Binomial Distribution X Bin(n, p)
Condence Intervals
Given n trials with independent, dichotomous outcomes, with
probability p of success, the probability of x successes is:
b(x; n, p) =
n!
x
x!(nx)! p (1
p)
0
nx
x [0, 1, ., n]
otherwise
E[X] = np and V
Daymanuel Sampson
September 9, 2011
Homework #1
1.3 Orchard Contamination from Insecticides
a. The population of interest to the researchers is any extensively grown crop that
has the possibility of receiving pesticides.
b. The sample in this case is dorm
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 2
1) Section 2.1, Exercise 2 (10 points)
You have a cubic die with the following faces: cfw_1,1,1,2,2,3.
a) The sample space, showing all possible outcomes for one roll of
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 1
1) Section 1.2, Exercise 6 (5 points)
The point of this exercise is to get you to think about how outliers aect calculations
of variance and standard deviation. The answe
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 12
1) Section 10.5, Exercise 2 (4 points)
a) X is closer to the USL, so
Cpk =
USL X
15.60 15.52
=
= 1.518
3 0.05/2.847
3R/d2
b) Since Cpk > 1, the process capability is acc
36-220 Spring 2013 36-247
Solutions for Homework 5
1) Section 3.4, Exercise 14 (10 points)
We assume that l and d are independent measurements with uncertainties l = 0.1 and
d = 0.1.
a) We compute R directly:
R=k
14.0
l
= k 2 = 0.723k .
2
d
4.4
The uncert
36-220 Spring 2013 36-247
Solutions for Homework 9
1) Section 6.3, Exercise 6 (8 points)
We are given that n = 113, p = X/n = 65/113 = 0.575, and po = 0.50. The phrase
can you conclude.diers. implies that Ha is p = po , and thus that Ho is p = po .
Since
36-220 Spring 2013 36-247
Solutions for Homework 7
1) Section 4.11, Exercise 2 (8 points)
Let the rv X represent the thickness of each sheet of paper. Then X Dist(X =
2
0.08,X = 104 ), where the distribution shape is unspecied.
2
a) By the CLT, the sum Sn
36-220 Spring 2013 36-247
Solutions for Homework 11
1) Section 7.1, Exercise 8(a-c) (6 points)
a) One may compute these quantities with Navidis equation 7.3 (after correction),
but it is far simpler to use MINITAB. Stat > Basic Statistics > Correlation yi
36-220 Spring 2013 36-247
Solutions for Homework 8
1) Section 6.1, Exercise 6 (10 points)
We are given that n = 87, X = 15.2, and sX = 1.8, and that we are testing Ho : X 15
vs. Ha : X = 15.
a) The test statistic is
T =
X o
15.2 15
=
= 1.036 ,
sX / n
1.8
36-220 Spring 2013 36-247
Solutions for Homework 10
1) Section 6.6, Exercise 6 (12 points)
We are given that pX = X/nX = 77/365 = 0.211 (participated in cleaning) and pY =
Y /nY = 23/179 = 0.128 (did not participate in cleaning). Can you conclude the
alte
Notes Set 1
Visualizing and Describing Data
Reading Assignment: Navidi, Chapter 1
There are three types of lies:
lies, damn lies, and statistics.
attributed to Benjamin Disraeli
1.1
Basic Visualization of Simple Datasets
Well begin by showing how one mig
36-220 Spring 2013 36-247
Solutions for Homework 6
1) Section 4.5, Exercise 6 (11 points)
We are given that the rv T N (99.8,0.12 ).
a) We seek P (T > 100). We standardize the rv:
P (T > 100) = 1 P (T 100)
100 T
= 1P Z
T
100 99.8
= 1P Z
0.1
= 1 P (Z 2)
36-220 Spring 2013 36-247
Solutions for Homework 3
1) Section 2.4, Exercise 2 (10 points)
a) P (X 2) = F (3) = p(0) + p(1) + p(2) = 0.4 + 0.3 + 0.15 = 0.85, where p is the pmf
and F is the cdf.
b) P (X > 1) = 1 P (X 1) = 1 p(0) p(1) = 1 0.4 0.3 = 0.3.
c)
1. You have a sample of n = 100 observations from a Weibull distribution with mean X = 20 and
2
variance X = 25. (a) (2 points) What is the (approximate) distribution of X? (Name only; no
parameter values.) (b) (2 points) What rule of statistics allows yo
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 1
1) Section 1.2, Exercise 6 (5 points)
The point of this exercise is to get you to think about how outliers aect calculations
of variance and standard deviation. The answe
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 3
1) Section 2.4, Exercise 2 (15 points)
a) P (X 2) = F (3) = p(0) + p(1) + p(2) = 0.4 + 0.3 + 0.15 = 0.85, where p is the pmf
and F is the cdf.
b) P (X > 1) = 1 P (X 1) =
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 5
1) Section 3.4, Exercise 20 (10 points)
We assume that X and Y are independent measurements with uncertainties X = 0.2
and Y = 0.5. Then U is
2
U
X
U =
2
X +
U
Y
2
2
Y
an
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 9
1) Section 6.3, Exercise 4 (9 points)
We are given that n = 304, p = 0.62, and po = 0.50. Since npo > 10 and n(1 po ) > 10,
we can test the claim that more than half pref
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 7
1) Section 5.1, Exercise 8 (10 points)
We are given the n = 53, X = 21.6, and s = 3.2. We are in the large-sample limit, so
the distribution from which the data are sampl
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 8
1) Section 6.1, Exercise 6 (10 points)
We are given that n = 87, X = 15.2, and s = 1.8, and that we are testing Ho : 15
vs. Ha : = 15.
a) The test statistic is
Z =
X o
15
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 6
1) Section 4.5, Exercise 4 (9 points)
We are given that X N (2, 9), or that X = 2 and X = 3.
a) After transformation, the lower limit is zlo = (2 )/ = 0; so P (X 2) = P (
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 10
1) Section 6.6, Exercise 2 (12 points)
We are given that pX = X/nX = 150/180 = 0.833 and pY = Y /nY = 233/270 = 0.863.
a) We are to assume that vendor B (Y ) is no bette
36-220 Engineering Statistics and Quality Control Fall 2012
Solutions for Homework 11
1) Section 10.2, Exercise 2 (5 points)
The WECO rules are not ordered in the sense that if any rule is violated, it is evidence
that the process is out of control.
First
36-220 Spring 2013 36-247
Solutions for Homework 2
1) Section 2.1, Exercise 2 (10 points)
You have a cubic die with the following faces: cfw_1,1,1,2,2,3.
a) The sample space, showing all possible outcomes for one roll of the die, is S = cfw_1,2,3.
b) As s
Notes Set 4
Propagation of Error
Reading Assignment: Navidi, Chapter 3 (skip any sub-sections whose titles includes the
word Dependent)
4.1
Measurement Error
0.2
0.0
0.1
f(x)
0.3
0.4
Lets say that you are carrying out an experiment in which you take a sin
a
,M
EM
If}:
Class Notes
38247
2013 - I
pring;
36-220 S Class Notes
m3 :3: gawk
364220 - Spring; 2013 A 136-247 Class Notes
36220 7 Sring 2013 m 36247 Worksheet for Notes Set 6: Common Continuous Distributions
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