ME 335
Engineering Statistics
Recitation #3
Chapter 3
(continued)
Discrete Random Variables
ME 335 Fall 2012
Exercise 3.14
A shelf supports 20 ceramic resistors. Of the 20
resistors, three are faulty. Six resistors are sampled
from the shelf with replacem

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Random Variables
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Associate Professor
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The Uni

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Discrete Random Variables
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Continuous Random Variables
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Descriptive Statistics
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Associate Professor & Director
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ME 335
HO08
Engineering Probability and Statistics
Homework 4
11 February 2016
Announcements
This assignment is due by 5 PM on 19 February 2016. Please place your section number
at the top of your homework.
Readings
Sections 2.8, 3.1, 3.2, and 3.3 your te

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Operations Research

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Choosing a Good Candidate Solution
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Jointly Distributed Random Variables
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ME 335
ME 335
Quiz 8 Solution
11 March 2016
QUIZ
Suppose that the random variable X has a continuous uniform distribution over the
interval [2, 6]. Determine the cumulative distribution function.
Solution
A continuous uniform distribution of random variab

ME 335
Engineering Statistics
Recitation #2
Chapter 2
(continued)
Conditional Probability
ME 335 Fall 2012
Exercise 2.27
Consider two boxes:
Box A contains 1 black and 3 white balls
Box B contains 3 black and 2 white balls
A box is selected at random and

ME335
Engineering Statistics
Recitation 1
TA Information
l
l
l
Alexander (Alex) Zolan
Email: [email protected]
Office hours: M 12-2pm
Youngbum Hur
Email: [email protected]
Office hours: W 3:30-5:30pm
Wen Bao
Email: [email protected]
Office hours

Chapter 9 Sample Problem
Problem 1:
a)
Sample size = 54
B)
B = 1 power = 0.435
2)
Null: Mean >=100
Alternate: Mean <100
Prob<t = 0.0088 < 0.05 so we reject the null hypothesis. And mean is less than 100
Problem 3:
Prob > chisq = 0.5112 so we can not rejec

Homework 8:
Problem 1:
Mean (38.6845,41.3155)
Std. Dev (4.316464, 6.210966)
Problem 2:
Mean (79.10422, 82.25054)
Mean Price (79.10422 * 2.30 * 5000, 82.25054 * 2.30 * 5000) = (909699, 945881)
Problem 3:
Problem 4:
Average width ( 5.181979, 9.818021)
Probl

Chapter 6:
Problem 1:
a)
Range = 27-0 = 27
Meadian = 10
Mode = 6
Mean = 10.34
Variance = 35.54
B1 = Skewness2 = (0.7258917)2
B2 = Kurtosis + 3 = 0.083956 + 3
b) average time of 10 is not too far is not too far from the mean on the distribution of 10.34. S

Chapter 6
No need to know frequency polygon
If you dont divide by the n-1 then the estimator is biased
Better know jump to solve this chapter questions or else you are fucked.
Chapter 7
Not included in the open book test, but in closed book exam
Rand

Ch. 9 HW
Problem 1
HO: = 16 (The mean waiting time is equal to 16 minutes)
HA: > 16 (The mean waiting time is greater than 16 minutes)
= 0.05 =5
p is greater than 0.05 so we cannot reject the null hypothesis.
Problem 2
: = 15
a) Power= 0.32 b) n=156
Prob

Problem 1:
Problem2:
Normal distribution fits the histogram well, and Prob < W is 0.5183 which is really good.
Problem 3:
Gamma and Beta distributions fit very well, but gamma distribution is better since it has higher Prob > W2 than betas Prob > D.

ME 335 Chapter 3 Homework
Problem 1
a)Use JMP to find the probability of obtaining at least one 6 in four rolls of a fair die.
p(x1) = 1- p(0) = 1- 0.4823 = 0.5177
b)Now consider another game; throw a pair of dice 24 times and compute the probability of a

ME 335
HO11
Engineering Probability and Statistics
Homework 6
25 February 2016
Announcements
This assignment is due by 5 PM on March 11, 2016 (two weeks). Please note, that you
will have another homework, Homework 7, due at this time as well. So, dont wai

ME 335
HO10
Engineering Probability and Statistics
Homework 5
18 February 2016
Announcements
This assignment is due by 5 PM on February 26. Please place your section number at the
top of your homework. Turn your homework into the TA office, not Prof. Bick

ME 335
HO21
ME 335
Homework 9 Solutions
1 April 2016
Section 6.3: Frequency Distributions and Histograms
6.53, 6.56, 6.58
6.53
Histogram of Energy
Frequency
20
15
10
5
0
Engery Consumption
The data are skewed.
6.56
25
Frequency
20
15
10
5
0
20-30
30-40

ME 335
HO17
ME 335
Homework 6 Solutions
17 March 2016
Section 4.2: Probability Distributions and Probability Density Functions
4.1, 4.4, 4.11
4-1.
a) P (1 < X ) = e x dx = (e x )
1
1
= e 1 = 0.3679
2 .5
b) P (1 < X < 2.5) =
e
x
dx = (e x )
1
2.5
1
= e 1 e

ME 335
ME 335
Quiz 4 Solution
10 February 2016
Problem
The analysis of shafts for a compressor is summarized by conformance
to specifications:
444
22
520
14
If a shaft is selected at random, what is the probability that it conforms
to surface finish requi

ME 335
ME 335
Quiz 6 Solution
25 February 2016
Problem
Cumulative distribution function of a discrete random variable X:
F(x) =
P (3 <X 6) = _
!"#$!
!.&'"!(#$)
!.*!")(#$'
!.)+"'(#$+
!.',"+(#$,
&",(#
A. 0.16
B. 0.38
C. 0.22
D. 0.21
Solution:
B.
P (X=0) = f

ME 335
ME 335
Quiz 5 Solution
18 February 2016
QUIZ
You are given two coins, a coin A, with a head and tail and a coin B, with heads on
both side. You randomly select a coin and flip it once, obtaining a head. Once again, you
flip the same coin you select

ME 335
HO22
Engineering Probability and Statistics
Homework 11
07 April 2016
Announcements
This assignment is due by 5 PM on April 22, 2016 (two weeks). Please place your section
number at the top of your homework.
Readings
Sections 8.1, 8.2, 8.3, 8.4 and