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Notes on the Normal Probability Distribution:
You should all have studied the normal probability distribution in your statistics course,
so the material here should merely be a review.
A random variable with a normal probability distribution is characterized with the
following probability density function:
 ( x  ) 2 
f ( x) 
exp 
,

2
2
2
 2

1
  x 
The normal r.v. is actually a family of random variables, each having the same density
function but with different mean (your textbook used the greek letter β, instead of μ) and
standard deviation (σ).
This density function looks rather menacing. The exp(.) stands for the natural
exponential, and is the inverse of the natural logarithm (ln(x)). We will not be using this
function except to look at it. The expression -∞ ≤ x ≤ +∞ gives us the range of possible
values for the normal random variable. This says that a normal random variable can take
on any value on the real number line. However, the shape of the density function tells us
that values out into the tails are very unlikely. The density function, when graphed has
the familiar bell-shape, is centered at the mean (β), (and median and mode since
mean=median=mode), and the steepness (or flatness) is determined by the standard
deviation (σ)
total area = 1.0
area = 0.5
area = 0.5

x
To find the probability of a normal random variable falling in some given interval, it is
necessary to “standardize”. This involves subtracting the mean and dividing by the
standard deviation to create a new random variable, we call Z that has a mean of zero and
a standard deviation of 1.0. (Note that this also means that the variance is 1.0 since the
variance is the square of the standard deviation and 12 = 1)
Z = (x - β)/ σ where z ~N(0,1)
Z is a standard normal r.v.
X is a normal r.v.
0

z
X
By subtracting the mean, we move the distribution from being centered over the mean β,
to one being centered over a mean of zero. By dividing by the standard deviation (σ), we
move the distribution from one with a standard deviation of σ, to one with a standard
deviation of 1.0. We can now use the standard normal tables to find area under the curve.
For example, suppose we want to find P(X >10) when X~N(β = 5, σ = 2). You should
always draw a picture of what you are being asked. P(X>10) = P(Z>(10-5)/2) = P(Z>2.5)
This area is P( 0 < Z < 2.5)
This Area is P(Z > 2.5)
0
2.5
Z
Now, to use the table in the back of the book to find the appropriate areas, you need to
know how the table is set up. DO NOT MEMORIZE whether the appropriate probability
needs to be subtracted from 1.0 or from 0.50. Instead, ALWAYS DRAW A PICTURE
and shade the area you are looking for.
For the probability P(Z > 2.5), you need to look up the value 2.50 in the standard z table
(page 389 in textbook). The Z values appear along the left most column. Use the top row
to determine the second decimal for z. Here, with Z of 2.5, the second decimal is 0 so we
use the second column, which has a .00 at the top. You should find the value 0.4938.
What does this number represent? It is the area under the normal curve from 0 out to 2.5:
that is, it measures P(0 < z < 2.5). It is NOT P(Z > 2.5). But because the total right side
has an area of 0.5, the P(Z> 2.5) = 0.5 – 0.4938 = 0.0062.
From this one number of 0.4938, we know the following:
P( Z > 2.5) = 0.0068 (this is 05 – 0.4938)
Shade the area to the right of 2.5. (I
can’t shade with this program.)
-2.5
0
2.5
Z
P(0 < Z < 2.5) = 0.4938 (this what the table gives you)
Shade the are between 0 and 2.5
-2.5
0
2.5
Z
P( Z < 2.5) = 0.9938 (this is 0.5 + 0.4938)
-2.5
0
shade all area to the left of
2.5
2.5
Z
P(-2.5 < Z < 2.5) = 0.9872 (this is 0.4938 + 0.4938)
Shade the area between –2.5
and 2.5
-2.5
0
2.5
Z
Make sure that you can complete questions 2.7, 2.9 and 2.25. Answers are posted on the
web site.