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IE 230 — Probability & Statistics in Engineering I
Name __ < KEY > __
Closed book and notes. 60 minutes.
Cover page, four pages of exam.
Score ___________________________
The essence of five families of probability mass functions:
e−λ λx / x !
p (1−p )x −1
−1 r
Crx−1
p (1−p )x −r
CxK CnN−x−K / CnN
Cxn p x (1−p )n −x
Exam #2, Fall 2000
Schmeiser
IE 230 — Probability & Statistics in Engineering I
Name __ < KEY > __
Closed book and notes. 60 minutes.
1. True or false. (for each, 2 points if correct, 1 point if left blank.)
(a)
F
Each time an experiment is replicated, a random variable, say X , either
occurs or it does not occur.
(b)
F
The phrases "probability mass function" and "probability density function"
are synonyms.
(c)
T
For a discrete random variable X , the pmf f X (x ) = P(X = x ) for every real
number x .
(d)
T
(e)
F
The mean E(X ) is analogous to the physical concept of "moment of
inertia".
P(X = x ) ≤ P(X ≤ x ) for every real number x .
I
5M
J
J
= 10.
(f)
T
(g)
F
If
X
is
a
P(2 ≤ X ≤ 4) = P(1.5 < X < 4.5).
(h)
3O
L
F
continuous
random
variable,
then
The geometric distribution is a special case of the binomial distribution.
2. Consider an experiment with sample space S and an associated random variable X with
cdf F . Choose the correct answer.
(a) X is a function. Values of the function are
(i) outcomes (ii) events (iii) real numbers ← (iv) probabilities.
(b) X is a function. Arguments of the function are
(i) outcomes ← (ii) events (iii) real numbers (iv) probabilities.
(c) F is a function. Values of the function are
(i) outcomes (ii) events (iii) random variables (iv) probabilities ← .
(d) F is a function. Arguments of the function are
(i) outcomes (ii) events (iii) real numbers ← (iv) probabilities.
Exam #2, Fall 2000
Page 1 of 4
Schmeiser
IE 230 — Probability & Statistics in Engineering I
Name __ < KEY > __
3. (Spread-sheet question.) Consider a column of 100 spread-sheet cells each containing the
code "=if (rand()< .6, 1, 0)". Suppose that Cell A1 contains the sum of the 100 cells.
(a) What is the distribution name associated with Cell A1?
---------------------------------------------------------------------binomial
---------------------------------------------------------------------(b) What are the distribution parameter values for Cell A1.
---------------------------------------------------------------------n = 100, p = 0.6 (where n is the number of trials and
p is the probability of success for any one Bernoulli trial)
---------------------------------------------------------------------4. Result: If X is a discrete random variable, then f (x ) ≤ F (x ).
For each of the five lines of the proof, write the reason number.
(i)
Axiom 1
(ii) Axiom 2
(iii) Axiom 3
(iv) Definition of conditional probability
(v) Events are independent.
(vi) Events are complementary.
(vii) Probability of the same event.
(viii) Multiplication Rule
(ix) Total Probability
(x) Definition of probability mass function.
(xi) Definition of probability density function.
(xii) Definition of cumulative distribution function.
(xiii) Definition of expected value.
(xiv) Definition of standard deviation.
(xv) Simplification
F (x )
Exam #2, Fall 2000
= P(X ≤ x )
(a) __ < xii > __
= P(X = x or X < x )
(b) __ < vii > __
= P(X = x ) + P(X < x )
(c) __ < iii > __
= f (x ) + P(X < x )
(d)__ < x > __
≥ f (x )
(e)__ < ii > __
Page 2 of 4
Schmeiser
IE 230 — Probability & Statistics in Engineering I
Name __ < KEY > __
5. Sometimes a hypergeometric random variable is approximated by a binomial random
variable. Similarly, sometimes a binomial random variable is approximated by a
Poisson random variable.
(a) Do such approximations require a continuity correction?
yes
no ←
(b) Suppose that X is hypergeometric with µX = 6.5 and σX = 2.5. If we use a Poisson
approximation to P(X = 0), which Poisson distribution would be appropriate?
---------------------------------------------------------------------Set the Poisson mean to µX = 6.5.
---------------------------------------------------------------------6. (Montgomery and Runger, 5-8.) The probability density function of the length of a metal
rod is f (x ) = 2 for 2.3 < x < 2.8 meters and zero elsewhere.
(a) Suppose that specifications for this process require the length to be between 2.25 and
2.75 meters. What is the probability that a randomly selected bar meets
specifications?
---------------------------------------------------------------------P( "meets specifications" ) = P( 2.25 < X < 2.75 )
=∫
2.75
=∫
2.3
2.25
2.25
f (x ) dx
0 dx + ∫
2.75
2.3
2 dx
= 0.9←
←
---------------------------------------------------------------------(b) Assume that the probability density function is f (x ) = 2 for an interval of length 0.5
meter. Over what value should the density be centered to achieve the greatest
probability that a randomly selected bar is within specifications?
---------------------------------------------------------------------Ideally the pdf would be centered over the specifications, ( 2.25, 2.75 ) meters.
Therefore the ideal center is 2.5 meters. ←
(Then
P( "meets specifications" ) = P( 2.25 < X < 2.75 )
=∫
2.75
2.25
2 dx
= 1.0)
---------------------------------------------------------------------Exam #2, Fall 2000
Page 3 of 4
Schmeiser
IE 230 — Probability & Statistics in Engineering I
Name __ < KEY > __
7. (Montgomery and Runger, 4-63.) The probability of a successful optimal alignment in the
assembly of an optical data-storage product is 0.8. Assume that trials are independent.
(a) What is the probability that the first alignment is successful?
---------------------------------------------------------------------0.8←
←
---------------------------------------------------------------------(b) What is the probability that the first successful alignment requires exactly four
trials?
---------------------------------------------------------------------Let X = "the number of trials until a successful alignment.
Then X ∼ geometric ( p = 0.8 ).
Therefore, P(X = 4) = p (1 − p )4−1 = 0.8 (1 − 0.8)3 = 0.0064←
←
---------------------------------------------------------------------(c) What is the probability that exactly three trials are required to achieve two successful
alignments?
---------------------------------------------------------------------Let X = "the number of trials until a the second successful alignment.
Then X ∼ negative binomial (r = 2, p = 0.8 ).
Therefore,
−1 r
P(X = 3) = Crx−1
p (1 − p )x −r = C 12 0.82 (1 − 0.8)1 = 2 (0.64) (0.2) = 0.256←
←
---------------------------------------------------------------------(d) Let p denote the probability that an alignment is successful. (Now p might not be
equal to 0.8.) Let A 3 denote the event that the first three alignments are
successful. Find the value of p so that P(A 3) = 0.7.
---------------------------------------------------------------------3
0.7 = P(A 3) = p (since alignments are independent).
Therefore, p = (0.7)1/3 ≈ 0.888←
←
----------------------------------------------------------------------
Exam #2, Fall 2000
Page 4 of 4
Schmeiser