Download Document

Statistics for
Business and Economics
Chapter 5
Continuous Random Variables
and Probability Distributions
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-1
Probability Distributions
Probability
Distributions
Ch. 4
Discrete
Probability
Distributions
Continuous
Probability
Distributions
Binomial
Uniform
Ch. 5
Normal
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-2
5.1
Continuous Probability Distributions
n 
A continuous random variable is a variable that
can assume any value in an interval
n 
n 
n 
n 
n 
thickness of an item
time required to complete a task
temperature of a solution
height, in inches
These can potentially take on any value,
depending only on the ability to measure
accurately.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-3
Cumulative Distribution Function
n 
The cumulative distribution function, F(x), for a
continuous random variable X expresses the
probability that X does not exceed the value of x
F(x) = P(X ≤ x)
n 
Let a and b be two possible values of X, with
a < b. The probability that X lies between a
and b is
P(a < X < b) = F(b) − F(a)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-4
Probability Density Function
The probability density function, f(x), of random variable X
has the following properties:
1.  f(x) > 0 for all values of x
2.  The area under the probability density function f(x)
over all values of the random variable X is equal to
1.0
3.  The probability that X lies between two values is the
area under the density function graph between the two
values
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-5
Probability Density Function
(continued)
The probability density function, f(x), of random variable X
has the following properties:
4.  The cumulative density function F(x0) is the area under
the probability density function f(x) from the minimum
x value up to x0
x0
f(x 0 ) = ∫ f(x)dx
xm
where xm is the minimum value of the random variable x
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-6
Probability as an Area
Shaded area under the curve is the
probability that X is between a and b
f(x)
P (a ≤ X ≤ b)
= P (a < X < b)
(Note that the probability
of any individual value is
zero)
a
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
b
x
Ch. 5-7
The Uniform Distribution
Probability
Distributions
Continuous
Probability
Distributions
Uniform
Normal
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-8
The Uniform Distribution
n 
The uniform distribution is a probability
distribution that has equal probabilities for all
possible outcomes of the random variable
f(x)
Total area under the
uniform probability
density function is 1.0
a
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
b
x
Ch. 5-9
The Uniform Distribution
(continued)
The Continuous Uniform Distribution:
f(x) =
1
if a ≤ x ≤ b
b−a
0
otherwise
where
f(x) = value of the density function at any x value
a = minimum value of x
b = maximum value of x
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-10
Properties of the
Uniform Distribution
n 
The mean of a uniform distribution is
µ=
n 
a+b
2
The variance is
2
σ =
(b - a)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
2
12
Ch. 5-11
Uniform Distribution Example
Example: Uniform probability distribution
over the range 2 ≤ x ≤ 6:
1
f(x) = 6 - 2 = .25 for 2 ≤ x ≤ 6
f(x)
µ=
.25
2
2
6
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
x
σ =
a+b
2
=
(b - a)
12
2+6
2
2
=
=4
(6 - 2)
12
2
= 1.333
Ch. 5-12
5.2
n 
Expectations for Continuous
Random Variables
The mean of X, denoted µX , is defined as the
expected value of X
µX = E(X)
n 
The variance of X, denoted σX2 , is defined as the
expectation of the squared deviation, (X - µX)2, of a
random variable from its mean
2
X
2
σ = E[(X − µX ) ]
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-13
Linear Functions of Variables
n 
n 
Let W = a + bX , where X has mean µX and
variance σX2 , and a and b are constants
Then the mean of W is
µW = E(a + bX) = a + bµX
n 
the variance is
2
W
2
σ = Var(a + bX) = b σ
n 
2
X
the standard deviation of W is
σW = b σX
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-14
Linear Functions of Variables
(continued)
n 
An important special case of the previous results is the
standardized random variable
X − µX
Z=
σX
n 
which has a mean 0 and variance 1
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-15
5.3
The Normal Distribution
Probability
Distributions
Continuous
Probability
Distributions
Uniform
Normal
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-16
The Normal Distribution
(continued)
n  ‘Bell
Shaped’
n  Symmetrical
f(x)
n  Mean, Median and Mode
are Equal
Location is determined by the
mean, µ
Spread is determined by the
standard deviation, σ
The random variable has an
infinite theoretical range:
+ ∞ to - ∞
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
σ
µ
x
Mean
= Median
= Mode
Ch. 5-17
The Normal Distribution
(continued)
n 
n 
n 
n 
The normal distribution closely approximates the
probability distributions of a wide range of random
variables
Distributions of sample means approach a normal
distribution given a “large” sample size
Computations of probabilities are direct and elegant
The normal probability distribution has led to good
business decisions for a number of applications
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-18
Many Normal Distributions
By varying the parameters µ and σ, we obtain
different normal distributions
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-19
The Normal Distribution Shape
f(x)
Changing µ shifts the
distribution left or right.
σ
Changing σ increases
or decreases the
spread.
µ
x
Given the mean µ and variance σ we define the normal
distribution using the notation
X ~ N(µ,σ 2 )
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-20
The Normal Probability
Density Function
n 
The formula for the normal probability density
function is
f(x) =
Where
1
2πσ
2
e
− (x −µ) 2 /2σ 2
e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
µ = the population mean
σ = the population standard deviation
x = any value of the continuous variable, -∞ < x < ∞
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-21
Cumulative Normal Distribution
n 
For a normal random variable X with mean µ and
variance σ2 , i.e., X~N(µ, σ2), the cumulative
distribution function is
F(x0 ) = P(X ≤ x 0 )
f(x)
P(X ≤ x 0 )
0
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
x0
x
Ch. 5-22
Finding Normal Probabilities
The probability for a range of values is
measured by the area under the curve
P(a < X < b) = F(b) − F(a)
a
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
µ
b
x
Ch. 5-23
Finding Normal Probabilities
(continued)
F(b) = P(X < b)
a
µ
b
x
a
µ
b
x
a
µ
b
x
F(a) = P(X < a)
P(a < X < b) = F(b) − F(a)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-24
The Standardized Normal
n 
Any normal distribution (with any mean and
variance combination) can be transformed into the
standardized normal distribution (Z), with mean 0
and variance 1
f(Z)
Z ~ N(0 ,1)
1
0
n 
Z
Need to transform X units into Z units by subtracting the
mean of X and dividing by its standard deviation
X −µ
Z=
σ
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-25
Example
n 
If X is distributed normally with mean of 100
and standard deviation of 50, the Z value for
X = 200 is
Z=
n 
X −µ
σ
=
200 − 100
50
= 2.0
This says that X = 200 is two standard
deviations (2 increments of 50 units) above
the mean of 100.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-26
Comparing X and Z units
100
0
200
2.0
X
Z
(µ = 100, σ = 50)
( µ = 0 , σ = 1)
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)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-27
Finding Normal Probabilities
b − µ ⎞
⎛ a − µ
P(a < X < b) = P⎜
<Z<
⎟
σ ⎠
⎝ σ
⎛ b − µ ⎞
⎛ a − µ ⎞
= Φ⎜
⎟ − Φ⎜
⎟
⎝ σ ⎠
⎝ σ ⎠
f(x)
a
a −µ
σ
µ
b
x
0
b −µ
σ
Z
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-28
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
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
X
Ch. 5-29
Appendix Table 1
n 
n 
The Standardized Normal table in the textbook
(Appendix Table 1) shows values of the
cumulative normal distribution function
For a given Z-value a , the table shows Φ(a)
(the area under the curve from negative infinity to a )
Φ (a) = P(Z < a)
0
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
a
Z
Ch. 5-30
The Standardized Normal Table
§  Appendix Table 1 gives the probability Φ(a) for
any value a
.9772
Example:
P(Z < 2.00) = .9772
0
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
2.00
Z
Ch. 5-31
The Standardized Normal Table
(continued)
§  For negative Z-values, use the fact that the
distribution is symmetric to find the needed
probability:
.9772
.0228
Example:
P(Z < -2.00) = 1 – 0.9772
= 0.0228
0
2.00
Z
.9772
.0228
-2.00
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
0
Z
Ch. 5-32
General Procedure for Finding
Probabilities
To find P(a < X < b) when X is
distributed normally:
n 
Draw the normal curve for the problem in
terms of X
n 
Translate X-values to Z-values
n 
Use the Cumulative Normal Table
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-33
Finding Normal Probabilities
Suppose X is normal with mean 8.0 and
standard deviation 5.0
n  Find P(X < 8.6)
n 
X
8.0
8.6
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-34
Finding Normal Probabilities
(continued)
n 
Suppose X is normal with mean 8.0 and
standard deviation 5.0. Find P(X < 8.6)
Z=
X −µ
σ
=
8.6 − 8.0
5.0
= 0.12
µ=8
σ = 10
8 8.6
µ=0
σ=1
X
P(X < 8.6)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
0 0.12
Z
P(Z < 0.12)
Ch. 5-35
Solution: Finding P(Z < 0.12)
Standardized Normal Probability
Table (Portion)
z
Φ(z)
.10
.5398
.11
.5438
.12
.5478
.13
.5517
P(X < 8.6)
= P(Z < 0.12)
Φ(0.12) = 0.5478
0.00
Z
0.12
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-36
Upper Tail Probabilities
Suppose X is normal with mean 8.0 and
standard deviation 5.0.
n  Now Find P(X > 8.6)
n 
X
8.0
8.6
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-37
Upper Tail Probabilities
(continued)
n 
Now Find P(X > 8.6)…
P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12)
= 1.0 - 0.5478 = 0.4522
0.5478
1.000
Z
0
0.12
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
1.0 - 0.5478
= 0.4522
Z
0
0.12
Ch. 5-38
Finding the X value for a
Known Probability
n 
Steps to find the X value for a known
probability:
1. Find the Z value for the known probability
2. Convert to X units using the formula:
X = µ + Zσ
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-39
Finding the X value for a
Known Probability
(continued)
Example:
n  Suppose X is normal with mean 8.0 and
standard deviation 5.0.
n  Now find the X value so that only 20% of all
values are below this X
.2000
?
?
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
8.0
0
X
Z
Ch. 5-40
Find the Z value for
20% in the Lower Tail
1. Find the Z value for the known probability
Standardized Normal Probability
Table (Portion)
z
Φ(z)
.82
.7939
.83
.7967
.84
.7995
.85
.8023
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
n 
20% area in the lower
tail is consistent with a
Z value of -0.84
.80
.20
?
8.0
-0.84 0
X
Z
Ch. 5-41
Finding the X value
2. Convert to X units using the formula:
X = µ + Zσ
= 8.0 + ( −0.84 )5.0
= 3.80
So 20% of the values from a distribution
with mean 8.0 and standard deviation
5.0 are less than 3.80
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-42
Normal Approximation for the
average of Bernoulli random var.
5.4
n 
n 
Bernoulli random variable Xi :
By Central Limit Theorem,
n 
n 
Xi =1 with probability p
Xi =0 with probability 1-p
E(Xi ) = p
n 
Var(Xi ) = p(1-p)
By Central Limit Theorem,
1
n
n
∑ (X
i
− E(Xi )) → N ( 0,Var(Xi ))
i=1
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-43
Normal Approximation for the
average of Bernoulli random var.
(continued)
n 
The shape of the average of independent Bernoulli
is approximately normal if n is large
n
$ p(1− p) '
1
X − p = ∑ ( Xi − p) → N & 0,
)
%
(
n i=1
n
n 
Standardize to Z from the average of Bernoulli
random variable:
X − E(X)
X −p
Z=
=
p(1− p)/n
Var(X)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-44
Normal Distribution Approximation
for Binomial Distribution
n 
Let Y be the number of successes from n independent
trials, each with probability of success p.
n
Y = ∑ Xi = n X
i=1
Y − E(Y ) Y − nE(X)
Y − np
=
=
→ N ( 0,1)
Var(Y )
np(1− p)
Var(nX)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-45
Example
n 
40% of all voters support ballot proposition A. What
is the probability that between 0.36 and 0.40 fraction
of voters indicate support in a sample of n = 100 ?
E(X) = p = 0.40
Var(X) = p(1-p)/n = (0.40)(1-0.40)/100 = 0.024
# 0.36 − 0.40
0.40 − 0.40 &
P(0.36 < X < 0.40) = P %
≤Z≤
(
(0.4)(1− 0.4)/100 '
$ (0.4)(1− 0.4)/100
= P( − 0.82 < Z < 0)
= Φ(0) − Φ( − 0.82)
= 0.5000 − 0.2939 = 0.2061
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-46
5.6
n 
n 
Joint Cumulative Distribution
Functions
Let X1, X2, . . .Xk be continuous random variables
Their joint cumulative distribution function,
F(x1, x2, . . .xk)
defines the probability that simultaneously X1 is less
than x1, X2 is less than x2, and so on; that is
F(x1 , x 2 , …, x k ) = P({X1 < x1}∩ {X 2 < x 2 }∩ !{X k < x k })
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-47
Joint Cumulative Distribution
Functions
(continued)
n 
n 
The cumulative distribution functions
F(x1), F(x2), . . .,F(xk)
of the individual random variables are called their
marginal distribution functions
The random variables are independent if and only if
F(x1, x2,…, xk ) = F(x1)F(x2 )!F(xk )
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-48
Covariance
n 
n 
Let X and Y be continuous random variables, with
means µx and µy
The expected value of (X - µx)(Y - µy) is called the
covariance between X and Y
Cov(X, Y) = E[(X − µx )(Y − µy )]
n 
An alternative but equivalent expression is
Cov(X, Y) = E(XY) − µxµy
n 
If the random variables X and Y are independent, then the
covariance between them is 0. However, the converse is not true.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-49
Correlation
n 
Let X and Y be jointly distributed random variables.
n 
The correlation between X and Y is
Cov(X, Y)
ρ = Corr(X, Y) =
σ Xσ Y
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-50
Sums of Random Variables
Let X1, X2, . . .Xk be k random variables with
means µ1, µ2,. . . µk and variances
σ12, σ22,. . ., σk2. Then:
n 
The mean of their sum is the sum of their
means
E(X1 + X2 + !+ Xk ) = µ1 + µ2 + !+ µk
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-51
Sums of Random Variables
(continued)
Let X1, X2, . . .Xk be k random variables with means µ1,
µ2,. . . µk and variances σ12, σ22,. . ., σk2. Then:
n 
If the covariance between every pair of these random
variables is 0, then the variance of their sum is the
sum of their variances
Var(X1 + X2 + ! + Xk ) = σ12 + σ 22 + ! + σk2
n 
However, if the covariances between pairs of random
variables are not 0, the variance of their sum is
K −1 K
Var(X1 + X2 + ! + Xk ) = σ12 + σ 22 + ! + σk2 + 2∑ ∑ Cov(Xi , X j )
i=1 j=i+1
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-52
Differences Between Two
Random Variables
For two random variables, X and Y
n 
The mean of their difference is the difference of their
means; that is
E(X − Y) = µX − µY
n 
If the covariance between X and Y is 0, then the
variance of their difference is
Var(X − Y) = σ 2X + σ 2Y
n 
If the covariance between X and Y is not 0, then the
variance of their difference is
Var(X − Y) = σ 2X + σ 2Y − 2Cov(X, Y)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-53
Linear Combinations of
Random Variables
n 
A linear combination of two random variables, X and Y,
(where a and b are constants) is
W = aX + bY
n 
The mean of W is
µW = E[W] = E[aX + bY] = aµX + bµY
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-54
Linear Combinations of
Random Variables
(continued)
n 
The variance of W is
σ 2W = a2σ 2X + b2σ 2Y + 2abCov(X, Y)
n 
Or using the correlation,
2
W
2
2
X
2
2
Y
σ = a σ + b σ + 2abCorr(X, Y)σ Xσ Y
n 
If both X and Y are joint normally distributed random
variables then the linear combination, W, is also
normally distributed
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-55
Portfolio Analysis
n 
A financial portfolio can be viewed as a linear
combination of separate financial instruments
⎛ Proportion of ⎞
⎛ Proportion of ⎞
⎜
⎟
⎜
⎟ ⎛ Stock 2 ⎞
⎛ Return on ⎞
⎛ Stock 1⎞
⎜⎜
⎟⎟ = ⎜ portfolio value ⎟ × ⎜⎜
⎟⎟ + ⎜ portfolio value ⎟ × ⎜⎜
⎟⎟
⎝ portfolio ⎠ ⎜ in stock1
⎟ ⎝ return ⎠
⎜ in stock2
⎟ ⎝ return ⎠
⎝
⎠
⎝
⎠
⎛ Proportion of ⎞
⎜
⎟ ⎛ Stock N ⎞
⎟⎟
! + ⎜ portfolio value ⎟ × ⎜⎜
⎜ in stock N
⎟ ⎝ return ⎠
⎝
⎠
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-56
Portfolio Analysis Example
n 
Consider two stocks, A and B
n 
n 
n 
n 
The price of Stock A is normally distributed with mean
12 and standard deviation 4
The price of Stock B is normally distributed with mean
20 and standard deviation 16
The stock prices have a positive correlation, ρAB = .50
Suppose you own
n 
10 shares of Stock A
n 
30 shares of Stock B
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-57
Portfolio Analysis Example
(continued)
n 
The mean and variance of this stock portfolio
are: (Let W denote the distribution of portfolio value)
µW = 10µA + 20µB = (10)(12) + (30)(20) = 720
σ 2W = 10 2 σ 2A + 30 2 σ B2 + (2)(10)(30)Corr(A,B)σ A σ B
= 10 2 (4)2 + 30 2 (16)2 + (2)(10)(30)(.50)(4)(16)
= 251,200
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-58
Portfolio Analysis Example
(continued)
n 
What is the probability that your portfolio value is
less than $500?
µW = 720
σ W = 251,200 = 501.20
n 
500 − 720
= −0.44
The Z value for 500 is Z =
501.20
n 
P(Z < -0.44) = 0.3300
n 
So the probability is 0.33 that your portfolio value is less than $500.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-59