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Homework #11. Due: Friday, November 14, 2003.
< draft KEY > IE 230
Textbook: D.C. Montgomery and G.C. Runger, Applied Statistics and Probability for
Engineers, John Wiley & Sons, New York, 2003. Sections 5.5-5.7.
The topics are measuring dependence with covariance and correlation, the bivariate normal
distribution, and linear combinations of random variables. As always, I encourage you to
read the textbook before beginning the assignment.
1. (Montgomery and Runger, Equation 5–37)
Result: E[c 0 + Σip=1 ci Xi ] = c 0 + Σip=1 ci E(Xi )
E[c 0 + Σip=1 ci Xi ]
∞
∞
∞
∞
∫−∞ . . . ∫−∞ [c 0 + Σi =1 ci xi ] f X ,...,X
=
p
1
∫−∞ . . . ∫−∞ c 0 f X ,...,X
=
+
1
∞
p
..
i =1 −∞
∞
∞
Σ ∫
.
p
∞
p
(x 1, . . . , xp ) dx 1...dxp
(a)_ < def of E > _
(x 1, . . . , xp ) dx 1...dxp
∫−∞ (ci xi ) f X ,...,X
1
p
(x 1, . . . , xp ) dx 1...dxp
(b)_ < calculus > _
= c 0 ∫ . . . ∫ f X ,...,X (x 1, . . . , xp ) dx 1 . . . dxp
−∞
+
Σ
−∞
∞
p
c
i =1 i −∞
∫
= c 0 + Σip=1 ci
1
...
∞
p
∞
∫−∞
xi f X ,...,X (x 1, . . . , xp ) dx 1...dxp
1
∞
∫−∞ . . . ∫−∞
p
xi f X ,...,X (x 1, . . . , xp ) dx 1...dxp
1
p
= c 0 + Σip=1 ci E(Xi )
(c)_ < calculus > _
(d)_ < f X 1,...,Xp is a density > _
(e)_ < def of E > _
2. (Montgomery and Runger, Equation 5–28)
Result: cov(X , Y ) = E(XY ) − µX µY
cov(X , Y )
= E[(X − µX ) (Y − µY )]
(a)_ < def of covariance > _
= E[XY − µX Y − µY X + µX µY ]
(b)_ < algebra > _
= E(XY ) − µX E(Y )− µY E(X ) + µX µY
(c)_ < Problem 1 > _
= E(XY ) − µX µY − µY µX + µX µY
(d)_ < notation > _
= E(XY ) − µX µY
(e)_ < simplify > _
3. Result: If X and Y are independent random variables, then E(XY ) = E(X ) E(Y ).
E(XY )
=
=
∞
∞
∫−∞ ∫−∞ xy
∞ ∞
∫−∞ ∫−∞ xy
f XY (x , y ) dx dy
(a)_ < def of expected value > _
f X (x ) f Y (y ) dx dy
(b)_ < independence > _
∞
∞
=
∫−∞ y f Y (y ) ∫−∞ x
=
∫−∞ y f Y (y ) E(X ) dy
∞
E(X ) ∫ y f Y (y ) dy
−∞
=
∞
I
L
M
f X (x ) dx O dy
(c)_< calculus > _
(d)_ < def of expected value > _
(e)_ < calculus > _
= E(X ) E(Y )
(f)_ < def of expected value > _
– 1 of 5 –
Schmeiser
Homework #11. Due: Friday, November 14, 2003.
< draft KEY > IE 230
4. (Montgomery and Runger, Problem 5–78)
Result: If X and Y are independent random variables, then cov(X , Y ) = 0.
cov(X , Y )
= E(XY ) − µX µY
(a)_ < Problem 2 > _
= E(X ) E(Y ) − µX µY
(b)_ < Problem 3 > _
= µX µY − µX µY
(c)_ < notation > _
= 0
(d)_ < simplify > _
5. (Montgomery and Runger, Figure 5–13(d))
Result: cov(X , Y ) = 0 does not imply that X and Y are independent.
We consider an example for which X and Y are dependent, yet cov(X , Y ) = 0.
Suppose that a spinning disk has a mark on its outer edge. (An example is the mark used
to time the ignition on an engine.) For simplicity, let (0, 0) denote the center of the
spinning disk and let the radius be one. The experiment is to choose a random position
of the disk. Let the random variable W denote the angle of the mark from the center. If
the position is chosen at a random time, then we can assume that W is uniformly
distributed between zero and 2 π.
Let X and Y denote the Cartesian coordinates of the mark. That is, X = sin(W ) is the
horizontal position and Y = cos(W ) is the vertical position.
(a) Argue that X and Y are dependent by showing that f X Y =y is not equal to f X .
|
(It is sufficient to show that the range of f X | Y =y differs from the range of f X .)
_______________________________________________________________
The range of f X Y =y is composed of the two values √1ddddd
− y 2 and − √1ddddd
− y 2.
|
The range of f X is the interval [−1, 1].
_______________________________________________________________
(b) Complete the reasons showing that cov(X , Y ) = 0.
cov(X , Y )
= E(XY ) − µX µY
(a)_ < Problem 2 > _
= E(XY )
(b)_ < symmetric about zero > _
= E[sin(W ) cos(W )]
(c)_ < def of X and Y > _
2π
∫0
sin(w ) cos(w ) f W (w ) dw
(d)_ < def of expected value > _
=
∫0
1
sin(w ) cos(w ) ( hhh ) dw
2π
(e)_ < W is U (0, 2π) > _
=
hhh
=
=
=
2π
1
2π
1
hhh
2π
∫0
π
∫0
I
2π
1
hhh
L
2π
L
π
∫0
I
(f)_ < calculus > _
sin(w ) cos(w ) dw
sin(w ) cos(w ) dw + ∫
2π
π
M
(g)_ < calculus > _
sin(w ) cos(w ) dw O
π
sin(w ) cos(w ) dw + ∫ sin(w ) [− cos(w )] dw O
0
= 0
M
(h)_ < property of cos > _
(i)_ < simplify > _
– 2 of 5 –
Schmeiser
Homework #11. Due: Friday, November 14, 2003.
< draft KEY > IE 230
6. (Montgomery and Runger, Problem 5–95)
Assume that the weights of individuals are independent and normally distributed with a
mean of 160 pounds and a standard deviation of 30 pounds. Suppose that 25 persons
squeeze into an elevator that is designed to hold 4300 pounds. We consider the
elevator’s load, Y = Σi25=1 Xi .
(a) Determine the expected load.
_______________________________________________________________
25
= E( Σ Xi )
E(Y )
i =1
25
Σ E(Xi )
=
i =1
= 25 × 160
= 4000
simplify
_______________________________________________________________
(b) Determine the standard deviation of the load.
_______________________________________________________________
25
V(Y )
= V( Σ Xi )
i =1
25
=
Σ V(Xi )
independence
i =1
= 25 × 30
2
= 22,500
simplify
d ddddd = 150 (pounds).
Therefore, std(Y ) = √22,500
_______________________________________________________________
– 3 of 5 –
Schmeiser
Homework #11. Due: Friday, November 14, 2003.
< draft KEY > IE 230
(c) Determine the probability that the load exceeds the design limit.
_______________________________________________________________
Let Xi denote the weight (in pounds) of the i th person.
Let Y = Σi25=1 Xi , the total weight.
We are given that Xi ∼ Normal(160,302).
From normality and independence of the Xi s,
we know that Y ∼ Normal(25 × 160, 25 × 302) = Normal(4000, 1502).
Therefore,
P(Y > 4300)
I
Y − 4000
4300 − 4000 M
= P J hhhhhhhhh > hhhhhhhhhhh J
150
150
O
L
I
= P JZ >
L
=
=
=
=
4300 − 4000
M
150
O
hhhhhhhhhhh
Standardize
Definition of Z
J
Simplify
Simplify
Table II
Simplify
P(Z > 2)
1 − P(Z ≤ 2)
1 − 0.977
0.023 ←
_______________________________________________________________
(d) Determine the design limit that is exceeded by the load with probability 0.0001.
_______________________________________________________________
Still Y ∼ Normal(25 × 160, 25 × 302) = Normal(4000, 1502).
→
→
→
→
→
→
→
→
→
→
→
→
P(Y > y ) = 0.0001
Y − µY
y − µY
P( hhhhhhh > hhhhhhh ) = 0.0001
σY
σY
y − µY
P(Z > hhhhhhh ) = 0.0001
σY
y − 4000
P(Z > hhhhhhhh ) = 0.0001
150
y
−
4000
P(Z ≤ hhhhhhhh ) = 0.9999
150
y − 4000
hhhhhhhh
= 3.72
150
y = 4558 ←
Desired property
Standardize
Definition of Z
Substitute known values
Complement
Table II
Complement
_______________________________________________________________
(e) What is the experiment and sample space that underlies this problem?
_______________________________________________________________
The experiment is to choose a group of 25 persons from some population.
The sample space is the set of all such groups.
_______________________________________________________________
– 4 of 5 –
Schmeiser
Homework #11. Due: Friday, November 14, 2003.
< draft KEY > IE 230
7. (Montgomery and Runger, Problem 5–80)
Let X and Y denote two dimensions (in inches) of an injection-molded part. Suppose
that X and Y have a bivariate normal distribution with µX = 3.00, σX = 0.04, µY = 7.70,
σY = 0.08, and correlation ρ.
(a) If ρ = 0, determine P(2.95 < X < 3.05, 7.60 < Y < 7.80).
_______________________________________________________________
P(2.95 < X < 3.05, 7.60 < Y < 7.80)
= P(2.95 < X < 3.05) P(7.60 < Y < 7.80)
2.95 − 3
3.05 − 3
7.60 − 7.70
7.80 − 7.70
= P( hhhhhhhh < Z < hhhhhhhh ) P( hhhhhhhhhh < Z < hhhhhhhhhh )
0.04
0.04
0.08
0.08
= P(−1.25 < Z < 1.25) P(−1.25 < Z < 1.25)
2
= [P(−1.25 < Z < 1.25)]
2
= [P(Z < 1.25) − P(X < −.125)]
2
= [Φ(1.25) − Φ(−1.25)]
2
≈ [0.89435 − 0.10565]
2
= [0.7887]
= 0.622
Independence
Standardize
Simplify
Simplify
Axiom 3
Notation
Table II
Simplify
Simplify
_______________________________________________________________
(b) (To be submitted electronically.) If ρ =/ 0, then joint bivariate normal probabilities
are difficult to determine.
Here we estimate
the value of
P(2.95 < X < 3.05, 7.60 < Y < 7.80) using Monte Carlo simulation.
(i) From the course web page, "gilbreth.ecn.purdue.edu/˜ie230/", copy the Bivariate
Normal spreadsheet to your computer. Replace my header information with
your header information.
(ii) In the three columns to the right of the (X , Y ) values, compute the indicator
random variables for the three events 2.95 < X < 3.05, 7.60 < Y < 7.80, and
2.95 < X < 3.05, 7.60 < Y < 7.80. (You can use "=if(__,1,0)".)
(iii) Above these three columns, create cells that contain the fraction of the
indicator random variables that are ones (use "=average"). The third average
is the Monte Carlo point estimator of P(2.95 < X < 3.05, 7.60 < Y < 7.80). What
do the first two estimate? (Answer on the second page of the spreadsheet.)
(iv) Enter the model parameter values for Part (a). Hit F9 a few times to verify that
your code is correct. (That is, be sure that the Monte Carlo point estimates are
close to your answer from Part (a).)
(v) Create a plot that shows P(2.95 < X < 3.05, 7.60 < Y < 7.80) as a function of ρ.
(Answer on the third page of the spreadsheet. Use at least seven values of ρ,
ranging from -1 to 1. Obtain these values from the Monte Carlo simulation
and hard code them in two columns. Highlight the data values, and in the
chart wizard choose "scatter plot".)
Not to submit, but I encourage you to work some textbook problems.
Good examples are 5–67 and 5–87.
– 5 of 5 –
Schmeiser