Unit 5 Chapter
Normal Distribution
Chapter 5 deals with the Normal Distribution which is a continuous
distribution. This is probably the most important probability
distribution.
Graphs of Normal Probability Distribution
Properties of a normal curve
1. The curve is bell-shaped
with highest point over
the mean µ.
2. The curve is
symmetrical about a
vertical line through µ.
3. The curve approaches
the horizontal axis but
never touches or
crosses it.
4. The inflection
(transition) points
between cupping up
and down occur at µ - σ
and µ + σ.
5. The total area under the
curve is 1.
The mean µ and standard deviation
σ control the shape of the normal
curve. The curves on the right show
examples of normal curves of
different height.
Empirical Rule. For a distribution that is symmetrical and bell-shaped
1. Approximately 68% of the data values will lie within one
standard deviation on each side of the mean.
2. Approximately 95% of the data values will lie within two
standard deviation on each side of the mean.
3. Approximately 99.7% of the data values will lie within three
standard deviation on each side of the mean.
Examples showing the empirical rule:
1. The mean value of equipment from an inventory sample is µ =
$1500 with a standard deviation σ = $200. The data set is a
normal distribution. Estimate the percent of the equipment
whose values are between $1300 and $1700.
The interval is 1500 – 200 and 1500 + 200 which translates to
µ - 1σ and µ + 1σ. Using the empirical rule, 68% of the
equipment falls in this interval.
2.
Using µ = $1500 with a standard deviation σ = $200, between
what 2 values does 95% of the values fall. From the rule this
would be µ - 2σ and µ + 2σ. The calculations would be 1500 –
2(200) and 1500 + 2(200) or $1100 and $1900.
Standard units and areas under the standard normal distribution.
Since there are infinitely many normal distribution each with its own mean
and standard deviation, a standard normal distribution with mean µ = 0
and standard deviation σ = 1 can be used to transform all the other
distributions into a comparable format. The standard normal distributions use
the variable z to represent the random variable. The other normal
distributions may be denoted as x.
Values on any normal distribution x can be transformed using the formula
z
x

. This is usually referred to as a z-score or z value. Percents
(areas under the curve) can be derived using the standard normal. Before
working with the conversions for the x normal distribution, you must be able
to work with the standard normal distribution table (Table in the appendix or
the pull-out card) The table shows negative values which are to the left of
the mean of 0. The other side of the table shows positive values which are to
the right of the mean of 0.
Partial negative table
Partial positive table
Examples using the normal table. (Remember the mean = 0 and
cuts the graph in 2 equal sides. The area under the curve is 1 or
100%. Either side of the mean contains 50% or .5 of the area or
probability. Area and probability are interchangeable for terminology.)
Find the area as specified.
Between -1.6 and 0
Calculations:
z = -1.6 gives
.0548
z=0 gives
.5000
To the left of 1.25 (-5 is the same as ∞ for the normal)
.5000 - .0548
=
.4452 or
44.52% of the
data falls
between -1.6
and 0
Calculations:
z = 1.25 give
.8944 = 89.4%
of the data falls
below 1.25
Between -1.96 and 1.96
Calculations:
z = -1.96 gives
.025
z = 1.96 gives
.975
.975 - .025 =
.95 or 95% of
the data falls
between -1.96
and 1.96
From mean 0 to 1.52
Calculations:
z = 1.52
gives .9357
which
includes the
area to the
left of 0. So
that half of
the area
need to be
subtracted
.9357 - .5000
=
.4357 or
43.57%
Calculations:
z = 1.52
gives .9357
z = -1.23
gives .1093
.9357 - .1093
= .8264 0r
82.64%
Calculations:
z = 2.3 gives
.9893
z = 1.2 gives
.8849
.9893 - .8849
= .1044 or
10.44%
Use the standard normal distribution and table to calculate probabilities from
an x normal distribution.
The fall deer population in Mesa Verda National Park is approximately
normally distributed with mean µ = 4400 deer and standard deviation σ =
620 deer. Covert each of the following x intervals to z intervals and find the
associated probability.
a) x > 3300
z
3300  4400
 1.77 (round to 2 decimals since
620
the normal table only gives 2 decimals. So z > -1.77 which
would give a probability 1 - .0384 = .9616 or 96.16% of the
dear population is above 3300.
b) x < 5400
z
5400  4400
 1.61 So z < 1.61 which would give
620
a probability of .9463 or 94.63 % of the populations are below
5400.
c) 3500 < x < 5300 Both endpoints need to be converted.
z
3500  4400
5300  4400
 1.45 and z 
 1.45 .
620
620
So -1.45 < z < 1.45 which would convert to .9265 - .0735 =
.853 or 85.3% of the deer populations are between 3500 and
5300.
Areas under the Normal Curve
Examples:
1. A survey of the height of US men was conducted. Heights are normally
distributed with a mean µ = 69.6 and a standard deviation σ = 3.0. A
study participant is selected randomly.
a) Find the probability that his height is less than 66 inches.
P(x < 6) = P ( z 
66  69.6
) = P(z < -1.2) = .1151
3
b) Find the probability that his height is between 66 and 72 inches.
72  69.6 
 66  69.6
z
=
3
3


P( 66 < x < 72) = P
P( -1.2 < z < .8) = .7881 - .1151= .673
c) Find the probability that his height is more than 72 inches.


P(x > 72) = P z 
72  69.6 
 = P(z > .8) = 1 - .7881 = .2119
3

2. The weight of adult cocker spaniel are normally distributed wit a mean
µ = 25 lb and a standard deviations σ = 3 lb. Select a cocker
randomly.
a) Find the probability that the cocker’s weight is less than 23 lb.


P( x < 23) = P z 
23  25 
 = P(z < -.67) = .2514
2 
b) Find the probability that the weight is between 20 lb and 27 lb.
27  25 
 20  25
z
 =
3 
 3
P( 20 < x < 27) = P
P( -1.67 < z <.67) = .7486 - .0475 = .7011
c) Find the probability that the weight is more than 29 lb.


P( x > 29) = P z 
29  25 
 = P( z > 1.33)
3 
= 1 - .9082 = .0918
d) If 50 cockers are randomly selected, about how many would
you expect to weigh less than 22 lb.
Find the percent first and then applies this to the number of
cockers in the sample.


P( x < 22) = P z 
22  25 
 = P( z < -1) = .1587
3 
50 (.1587) = 7.933 ≈ 8 cockers