Transcript Business Statistics: A Decision

Business Statistics:
A Decision-Making Approach
8th Edition
Chapter 6
Introduction to Continuous
Probability Distributions
6-1
Chapter Goals
After completing this chapter, you should be
able to:

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
6-2
Continuous Probability Distributions

A continuous probability distribution differs
from a discrete probability distribution in
several ways.



A continuous random variable can take on an
infinite number of values. That is, the exact
probability for a specific variable is always zero.
As a result, a continuous probability distribution
cannot be expressed in tabular form.
Instead, an equation is used to describe a
continuous probability distribution.
6-3
Probability Density Function

The equation for describing a continuous probability
distribution is called a probability density function
and noted as “f(x).”



Sometimes refer to as Density function
pdf or PDF
1
 ( x   ) 2 / 2 2
f ( x) 
2
e
Excellent explanation about “probability density function” on the
Khan academy website.
6-4
Probability Density Function Example

The shaded area in the graph
represents the probability that the
random variable X is less than or
equal to a.



Cumulative probability
Always to the left
The probability that the random
variable X is exactly equal to a
would be zero.
6-5
Types of Continuous Distributions

Three types



Normal
Uniform
Exponential
A
B
Involves determining the probability for a RANGE
of values rather than 1 particular incident or
outcome
6-6
The Normal Distribution
‘Bell Shaped’
 Symmetrical
 Mean=Median=Mode

Location is determined by the
mean, μ
Spread is determined by the
standard deviation, σ
f(x)
σ
x
μ
Mean
Median
Mode
6-7
Position and Shape of the Normal Curve
6-8
Many Normal Distributions
By varying the parameters μ and σ, we obtain
different normal distributions
6-9
Finding Normal Probabilities
Probability is measured by the area
under the curve
f(x)
P (a  x  b)
a
b
x
6-10
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
x
P (    x   )  1 .0
6-11
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
6-12
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
6-13
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
6-14
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.
6-15
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)
6-16
Why standardizing?


To measure variable with different means and/or
standard deviations on a single standardized scale.
Example: Statistically, are below examples really or
fundamentally different matters?


SAT mean 1020 with standard deviation of 160. You scored
1260. What is your percentile?
Average adult male height is 70 inches with standard deviation
of 2 inches. You are 73 inches tall. What is your height
percentile?
6-17
Z-scores for Two Problems
x  μ 1260  1020
z

 1.5
σ
160
x μ
73  70
z

 1.5
σ
2
Both are 1.5 Std Dev above the mean….
Solve this problem using Excel:
•Download the Normal Dist. using Excel file from the class
website and then select the first tab “Example 1”….
6-18
The Standard Normal Table



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
0.4772
Example:
P(0 < z < 2.00) = 0.4772
0
2.00
z
6-19
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
6-20
General Procedure for
Finding Probabilities
1. Determine  and 
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
6-21
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)
6-22
z Table Example
(continued)

Suppose x is normal with mean 8.0 and
standard deviation 5.0. Find P(8 < x < 8.6)
=8
=5
8 8.6
P(8 < x < 8.6)
=0
=1
x
0 0.12
z
P(0 < z < 0.12)
6-23
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
6-24
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
6-25
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
6-26
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
6-27
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
6-28
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
6-29
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
6-30
Excel Functions

NORMDIST(x, mean, standard_dev, cumulative)

NORMSDIST(z)

NORMINV(probability, mean, standard_dev)

NORMSINV(probability)

STANDARDIZE(x, mean, standard_dev)
6-31
Formula Table
We have
X
We want
X
Z
p
z
Z
p
X=μ+zσ
Norminv(p,μ,σ)
x

Normsinv(p)
z  standardiz e( x,  ,  )
Normdist(x,μ,σ,1)
Normsdist(z)
6-32
Excel Functions

NORMDIST(x, mean, standard_dev, cumulative)

Find the probability that a number falls at or below a given value
of a normal distribution




x -- The value you want to test.
mean -- The average value of the distribution.
standard_dev -- The standard deviation of the distribution.
cumulative -- If FALSE or zero, returns the probability that x will
occur; if TRUE or non-zero, returns the probability that the value
will be less than or equal to x.
6-33
Excel Functions


The distribution of heights of American women aged 18
to 24 is normally distributed with a mean of 65.5 inches
(166.37 cm) and a standard deviation of 2.5 inches (6.35
cm).
What percentage of these women is taller than 68
inches (172.72 cm)?


NORMDIST(68, 65.5, 2.5, TRUE) = 84.13%: cumulative
probability (to the left on the normal curve)
Thus, 1- 84.13% = 15.87%
6-34
Excel Functions

NORMSDIST(z)
 Translate the number of standard deviations (z) into
cumulative probabilities
 z -- The value for which you want the distribution.

NORMSDIST(1) = 84.13%
6-35
Excel Functions

NORMINV(probability, mean, standard_dev)
Calculate the x variable given a probability
How tall would a woman need to be if she wanted to be
among the tallest 75% of American women (find 25%
shorter instead as the figure shows)?



NORMINV(0.25, 65.5, 2.5) = 63.81 inches
6-36
Excel Functions

NORMSINV(probability)
Given the probability that a variable is within a certain distance
of the mean, it finds the z value
Find the z values that mark the boundary that is 25% less than the
mean and 25% (75%) more than the mean.




NORMSINV(0.25) = - 0.674
NORMSINV(0.75) = 0.674
6-37
“STANDARDIZE” Excel Function

STANDARDIZE(x, mean, standard_dev)

Find the z value for a specified value, mean, and standard
deviation.
From the previous example, we found that a woman would need to
be at least 63.81 inches tall to avoid the bottom 25% of the
population, by height. The STANDARDIZE function tells us that the
z value for 63.81 inches is:
 STANDARDIZE(63.81, 65.5, 2.5) = -0.6745
We can check this number by using the NORM.S.DIST function:
 NORMSDIST(-0.6745) = 25%
 That is, a z value of -.6745 has a probability of 25%.


6-38
Normal Probabilities in PHStat

We can use Excel and PHStat to quickly
generate probabilities for any normal
distribution

We will find P(8 < x < 8.6) when x is
normally distributed with mean 8 and
standard deviation 5
6-39
PHStat Dialogue Box
Select desired options
and enter values
6-40
PHStat Output
6-41
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
x
68.26%
6-42
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
x
99.72%
6-43
Practice

Exercise





6-2
6-5
6-8
6-14
6-18
6-44
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?)
6-45
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
6-46
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
6-47
Example

Exercise 6-38





If buses stop every 20 minutes then the time you will
have to wait can be described by a uniform
distribution with a = 0 and b = 20. To find the
probability that you will have to wait for 10 minutes or
more
f(x) = 1/20-0 = 0.05
P(x > 10) = 1-P(x < 10) = 1 - 0.05(10-0) = 0.50.
P(x < 6) = 0.05(6-0) = 0.30
P (8 < x < 15) = 0.05(15-8) = 0.35.
Try exercise 6-40
6-48
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
6-49
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
6-50
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
6-51
The Exponential Distribution

Used to measure the time that elapses
between two occurrences of an event (the
time between arrivals)


Examples:
 Time between trucks arriving at a dock
 Time between transactions at an ATM Machine
 Time between phone calls to the main operator
Recall l = mean for Poisson (see Chpt. 5)
6-52
The Exponential Distribution

The probability that an arrival time is equal to or
less than some specified time a is
P(0  x  a)  1  e
 λa
where 1/l is the mean time between events and e = 2.7183
NOTE: If the number of occurrences per time period is Poisson
with mean l, then the time between occurrences is exponential
with mean time 1/ l and the standard deviation also is 1/l
6-53
Exponential Distribution
(continued)

Shape of the exponential distribution
f(x)
l = 3.0
(mean = .333)
l = 1.0
(mean = 1.0)
l= 0.5
(mean = 2.0)
x
6-54
Example
Example: Customers arrive at the claims counter at
the rate of 15 per hour (Poisson distributed). What
is the probability that the arrival time between
consecutive customers is less than five minutes?
 Time between arrivals is exponentially distributed
with mean time between arrivals of 4 minutes (15
per 60 minutes, on average)

1/l = 4.0, so l = .25

P(x < 5) = 1 - e-la = 1 – e-(.25)(5) = 0.7135
There is a 71.35% chance that the arrival time between
consecutive customers is less than 5 minutes
6-55
Using PHStat
6-56
Chapter Summary

Reviewed key continuous distributions



normal
uniform
exponential

Found probabilities using formulas and tables

Recognized when to apply different distributions

Applied distributions to decision problems
6-57
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recording, or otherwise, without the prior written permission of the publisher.
Printed in the United States of America.
6-58
Example 1

An average light bulb manufactured by the
Acme Corporation lasts 300 days with a
standard deviation of 50 days. Assuming that
bulb life is normally distributed, what is the
probability that an Acme light bulb will last at
most 365 days?
6-59
Example 2

Suppose scores on an IQ test are normally
distributed. If the test has a mean of 100 and a
standard deviation of 10, what is the probability
that a person who takes the test will score
between 90 and 110?
6-60
Example 3

Molly earned a score of 940 on a national
achievement test. The mean test score was 850
with a standard deviation of 100. What
proportion of students had a higher score than
Molly? (Assume that test scores are normally
distributed.)
6-61