ISE 130 Quiz #1 Fall 2016
Instructions: Closed Notes and Book
Total Points = 20
Name: SoLh u
Points
1") At a certain event, 20 people attend, and 3 will be chosen to receive door prizes. (4)
How many different groups of 3 people can be chosen if the prize
From Tau Beta Pi SJSU Exam Library: http:/exams.tbpsjsu.org
From Tau Beta Pi SJSU Exam Library: http:/exams.tbpsjsu.org
From Tau Beta Pi SJSU Exam Library: http:/exams.tbpsjsu.org
From Tau Beta Pi SJSU Exam Library: http:/exams.tbpsjsu.org
From Tau Beta P
ISE 130 Test #1 Fall 2016
Instructions: Closed Book & Notes
Show All Work
Total Points = 100
Name: Egg id !
Major: Level: Undergraduate Graduate (Circle one)
Points
1a) A company hired 15 new employees, and must assign 6 to the day shift, 5 to the (6)
gr
HW #1 Solution
2-6.
An ammeter that displays three digits is used to measure current in milliamperes.
S = cfw_(x, y, z)x = 0, 1, 2, , 9,; y = 0, 1, 2, , 9; z = 0, 1, 2, , 9)
A vector with three components can describe the three digits of the ammeter. Each
[SE 130 Final Exam Summer 2015
Instructions: Closed Book & Notes
Show All Work
Total Points = 100
Name: 50 W4;
Points
1) The wall thickness of 25 glass 2-liter bottles was measured by a qualitycontrol engineer (12)
The sample mean 9? = 4.05, millimeters,
ISE 130 _ Quiz #3 . Fall 2015
Instructions: Closed Notes and Book
Total Points = 20
Name: gﬁ (M;
Points
1) Consider the following computer output (5)
Test ofHo: u = 26 against H1: [.1 at 26
Assumed population standard deviation is o = 1.5
_
? i-cossb
[SE 130 Test #2 Fall 2015
Instructions: Closed Book & Notes
Show All Work
Total Points = 100
q
Name: gnlﬂ‘égau
Points
1a) Suppose that X has a lognormal distribution with parameters 0 = 0 and m2 = 4. (7)
Determine the following
(i) mean and variance of X.
ISE 130 Test #1 Fall 2015
Instructions: Closed Book & Notes
Show All Work
Total Points = 100
Name:
Points
1a) A group of 10 people have gotten together to play basketball. They will begin by dividing (6)
themselves into two teams of 5 players each. In ho
ISE 130
Final Exam
Spring2007
Instructions: Closed Book & Notes
Show All Work
Total Points = 100
Name: _
Points
1a) To estimate the average time it takes to assemble a certain number computer
(10)
component, the industrial engineer at an electronics firm
ISE 130
Test #1
Spring 2005
Instructions: Closed Book & Notes
Show All Work
Total Points = 100
Name: _
Points
1a) A computer password consists of eight characters. How many different
(15)
passwords are possible if each character may be any lowercase lette
Homework #2 Solution
2-54.
A hospital operating room needs to schedule three knee surgeries and two
hip surgeries in a day. Suppose that an operating room needs to handle three knee, four
hip, and five shoulder surgeries.
(a) How many different sequences
Homework#3_Solution
3-4.
The range of X is cfw_0, 1, 2, 3, 4, 5
3-16.
f X (0) P( X 0) 1 / 6 1 / 6 1 / 3
f X (1.5) P ( X 1.5) 1 / 3
f X (2) 1 / 6
f X (3) 1 / 6
a) P(X = 1.5) = 1/3
b) P(0.5< X < 2.7) = P(X = 1.5) +P(X = 2) = 1/3 + 1/6 = 1/2
c) P(X > 3) = 0
Homework #4 Solution
3-126.
X = number of opponents until the player is defeated.
p=0.2, the probability of the opponent defeating the player.
(a) f(x) = (1 p)x 1p = 0.8(x 1)(0.2), x = 1, 2, 3,
(b) P(X>2) = 1 P(X = 1) P(X = 2) = 0.64
(c) = E(X) = 1/p = 5
Homework #6 Solution
4-139. a) The time until the tenth call is an Erlang random variable with
5
calls per minute
and r = 10.
b) E(X) = 10/5 = 2 minutes. V(X) = 10/25 = 0.4 minutes.
c) Because a Poisson process is memoryless, the mean time is 1/5 = 0.2 m
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Homework #5 Solution
4-53.
a) The distribution of X is f(x) = 10 for 0.95 < x < 1.05. Now,
0,
FX ( x) 10 x 9.5,
1,
b)
x 0.95
0.95 x 1.05
1.05 x
P ( X 1.02) 1 P ( X 1.02) 1 FX (1.02) 0.3
c) If P(X > x) = 0.90, then 1 F(X) = 0.90 and F(X) = 0.10. Therefore
Homework #8 Solution
(7-4 Not assigned) Included for your reference
7-4.
X i ~ N (100,102 )
X 100 X
n 25
10
2
n
25
P[(100 1.8(2) X (100 2)] P(96.4 X 102) P
96 .4 100
2
X/ n 102 2100
P( 1.8 Z 1) P( Z 1) P( Z 1.8) 0.8413 0.0359 0.8054
7-6.
n=6
n = 49
X
Homework #7 Solution
(7-4 Not assigned) Included for your reference
7-4.
X i ~ N (100,10 2 )
X 100 X
n 25
10
2
n
25
P[(100 1.8( 2) X (100 2)] P(96.4 X 102) P
96 .4 100
2
X/ n 102 2100
P ( 1.8 Z 1) P( Z 1) P( Z 1.8) 0.8413 0.0359 0.8054
7-6.
n=6
n = 49
Homework #2 Solution
2-105. The following table summarizes the number of deceased beetles under autolysis (the
destruction of a cell after
its death by the action of its own enzymes) and putrefaction (decomposition of organic matter,
especially protein, b
ISE 130 Quiz #3 _ Fall 2014
Instructions: Closed Notes and Book
Total Points = 20
Name: 30 ﬂux
1) Consider a normal population with an unknown population variance. (5)
A sample of size 16 is taken from this population, and the sample results are
a? = 52 a
HW #10
9-4
a)
H 0 : 25 Newtons, H 1 : 25 Newtons
b) No, this result only implies that we do not have enough evidence to support H1.
9-10
a) = P(
X
98.5) + P(
X
> 101.5)
=
+
X 100 98.5 100 X 100 101.5 100
P
2 / 9 2 / 9 P 2 / 9 2 / 9
= P(Z 2.25) +
Homework #11
9-81
a) 0.1 < 1-P < 0.5 then 0.5 < P-value < 0.9
b) 0.1 < 1-P< 0.5 then 0.5 < P-value < 0.9
c) 0.99 <1-P <0.995 then 0.005 < P-value < 0.01
9-87
a) In order to use the 2 statistic in hypothesis testing and confidence interval construction, we
Homework #9 Solution
8-3
a) A z = 1.29 would result in a 90% one-sided confidence interval.
b) A z = 1.65 would result in a 95% one-sided confidence interval.
c) A z = 2.33 would result in a 99% one-sided confidence interval.
8.4
a) 95% CI for
,
n 10, 20
HW #7 Solution
6-6.
The mean is increased by 10 and the standard deviation is not changed. Try it!
6-8.
Sample average:
19
x
i
x i 1
19
272.82
14.359 min
19
Sample variance:
19
x
272.82
i
i
1
19
x
2
i
10333.8964
i
1
2
n
xi
n
272.82 2
2
xi i 1
10333.
HW #7 Solution
6-6.
The mean is increased by 10 and the standard deviation is not changed. Try it!
6-8.
Sample average:
19
x
x
i
i
1
19
272.82
14.359 min
19
Sample variance:
19
x
272.82
i
i
1
19
x
2
i
10333.8964
i
1
2
n
xi
n
272.82 2
2
xi i 1
10333.
ISE 130 Test #1
Instructions: Closed Book & Notes
Show All Work
Total Points = 100
Last Name: First Name: 50 W
Summer 2017
Points
1a) Decide whether a discrete or continuous random variable is the best model for (4)
each of the following variables:
(i
ISE 130 Test #2 Summer 2015
Instructions: Closed Book & Notes
Show All Work
Total Points = 100
Name: 309 ll 94 QM!
Points
1a) The size of silver particles in photographic emulsion is known to have a (12)
a log normal with parameters 0 = -7.7 and co = 1.26
ISE 130 Probability and Statistics for Engineers
Obtaining probabilities from a cumulative distribution function (CDF)
Let X be a discrete random variable and F(x) be its cumulative distribution function.
Then, for any real numbers a and b such that a < b
Homework #4 Solution
4-53.
a) The distribution of X is f(x) = 10 for 0.95 < x < 1.05. Now,
0,
FX ( x) 10 x 9.5,
1,
b)
x 0.95
0.95 x 1.05
1.05 x
P ( X 1.02) 1 P ( X 1.02) 1 FX (1.02) 0.3
c) If P(X > x) = 0.90, then 1 F(X) = 0.90 and F(X) = 0.10. Therefore
HW #2 Solution
3-52.
The sum of the probabilities is 1 and all probabilities are greater than or equal to zero;
f(1/8) = 0.2, f(1/4) = 0.7, f(3/8) = 0.1
a) P(X1/18) = 0
b) P(X1/4) = 0.9
c) P(X5/16) = 0.9
d) P(X>1/4) = 0.1
e) P(X1/2) = 1
3- 58.
Mean and Va
Homework #4 Solution
6-6.
The mean is increased by 10 and the standard deviation is not changed. Try it!
6-8.
Sample average:
19
x
x
i
i 1
19
272.82
14.359 min
19
Sample variance:
19
x
272.82
i
i 1
19
x
2
i
10333.8964
i 1
2
n
xi
n
272.82 2
2
xi i 1
ISE 130 Quiz #2 Summer 2017
Instructions: Closed Notes and Book
Total Points = 20 .
Name: SM w
Points
1) The life of a recirculating pump follows a Weibull distribution with par meters (3)
B = 2 and 5 = 700 hours.
Determine the probability that a pump wil