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IE 230
Seat # ________
Name ___ < KEY > ___
Please read these directions.
Closed book and notes. 60 minutes.
Covers through the normal distribution, Section 4.6 of Montgomery and Runger, fourth edition.
Cover page and four pages of exam.
Pages 8 and 12 of the Concise Notes.
A normaldistribution cdf table.
No calculator. No need to simplify beyond probability concepts.
For example, unsimplified factorials, integrals, sums, and algebra receive full credit.
Throughout,
f
denotes probability mass function or probability density function
and
F
denotes cumulative distribution function.
For one point of credit, write your name neatly on this cover page. Circle your family name.
For one point of credit, write your name on each of pages 1 through 4.
Score ___________________________
Exam #2, October 19, 2010
Schmeiser
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Name _______________________
Closed book and notes. 60 minutes.
1. Suppose that the random variable
X
has the discrete uniform distribution over the set
{0, 1,.
.., 10} and that the random variable
Y
has the continuous uniform distribution over
the set [0,10].
(a)
(3 pt) T
←
F
E(
X
)
=
E(
Y
).
(b)
(3 pts) T
F
←
V(
X
)
=
V(
Y
).
(c)
(3 pts) T
F
←
F
X
(5)
=
F
Y
(5).
(d)
(3 pts) T
F
←
f
X
(5)
=
f
Y
(5).
2. (from Montgomery and Runger, 3–113) In 1898 L. J. Bortkiewicz published a book
entitled
The Law of Small Numbers
. He used data collected over twenty years to show
that the number of soldiers killed by horse kicks each year in each corps in the Prussian
cavalry followed a Poisson distribution with a mean of 0.61.
(a) (7 pts) For a particular corps over one year, determine the value of the probability of
more than one death.
____________________________________________________________
Deaths occur with rate
λ =
0.61 deaths per year per corps.
Let
N
=
"number of deaths in one year in a particular corps".
Then
N
is Poisson with rate
μ = λ
t
=
(0.61)(1)
=
0.61 deaths.
Therefore, P(
N >
1)
=
1
−
[P(
N
=
0)
+
P(
N
=
1)]
=
1
−
R
J
Q
0!
e
−μ
μ
0
hhhhh
+
1!
e
−μ
μ
1
hhhhh
H
J
P
=
1
−
e
−
0.61
[1
+
0.61]
←
____________________________________________________________
(b) (7 pts) For a particular corps over five years, determine the value of the probability of
no death.
____________________________________________________________
Deaths still occur with rate
λ =
0.61 deaths per year per corps.
Now let
N
denote the number of deaths during five years.
Then
N
is Poisson with mean
μ = λ
t
=
(0.61)(5)
=
3.05.
Therefore, P(
N
=
0)
=
0!
e
−μ
μ
0
hhhhh
=
e
−
3.05
←
____________________________________________________________
Exam #2, October 19, 2010
Page 1 of 4
Schmeiser
Name _______________________
3. (from Montgomery and Runger, 4–61) The lifetime of a semiconductor laser at a constant
power is normally distributed with a mean of 6000 hours and standard deviation of 500
hours.
(a) (8 pts) Sketch the corresponding normal pdf. Label and scale both axes.
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 Fall '08
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