Random Sampling: Independent & Identical
Norm(mu, sigma^2/square root of n)
Sample Size does NOT matter
pros and cons of different
sampling methods
Stratified
Cluster
For Normal Population,
Sample Mean is
independent of Sample
Variance
Random Sampling
CLT

Background and specialty for
each distribution, the relations
among these distributions, the
meanings of random variables
in each distribution, Mean &
Variance
With/Without replacement,
With/Without order
Each outcome has equal chance
Counting
Uniform
Per

Public Source, Survey,
Experiments, Observation
1. Collect Data
Mean ( average value)
Pros and Cons
Median ( middle value)
Source of Data
Type of Data
Measures of Central Tendency
Mode ( most-frequent value)
Quantitative & Qualitative
Descriptive Statisti

Example 1. Based on past experience, a toy retailer forecasts the potential market
share during a sale season is given by f(x).
Then, we can find the corresponding cumulative distribution, that is,
0, if x < 0;
x
4 ydy, if 0 x < 0.5;
0
F ( x) = 0.5
x
4

Question: Please show that V ( X ) =
N n k
k
n 1 .
N 1 N N
Proof. Note that V ( X ) = E ( X 2 ) 2 , where = E ( X ) . From the proof in the
textbook, = nk / N . So, the key is to find E(X2).
Based on the definition of mean, we have
n
E( X 2 ) = x2
x =0

Question: Please show that E ( X ) =
k
k (1 p )
,V (X ) =
for negative binomial
p
p2
random variable.
Proof. By the definition of mean, we have
E( X ) = x
x=k
=
( x 1)!
( x + 1 1)!
k
p k q xk =
p k +1 q x +1( k +1)
(k 1)!( x k )!
p x = k (k + 1 1)!( x +

Question: Please show that 1 XY 1 .
Proof. 1) For any two random variables X,Y
For any t R, we denote f (t ) = [ E (tX Y )]2 .
Then, f (t ) = [ E (tX Y )]2 = t 2 [ E ( X )]2 2 E ( XY )t + [ E (Y )]2
Note that f(t)0 for any t R. Thus, we must have
= 4[ E

IELM 151: Engineering Probability and Statistics
(Spring 2008)
Solution of Problem Set 3
1. (a) = 2.85, E ( x 2 ) = 10.35, 2 = E ( x 2 ) 2 = 2.2275, = 1.49
(b) Let y be the weekly profit in $. As y = 1200x 1500, the probability distribution of y is
given

IELM 151: Engineering Probability and Statistics
(Spring 2008)
Solution of Problem Set 2 (For problem10,11,15,18, please refer to page 4&5)
1. The total number of ways to receive 2 or 3 defective sets among 5 that are purchased is
3C2 * 9C3 + 3C3 * 9C2 =

IELM 151: Engineering Probability and Statistics
(Spring 2008)
Solution of Problem Set 1 (Problems 8, 10, 14, 15 are in the last 2 pages)
1. (a) sample mean = 970.25, sample median = 967.5,
and sample standard deviation = 35.8166.
(b)
2. (a) Stem-and-Leaf

IELM 151: Engineering Probability and Statistics
(Spring 2006)
Problem Set 4
Normal Distribution
1. The life of a nickel-cadmium battery produced by a manufacturer is normally distributed with a mean
of 20 hours and a standard deviation of 4 hours.
(a) Wh

IELM 151: Engineering Probability and Statistics
(Spring 2006)
Problem Set 3
Mean & Variance
1. The number of orders for installation of a computer information system arriving at an agency per week
is a random variable x with the following probability dis

IELM 151: Engineering Probability and Statistics
(Spring 2007)
Problem Set 2
Basic Probability
1. (Exercise 13 in Text Page 62) A shipment of 12 television sets contains 3 defective sets. In how
many ways can a hotel purchase 5 of these sets and receive a

IELM 151: Engineering Probability and Statistics
(Spring 2007)
Problem Set 1
1. Volumes of 20 bottles are as follows (in ml):
985, 940, 975, 1020, 940, 975, 1060, 945, 920, 980,
920, 990, 1010, 1010, 935, 945, 980, 960, 955, 960.
(a) Compute the sample me

Lecture 9 Sampling Distributions
Review
Random sampling
Sampling distribution
Sample Mean
Sample Variance
Three distributions in inferential statistics
Readings:
Textbook: Chapter 8 (Section 8.3 is optional)
IELM151/ Stuart X. Zhu
1
Review: Types

SamplingMethods
Stratified Random Sampling
Stratified
Cluster Sampling
Cluster
Convenience Sampling
Convenience
Judgment Sampling
Judgment
IELM151/StuartX.Zhu
1
Stratified Random Sampling
The population is first divided into groups of
The population is fi

Solution to Slide 16:
Define X= the demand during the replenishment leadtime
We need to find a new reorder point s to make the stockout probability no more than 5%.
Because it is costly to hold too much inventory, we just need to find a s so that P(X>s) =

Lecture 8 Continuous Probability
Distributions
Uniform
Normal
Exponential
Gamma
Readings:
Textbook: Section 6.1~ 6.7
IELM151/ Stuart X. Zhu
1
Continuous Uniform Distribution
X~ uniformly over the interval [a,b] , and
f(x)=1/(b-a), where a x b.
Mea

Solution to Example in Slide 10:
Define X = number of defective products in n trials
X ~ Bin(n, p), where n = 10, p= 0.05
(1) P (acceptance) = P( X = 0) + P( X = 1) = C10 p (1 p ) + C10 p (1 p )
0
(2) By definition,
0
10
1
1
9
= 2, 2 = 1.6 . Then, 2 = 0.

Lecture 7 Discrete Probability
Distributions
Uniform
Uniform
Bernoulli, Binomial and Multinomial
Bernoulli, Binomial and
Hypergeometric and Multivariate
Hypergeometric and
Hypergeometric
Negative Binomial and Geometric
Negative
and
Poisson as a limit

Lecture 6 Mean & Variance
Mean
Variance
Covariance and correlation coefficient
Readings:
Textbook: Sects 4.1,4.2,4.3
IELM151/ Stuart X. Zhu
1
Mean
Sum of two fair coins: Throwing them 1,000
times, Head =1, Tail =0
How to find the sample mean?
nx
1n
1
No

Solution to the example in Slide 7
11
E (Y / X ) =
00
1
y
5
x(1 + 3 y 2 )dxdy = y (1 + 3 y 2 )dy =
x
4
0
Solution to the example in Slide 15
We shall first find the marginal probability density function for both X and Y.
1 x
1 y
0
0
g ( x) = 8 xydy = 4 x