STAT 2120 Probability and Statistics for Engineers
Instructor: E.M. Carter
e-mail: [email protected]
Phone: x53569. Office: MacNaughton 516.
Place: Rozh 103
Time: MWF 8:30-9:20
Office Hours: W and F 9:30-10:30
Textbook: Miller and Freunds Probability an
1. The mean of 1,4,5,6, 8 is
A) 4.2
B) 4.4
C) 4.6
D) 4.8
E) 5.0
2. The median of 1,4,5,6,8 is
A) 4
B) 5
C) 6
D) 7
E) 8
3. IF P(A) =0.4 and P(B)=0.5 and A and B are independent events P( A or B) =
A) 0.7
B) 0.8
C) 0.9
D) 1.0
E) none of the above
4. If the
Stat2120
Assignment 4
Name:_
Due: Wednesday March 6 Noon.
1. A sample of 40 beams had a breaking strength of 1020 psi and a standard deviation of 100 psi.
A new process produced a sample of 60 beams with a mean breaking strength of 1000 psi and a
standard
Stat2120
Assignment 4
Name:_
Due: Wednesday March 6 Noon.
1. A sample of 40 beams had a breaking strength of 1020 psi and a standard deviation of 100 psi.
A new process produced a sample of 60 beams with a mean breaking strength of 1000 psi and a
standard
Stat2120
Assignment 3
Name:_
Due: Wednesday February 27 Noon.
1. For Z a standard normal random variable:
a) P(Z < 1) = _0.8413_ P(Z>-0.5)= _0.6915
c) P(-1.5 < Z < 1 ) = 0.8413 - 0.0668 = 0.7745 P( 0.5 < Z < 1) = 0.1498
2. Suppose scores on a test are nor
Stat2120
Assignment 3
Name:_
Due: Wednesday February 27 Noon.
1. For Z a standard normal random variable:
a) P(Z < 1) = _ P(Z>-0.5)= _
c) P(-1.5 < Z < 1 ) = _ P( 0.5 < Z < 1) = _
2. Suppose scores on a test are normally distributed and have a mean of 70 a
Stat2120
Assignment 2
Name:_
Due: Wednesday February 6 Noon.
1. Roll two four sided dice. Calculate the sum of the numbers.
a) What is the sample space?
S=
b) Let A be the event the sum is 5.
What is P(A);_
c) Let B be the event the sum is less than 6.
Th
Stat2120
Assignment 2
Name:_Answers_
Due: Wednesday February 6 Noon.
1. Roll two four sided dice. Calculate the sum of the numbers.
a) What is the sample space?
S= cfw_(1,1)(1,2)(1,3)(1,4)(2,1)(2,2)(2,3)(2,4)
(3,1)(3,2)(3,3)(3,4)(4,1)(4,2)(4,3)(4,4)
b) Le
Comparison of Two Means
Chapter 8
Control versus a new Treatment
1 is the mean response of the treatment
and 2 is the mean response of the control.
Estimation of 1 - 2 with 95% confidence
interval
Test of hypothesis: H0: 1 - 2 = 0.
Design of Experiments
Lecture 15
t and z-tests and sample size
determination
Example
Question 7.62 page 235.
A sample of n=6 steel beams had a
compressive strength of = 58,392 psi with a
standard deviation of s=648 psi. It is
hypothesized that the beams came from a
process w
Lecture 14
Test of Hypothesis
5 Steps
There are 5 steps in hypothesis testing.
Step 1: The null hypothesis.
We consider the case of a single population
mean we wish to make a statement about.
Step 1
So assume that the historic output under a
process h
Lecture 12
Estimation
Review
If p was 0.4 what would the mean and
variance of X be .
E(X) = np = 400. 2 = np(1-p) = 240.
So =15.5.
Hypothesis Testing
We should expect X to be within 1.96 standard
deviations of its mean 95% of the time.
So X should be
Lecture 11
Binomial and Math
Binomial Distribution
X has a binomial distribution with mean =np
And variance 2 = np(1-p).
If np and n(1-p) are both greater than 5 then
we can approximate the binomial with a
normal curve.
Mean and Variance
So take n=16
Lecture 10
Sampling Distributions
Uniform
Consider a population with three values
0 1 and 2 with equal probability
The distribution mean and variance are in the
following table.
Distribution
x
0
1
2
Sum
p(x)
1/3
1/3
1/3
1
xp(x)
0
1/3
2/3
=1
(x-)
1
0
1
Lecture 9
Continuous Distributions
and the normal curve
Examples
Continuous data:
Examples: weight, time, height.
We talk about the probability of between 50
and 60 kg. rather than exactly 5.11234567 kg.
So someone who is five foot six inches tall is
Lecture 8
Other Distributions
Examples
Consider three outcomes x=0,1,2 with equal
probability. The resulting probability function
is given.
We need to calculate the mean and variance
of the distribution.
Uniform
x
0
1
2
Sum
p(x)
1/3
1/3
1/3
1
xp(x)
0
1/
Lecture 7
Binomial Distribution
Random Variables.
Consider n=2 people sampled and asked if they
agree or disagree with a statement. Response is
Yes or No basically. Assume independent
responses.
P(Y) = p
P(N) = q = 1-p
Outcomes
S = cfw_(N,N),(N,Y),(Y,N)
Lecture 6
Probability Distribution
Two Way Table
Consider a test (Positive or negative) and an
outcome Yes or No (disease, earthquake,
failure etc).
Test\Disease
B (Yes)
(No)
A (yes)
98
1
(No)
2
99
Total
100
100
Total
200
Sensitivity and Specificity
P
Lecture 6
Conditional Probability
Independent Events
A and B are independent events if and only if P(A
and B)=P(A)P(B).
Example: a mother has two children. Assume the
probability of a boy is 0.5.
Let A be the event that the first child is a boy and
B be
Lecture 6
Conditional Probability
Independent Events
A and B are independent events if and only if P(A
and B)=P(A)P(B).
Example: a mother has two children. Assume the
probability of a boy is 0.5.
Let A be the event that the first child is a boy and
B be
Lecture 4
Counting
How many three letter words from three letters
A,B,C only using the letters once.
A - B - C (A,B,C)
A - C - B (A,C,B)
B - A - C (B,A,C)
B - C - A (B,C,A)
C - A - B (C,A,B)
C - B - A (C,B,A)
Counting
So 6 ways.
First letter we have
STAT2120
Probability and Statistics for
Engineers
Population and Sample
A unit is an object or person whose
characteristics are of interest.
Examples: people, wind turbines, animals,
plants etc.
A population of units is the complete
collection of units