Chapter 6. Continuous Probability
Distributions
Chapter Goals
◼
Convert values from any normal distribution to a
standardized z-score
◼
Find probabilities using a normal distribution table
◼
Apply the normal distribution to business problems
◼
Recognize when to apply the uniform and
exponential distributions
Continuous Probability
Distributions
◼
A continuous random variable is a variable that
can assume any value on a defined continuum
(can assume an uncountable number of values)
– see Chapter 5
thickness of an item
◼ time required to complete a task
◼
◼
These can potentially take on any value,
depending only on the ability to measure
accurately.
Types of Continuous Distributions
◼
Three types
Normal
◼ Uniform
◼
A
B
Involves determining the probability for a RANGE
of values rather than 1 particular incident or
outcome
The Normal Distribution
◼ ‘Bell Shaped’
Symmetrical
◼ Mean=Median=Mode
◼
Location is determined by the
mean, μ
Spread is determined by the
standard deviation, σ
The random variable has an
infinite theoretical range:
+  to − 
f(x)
σ
μ
Mean
Median
Mode
x
Many Normal Distributions
By varying the parameters μ and σ, we obtain
different normal distributions
The Normal Distribution Shape
f(x)
Changing μ shifts the
distribution left or right.
σ
μ
Changing σ increases
or decreases the
spread.
x
Finding Normal Probabilities
Probability is measured by the area
under the curve
f(x)
P (a  x  b)
a
b
x
Probability as
Area Under the Curve
The total area under the curve is 1.0, and the curve is
symmetric, so half is above the mean, half is below
f(x)
P(−  x  μ) = 0.5
0.5
P(μ  x   ) = 0.5
0.5
μ
P(−  x   ) = 1.0
x
The Standard Normal Distribution
◼
◼
◼
Also known as the “z” distribution
Mean is defined to be 0
Standard Deviation is 1
f(z)
1
0
z
Values above the mean have positive z-values
Values below the mean have negative z-values
The Standard Normal
◼
Any normal distribution (with any mean and
standard deviation combination) can be
transformed into the standard normal
distribution (z)
◼
Need to transform x units into z units
◼
◼
Where x is any point of interest
Can use the z value to determine probabilities
Translation to the Standard
Normal Distribution
◼
Translate from x to the standard normal (the “z”
distribution) by subtracting the mean of x and dividing
by its standard deviation:
x −μ
z=
σ
z is the number of standard deviations units that
x is away from the population mean
Example
◼
If x is distributed normally with mean of 100
and standard deviation of 50, the z value for
x = 250 is
x − μ 250 − 100
z=
=
= 3.0
σ
50
◼
This says that x = 250 is three standard
deviations (3 increments of 50 units) above
the mean of 100.
Comparing x and z units
μ = 100
σ = 50
100
0
250
3.0
x
z
Note that the distribution is the same, only the
scale has changed. We can express the problem in
original units (x) or in standardized units (z)
The Standard Normal Table
The Standard Normal table in the textbook
(Appendix D)
◼ Gives the probability from the mean (zero)
up to a desired value for z
◼
0.4772
Example:
P(0 < z < 2.00) = 0.4772
0
2.00
z
The Standard Normal Table
(continued)
◼
The Standard Normal Table gives the
probability between the mean and a certain z
value
◼
The z value ALWAYS refers to the area
between some value (-z or +z) and the mean
◼
Since the distribution is symmetrical, the
Standard Normal Table only displays
probabilities for ½ of the full distribution
The Standard Normal Table
(continued)
The column gives the value of
z to the second decimal point
z
The row shows
the value of z
to the first
decimal point
0.00
0.01
0.02
…
0.1
0.2
.
.
.
2.0
.4772
P(0 < z < 2.00)2.0
= 0.4772
The value within the
table gives the
probability from z = 0
up to the desired z
value
General Procedure for
Finding Probabilities
1. Determine m and s
2. Define the event of interest
e.g., P(x > x1)
3. Convert to standard normal
x −μ
z=
σ
4. Use the table to find the probability
z Table Example
◼
Suppose x is normal with mean 8.0 and
standard deviation 5.0. Find P(8 < x < 8.6)
Calculate z-values:
x −μ 8 −8
z=
=
=0
σ
5
x − μ 8.6 − 8
z=
=
= 0.12
σ
5
8 8.6
x
0 0.12
Z
P(8 < x < 8.6)
= P(0 < z < 0.12)
z Table Example
(continued)
◼
Suppose x is normal with mean 8.0 and
standard deviation 5.0. Find P(8 < x < 8.6)
m=8
s=5
8 8.6
P(8 < x < 8.6)
m=0
s=1
x
0 0.12
P(0 < z < 0.12)
z
Solution: Finding P(0 < z < 0.12)
Standard Normal Probability
Table (Portion)
z
.00
.01
P(8 < x < 8.6)
= P(0 < z < 0.12)
.02
0.0478
0.0 .0000 .0040 .0080
0.1 .0398 .0438 .0478
0.2 .0793 .0832 .0871
Z
0.3 .1179 .1217 .1255
0.00
0.12
Finding Normal Probabilities
Suppose x is normal with mean 8.0
and standard deviation 5.0.
◼ Now Find P(x < 8.6)
◼
◼
The probability of obtaining a value less than 8.6
P = 0.5
Z
8.0
8.6
Finding Normal Probabilities
(continued)
Suppose x is normal with mean 8.0
and standard deviation 5.0.
◼ Now Find P(x < 8.6)
◼
P(x < 8.6)
0.0478
0.5000
= P(z < 0.12)
= P(z < 0) + P(0 < z < 0.12)
= 0.5000 + 0.0478 = 0.5478
Z
0.00
0.12
Upper Tail Probabilities
Suppose x is normal with mean 8.0
and standard deviation 5.0.
◼ Now Find P(x > 8.6)
◼
Z
8.0
8.6
Upper Tail Probabilities
(continued)
◼
Now Find P(x > 8.6)…
P(x > 8.6) = P(z > 0.12) = P(z > 0) - P(0 < z < 0.12)
= 0.5000 - 0.0478 = 0.4522
0.5000
Z
0
0.12
0.0478
0.4522
Z
0
0.12
Lower Tail Probabilities
Suppose x is normal with mean 8.0
and standard deviation 5.0.
◼ Now Find P(7.4 < x < 8)
◼
Z
8.0
7.4
Lower Tail Probabilities
(continued)
Now Find P(7.4 < x < 8)…the probability
between 7.4 and the mean of 8
The Normal distribution is
symmetric, so we use the
same table even if z-values
are negative:
0.0478
P(7.4 < x < 8)
= P(-0.12 < z < 0)
Z
= 0.0478
8.0
7.4
Empirical Rules
What can we say about the distribution of values
around the mean if the distribution is normal?
f(x)
μ ± 1σ covers about
68% of x’s
σ
σ
Recall
Tchebyshev
from Chpt. 3
μ−1σ
μ
μ+1σ
68.26%
x
The Empirical Rule
(continued)
μ ± 2σ covers about 95% of x’s
μ ± 3σ covers about 99.7% of x’s
2σ
3σ
2σ
μ
95.44%
x
3σ
μ
99.72%
x
Importance of the Rule
◼
If a value is about 2 or more standard
deviations away from the mean in a normal
distribution, then it is far from the mean
◼
The chance that a value that far or farther
away from the mean is highly unlikely, given
that particular mean and standard deviation
The Uniform Distribution
The uniform distribution is a probability distribution
that has equal probabilities for all possible
outcomes of the random variable
◼
Referred to as the distribution of “little information”
◼
Probability is the same for ANY interval of the same width
◼
Useful when you have limited information about how the
data “behaves” (e.g., is it skewed left?)
The Uniform Distribution
(continued)
The Continuous Uniform Distribution:
f(x) =
1
b−a
if a  x  b
0
otherwise
where
f(x) = value of the density function at any x value
a = lower limit of the interval of interest
b = upper limit of the interval of interest
The Mean and Standard Deviation
for the Uniform Distribution
The mean (expected value) is:
a+b
E(x) = μ =
2
The standard deviation is
(b − a)2
σ=
12
where
a = lower limit of the interval from a to b
b = upper limit of the interval from a to b
-33
Steps for Using the
Uniform Distribution
1. Define the density function
2. Define the event of interest
3. Calculate the required probability
f(x)
x
Uniform Distribution
Example: Uniform Probability Distribution
Over the range 2 ≤ x ≤ 6:
1
f(x) = 6 - 2 = .25 for 2 ≤ x ≤ 6
f(x)
.25
2
6
x
Uniform Distribution
Example: Uniform Probability Distribution
Over the range 2 ≤ x ≤ 6:
2+6
E(x) = μ =
=4
2
(b − a)
(6 − 2)
σ=
=
= 1.1547
12
12
2
2
Chapter Summary
◼
Reviewed key continuous distributions
normal
◼ uniform
◼
◼
Found probabilities using formulas and tables
◼
Recognized when to apply different distributions
◼
Applied distributions to decision problems