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IE 230
Seat # ________
Name ___ < KEY > ___
Closed book and notes. 60 minutes.
Cover page and four pages of exam.
No calculators.
Score ___________________________
Exam #3a, Spring 2002
Schmeiser
IE 230 — Probability & Statistics in Engineering I
Name ___ < KEY > ___
Closed book and notes. 60 minutes.
1. True or false. (for each, 3 points if correct, 2 points if left blank.)
If X is normally distributed with mean µX and standard deviation σX ,
then P(X > µX ) = P(X ≥ µX ).
(a)
T ← F
(b)
T ← F
If X is normally distributed with mean µX and standard deviation σX ,
then 0.65 < P(µX − σX < X < µX + σX ) < .75.
(c)
T ← F
If Y is exponentially distributed with rate λ, then P(Y < 0) = 0.
(d)
T F ← If approximating a binomial-distribution probability with a Poissondistribution probability, the continuity correction is useful.
(e)
F
If f XY is a joint probability
←
f XY (x , y ) = P(X ≤ x , Y ≤ y ) for all real numbers x and y .
function,
then
(f)
F ←
If f XY is a joint probability density function,
f XY (x , y ) = P(X ≤ x , Y ≤ y ) for all real numbers x and y .
then
(g)
T
F ←
If X and Y are independent random variables, then
P(X ≤ 6, Y ≤ 8) = f X (6) f Y (8), regardless of whether f X and f Y are pmfs or pdfs.
(h)
T ← F
If (X 1, X 2, . . . , X 6) is a multinomial random vector, then X 1 + X 2 is a
binomial random variable.
(i)
T ← F
If Xi is the time between counts i − 1 and i in a Poisson process, the
distribution of X 1 + X 2 + X 3 is Erlang.
T
T
Exam #3a, Spring 2002
Page 1 of 4
mass
Schmeiser
IE 230 — Probability & Statistics in Engineering I
Name ___ < KEY > ___
2. (Montgomery and Runger, 6–61) The yield in pounds from a day’s production is normally
distributed with a mean of 1500 pounds and a variance of 10000 pounds squared.
Assume that the yields on different days are independent random variables.
Let Xi be the yield on day i for i = 1, 2,..., 5.
(a) Sketch the pdf of Xi . Label and scale both axes.
________________________________________________________
Horizontal and vertical axis. Label with x and f X (x ).
Sketch a bell curve, centered at µX = 1500.
The points of inflection should be one standard deviation from µX .
Here one standard deviation is the square root of the variance, σX = 100.
Scale the vertical axis at zero and somewhere else, probably the mode.
The pdf integrates to one, so the value at the mode should be about 0.004.
ddd ≈ 0.004.)
(The exact value is 1 / ( √2πσ)
________________________________________________________
(b) What is the probability that the yield exceeds 1400 pounds on a randomly selected
day? (Approximate as necessary.)
________________________________________________________
P(Xi > 1400) = P(Xi > µX − σX ) ≈ 0.84 ←
using the approximation that about 68% of the normal distribution
falls within one standard deviation of the mean.
(Half of 68% is 34% to the left of the mean.
The right half is 50%, for a total of 84%.)
________________________________________________________
(c) Let p denote your answer to Part (b). What is the probability that production exceeds
1400 pounds all five days of a randomly selected week? (That is, determine
P(X 1 > 1400, X 2 > 1400,..., X 5 > 1400).)
________________________________________________________
5
P(X 1 > 1400, X 2 > 1400,..., X 5 > 1400) = Π P(Xi > 1400) = p 5 ←
i =1
________________________________________________________
Exam #3a, Spring 2002
Page 2 of 4
Schmeiser
IE 230 — Probability & Statistics in Engineering I
Name ___ < KEY > ___
3. Result: E(XY ) = E(X ) E(Y ). The following proof assumes that X and Y are continuous.
Choose exactly one reason for each step.
(i) Axiom 1
(ii) Axiom 2
(iii) Axiom 3
(iv) Definition of expected value
(v) Density functions integrate to one.
(vi) Independence
(vii) Total Probability
(viii) Definition of pdf
(ix) Calculus (No probability result needed)
E(XY )
∞ ∞
=
∫−∞ ∫−∞ xy
=
∫−∞ ∫−∞ xy
=
∫−∞ x
∞ ∞
∞
f XY (x , y ) dx dy
___ < (iv) > ___
f X (x ) f Y (y ) dx dy
___ < (vi) > ___
f X (x ) dx
∞
∫−∞ y
f Y (y ) dy
= E(X ) E(Y )
___ < (ix) > ___
___ < (iv) > ___
4. The Weibull cdf is F (x ) = 1 − exp[−(x / δ)β] if x > 0 and zero elsewhere.
Write P(2 ≤ X ≤ 5) in terms of F .
________________________________________________________
The Weibull distribution is continuous, so
P(2 ≤ X ≤ 5) = P(2 < X ≤ 5) = P(X ≤ 5) − P(X ≤ 2) = F (5) − F (2) ←
________________________________________________________
Exam #3a, Spring 2002
Page 3 of 4
Schmeiser
IE 230 — Probability & Statistics in Engineering I
Name ___ < KEY > ___
5. Recall: The exponential cdf is F (x ) = 1 − e−λx if x > 0 and zero elsewhere. The pdf is
−λx
if x > 0 and zero otherwise. The associated mean value is 1 / λ.
f (x ) = λe
(From Montgomery and Runger, Problem 5–114) The CPU of a personal computer has a
lifetime that is exponentially distributed with a mean lifetime of six years.
(a) What is the value of λ, including its units.
________________________________________________________
λ = 1 / 6 (failure per year) ←
________________________________________________________
(b) Suppose that you have owned this CPU for three years. What is the probability that
the CPU fails sometime during the next four years?
________________________________________________________
P(X < 7 | X > 3)
= P(X <4)
memoryless property
= P(X ≤ 4)
X is continuous
= F (4)
definition of cdf
=1−e
−(1 / 6)(4)
=1−e
−2 / 3
X is exponential
simplify
________________________________________________________
6. Consider the joint pdf f XY (a , b ) = ab for 0 < a < c and 0 < b < c and zero elsewhere.
Determine the value of the constant c .
________________________________________________________
Every pdf must integrate to one, which gives us one equation with the unknown c :
Let’s rewrite the pdf using the more-common dummy variables x and y :
f XY (x , y ) = xy for 0 < x < c and 0 < y < c and zero elsewhere.
∞ ∞
∫−∞ ∫−∞ f XY (x , y ) dx dy
c c
∫0 ∫0
=1
xy dx dy = 1
4
c /4=1
c = √2d ←
________________________________________________________
Exam #3a, Spring 2002
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Schmeiser