
3.50
3.60
3.70
3.80
3.90
4.00
4.10
0.00
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.2881
0.3159
0.3413
0.3643
0.3849
0.4032
0.4192
0.4332
0.4452
0.4554
0.4641
0.4713
0.4772
0.4821
0.4861
0.4893
0.4918
0.4938
0.4953
0.4965
0.4974
0.4981
0.4987
0.4990
0.4993
0.4995
0.49966
0.49977
0.49984
0.49989
0.49993
0.49995
0.49997
0.49998
0.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.3665
0.3869
0.4049
0.4207
0.4345
0.4463
0.4564
0.4649
0.4719
0.4778
0.4826
0.4864
0.4896
0.4920
0.4940
0.4955
0.4966
0.4975
0.4982
0.4987
0.4991
0.4993
0.4995
0.49968
0.49978
0.49985
0.49990
0.49993
0.49995
0.49997
0.49998
0.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.3686
0.3888
0.4066
0.4222
0.4357
0.4474
0.4573
0.4656
0.4726
0.4783
0.4830
0.4868
0.4898
0.4922
0.4941
0.4956
0.4967
0.4976
0.4982
0.4987
0.4991
0.4994
0.4995
0.49969
0.49978
0.49985
0.49990
0.49993
0.49996
0.49997
0.49998
0.03
0.0120
0.0517
0.0910
0.1293
0.1664
0.2019
0.2357
0.2673
0.2967
0.3238
0.3485
0.3708
0.3907
0.4082
0.4236
0.4370
0.4484
0.4582
0.4664
0.4732
0.4788
0.4834
0.4871
0.4901
0.4925
0.4943
0.4957
0.4968
0.4977
0.4983
0.4988
0.4991
0.4994
0.4996
0.49970
0.49979
0.49986
0.49990
0.49994
0.49996
0.49997
0.49998
0.04
0.0160
0.0557
0.0948
0.1331
0.1700
0.2054
0.2389
0.2704
0.2995
0.3264
0.3508
0.3729
0.3925
0.4099
0.4251
0.4382
0.4495
0.4591
0.4671
0.4738
0.4793
0.4838
0.4875
0.4904
0.4927
0.4945
0.4959
0.4969
0.4977
0.4984
0.4988
0.4992
0.4994
0.4996
0.49971
0.49980
0.49986
0.49991
0.49994
0.49996
0.49997
0.49998
0.05
0.0199
0.0596
0.0987
0.1368
0.1736
0.2088
0.2422
0.2734
0.3023
0.3289
0.3531
0.3749
0.3944
0.4115
0.4265
0.4394
0.4505
0.4599
0.4678
0.4744
0.4798
0.4842
0.4878
0.4906
0.4929
0.4946
0.4960
0.4970
0.4978
0.4984
0.4989
0.4992
0.4994
0.4996
0.49972
0.49981
0.49987
0.49991
0.49994
0.49996
0.49997
0.49998
0.06
0.0239
0.0636
0.1026
0.1406
0.1772
0.2123
0.2454
0.2764
0.3051
0.3315
0.3554
0.3770
0.3962
0.4131
0.4279
0.4406
0.4515
0.4608
0.4686
0.4750
0.4803
0.4846
0.4881
0.4909
0.4931
0.4948
0.4961
0.4971
0.4979
0.4985
0.4989
0.4992
0.4994
0.4996
0.49973
0.49981
0.49987
0.49992
0.49994
0.49996
0.49998
0.49998
0.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.3078
0.3340
0.3577
0.3790
0.3980
0.4147
0.4292
0.4418
0.4525
0.4616
0.4693
0.4756
0.4808
0.4850
0.4884
0.4911
0.4932
0.4949
0.4962
0.4972
0.4979
0.4985
0.4989
0.4992
0.4995
0.4996
0.49974
0.49982
0.49988
0.49992
0.49995
0.49996
0.49998
0.49998
0.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2517
0.2823
0.3106
0.3365
0.3599
0.3810
0.3997
0.4162
0.4306
0.4429
0.4535
0.4625
0.4699
0.4761
0.4812
0.4854
0.4887
0.4913
0.4934
0.4951
0.4963
0.4973
0.4980
0.4986
0.4990
0.4993
0.4995
0.4996
0.49975
0.49983
0.49988
0.49992
0.49995
0.49997
0.49998
0.49999
0.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
0.3133
0.3389
0.3621
0.3830
0.4015
0.4177
0.4319
0.4441
0.4545
0.4633
0.4706
0.4767
0.4817
0.4857
0.4890
0.4916
0.4936
0.4952
0.4964
0.4974
0.4981
0.4986
0.4990
0.4993
0.4995
0.4997
0.49976
0.49983
0.49989
0.49992
0.49995
0.49997
0.49998
0.49999
Example 3: Find Pr(z  -1.75)
Solution:
From the sketch to the right, we see that what is required is
the area under the probability density curve from z = -1.75
rightwards to z = +. This area can be split into two pieces:
z
-1.75
0
-the region from z = 0 to z = +, which is half the
bell, and so represents an area of 0.5 units.
-the region from z = -1.75 to z = 0, which has area Pr(-1.75  z  0).
Page 6 of 10
The Normal Distribution
David W. Sabo (1999)
Now, the standard normal probability table does not give probabilities of this sort directly. However, since
the distribution is symmetric about the vertical line z = 0, we know that
Pr(-1.75  z  0) = Pr(0 z  1.75)
But, the probability on the right-hand side here is in the standard form of a table entry. From the row labeled
1.70 and the column labeled 0.05, we get that Pr(0 z  1.75)  0.4599. Thus, we conclude
Pr(z  -1.75) = 0.5 + Pr(0 z  1.75)  0.5 + 0.4599 = 0.9599.

Example 4: Find Pr(-0.87  z  1.26)
Solution:
The sketch to the right shows the area that is required. It can
be viewed as made up of two parts:
z
-the region between z = 0 and z = 1.26, which has an
area equal to Pr(0  z  1.26)  0.3962 from the table
-0.87 0
1.26
-the region between z = -0.87 and z = 0. By symmetry, the area of this region is the same as the
area of the region between z = 0 and z = +0.87. But this is just Pr(0  z  0.87)  0.3078 from the
table.
Thus, the required final answer is
Pr(-0.87  z  1.26) = Pr(-0.87  z  0) + Pr(0  z  1.26)
= Pr(0  z  0.87) + Pr(0  z  1.26)
 0.3078 + 0.3962
= 0.7040

Example 5: Find Pr(0.42  z  1.31)
Solution
The region of interest in this case is entirely to the right side
of the line of symmetry at z = 0. You can see from the
sketch, however, that it is possible to write this probability as
the difference of two areas:
z
0
Pr(0.42  z  1.31) = Pr(0  z  1.31) - Pr(0  z  0.42)
 0.4049 - 0.1628
= 0.2421

Example 6: Find Pr(z  1.27).
Solution
From the sketch, we see that
Pr(0  z  1.27) + Pr(z  1.27) = 0.5
z
because together these two terms include the right half of the
bell. Thus,
0 1.27
Pr(z  1.27) = 0.5 - Pr(0  z  1.27)  0.5 - 0.3980 = 0.1020.

David W. Sabo (1999)
The Normal Distribution
Page 7 of 10
These last six examples cover pretty well all of the variations possible in calculating standard normal
probabilities.
Again, you may encounter standard normal probability tables organized in different ways than the table
given in this document. In very general terms, the strategies for using them parallel the strategies illustrated
above.
Having mastered the task of determining standard normal probabilities, it is now straightforward to use
formula (ND-5) to compute probabilities for any other normal distribution. We illustrate this with a few short
examples.
Example 7: Suppose that the amount of vegetable soup dispensed by a machine into containers at a food
processing plant is an approximately normally-distributed random variable with a mean of 356 g and a
standard deviation of 23 g. Determine the probability that a container selected at random will contain
between 350 and 375 g of soup.
Solution
If we define
x = grams of soup dispensed by the machine into a container
then, according to this problem, x is an approximately normally-distributed random variable with a mean, ,
of 356 g, and a standard deviation, , of 23 g. The question is then asking for the probability:
Pr(350  x  375)
Using formula (ND-5), we first get:
375  356 
 350  356
Pr( 350  x  375 )  Pr 
z

23
23


= Pr(-0.26  z  0.83)
= Pr(0  z  0.26) + Pr(0  z  0.83)
 0.1026 + 0.2967 = 0.3993
z
Thus, the probability that a randomly selected container will
contain between 350 g and 375 g of soup is 0.3993. This
means as well that approximately 39.93% of the containers have between 350 g and 375 g of soup in them.

Example 8: Suppose that the label on these soup containers states that they contain 340 g of soup. What
percentage of the containers will actually contain less soup than stated on the label?
Solution:
The percentage of containers with less than 340 g of soup is
the same thing as the probability that a randomly selected
can of soup will contain less than 340 g of soup. Thus,
defining the random variable x in the same way as was done
in Example 7, we see that the question is really asking us to
determine
-0.70
Pr( x < 340)
But
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The Normal Distribution
David W. Sabo (1999)
340  356 

Pr( x  340 )  Pr  z 
  Pr( z   0.70 )
23


From the sketch, you can see that
Pr(z < -0.70) = 0.5 - Pr(0  z  0.70)
 0.5 - 0.2580
= 0.2420
Thus, since the probability of a randomly selected container containing less than 340 g of soup is 0.2420, we
conclude that 24.20% of the containers contain less than 340 g of soup.

Example 9: A technologist has a procedure for preparing culture tubes for screening tests. The
specifications require that the amount of nutrient solution be between 9.75 ml and 10.25 ml. Studies
indicate that the procedure results in a somewhat random amount of nutrient solution placed in each tube
which is approximately normally distributed with a mean of 10.05 ml and a standard deviation of 0.18 ml.
What percentage of these culture tubes do not conform to the nutrient solution amount specifications?
Solution
This question is asking for the percentage of culture tubes that either contain less than 9.75 ml of nutrient
solution or more than 10.25 ml of nutrient solution. Again, percentages of culture tubes containing a certain
amount of solution are equivalent to the probability of finding that the amount of nutrient solution in a
randomly selected culture tube, so this is really a probability problem.
To start, define
x = the ml of nutrient solution in a randomly selected culture tube
Then, according to the problem, x is an approximately normally distributed random variable with a mean of
10.05 ml and a standard deviation of 0.18 ml. We can compute:
9.75  10 .05 

Pr( x  9.75 )  Pr  z 
  Pr( z   1.67 )  0.5  Pr( 0  z  1.67 )
0.18


 0.5 - 0.4525 = 0.0475
If you have trouble following these calculations, try sketching a diagram, as was done above for Examples 1
through 7. Similarly,
10 .25  10 .05 

Pr( x  10 .25 )  Pr  z 
  Pr( z  1.11)  0.5  Pr(0  z  1.11)
0.18


 0.5 - 0.3665 = 0.1335
From these two results, we see that 4.75% of the culture tubes will have less than 9.75 ml of solution, and
13.35% of the tubes will have more than 10.25 ml of solution. Thus, a total of 4.75% + 13.35% = 18.1% of
the culture tubes will be outside of the specifications relating to amount of nutrient solution.

Remark
Microsoft Excel provides the functions NORMSDIST() for calculating standard normal probabilities, and
NORMDIST() for calculating cumulative probabilities for general normal distributions. Specifically,
NORMSDIST(b)  Pr(- < z  b) = 0.5 + Pr(0 < z  b)
Thus, for a given value 'b', Excel's function NORMSDIST(b) gives a result which is 0.5 greater than the
entries in the table of standard normal probabilities included in this document. Excel's NORMSDIST()
function returns what are truly cumulative standard normal probabilities. You should be able to figure out
David W. Sabo (1999)
The Normal Distribution
Page 9 of 10
how to use values of the NORMSDIST() function to compute standard normal probabilities for any of the
variations illustrated in Examples 1 - 6 above. The reason we opted for a table based on Pr(0 < z  b) for
hand calculations here is out opinion that such a table reduces the amount of arithmetic necessary in most
instances (unless you are willing to expand the table to include values of 'b' ranging over negative as well as
positive numbers).
The NORMDIST() function in Excel is a bit more complicated, but again gives a true cumulative probability:
NORMDIST(b, , , True)  Pr(- < x  b)
where x is a normally distributed random variable with mean  and standard deviation . For a cumulative
probability such as this, the fourth parameter must always be the word 'True'. (If you specify 'False' in this
position, NORMDIST gives the value of the probability density function, f(b), rather than a cumulative
probability). Recall formula (RV-6)
Pr(a < x < b) = F(b) - F(a)
(RV-6)
from the preceding document, which indicates how to calculate other probabilities from values of cumulative
probabilities such as produced by NORMDIST().
Obviously, if you were setting up normal probability calculations in an Excel spreadsheet (or other
spreadsheet applications that have similar functionality), you would use these built-in functions to get values
of required probabilities rather than reading the probabilities from a printed table and typing them in as
numbers.
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The Normal Distribution
David W. Sabo (1999)