Base Stock Model Essay

Base Stock Model Essay
The Base Stock Model
1
Assumptions
 Demand occurs continuously over time  Times between consecutive orders are stochastic but
independent and identically distributed (i.i.d.)  Inventory is reviewed continuously  Supply
leadtime is a fixed constant L  There is no fixed cost associated with placing an order  Orders
that cannot be fulfilled immediately from on–hand inventory are backordered
2
The Base–Stock Policy
 Start with an initial amount of inventory R. Each time a new demand arrives, place a
replenishment order with the supplier.  An order placed with the supplier is delivered L units of
time after it is placed.  Because demand is stochastic, we can have multiple orders (inventory on–
order) that have been ... Show more content on Helpwriting.net ...
 E[X ]  1  Pr( X  x )   x Pr( X  x )  1   x 1
18
Example 2 (Continued...)
The optimal base–stock level is the smallest integer R* that satisfies
Pr( X  R * )  b bh ln[ b ] b  h 1 ln[  ]
 1 
R * 1
b   R*  bh
b   ln[ ]  * bh  R   ln[  ]    
19
Computing Expected Backorders

 It is sometimes easier to first compute (for a given R),
E[I ] 

R x0
( R  x ) Pr( X  x )
and then obtain E[B]=E[I] + E[X] – R.  For the case where leadtime demand has the Poisson
distribution (with mean  = E(D)L), the following relationship (for a fixed R) applies E[B]=
Pr(X=R)+(–R)[1–Pr(X R)]
... Get more on HelpWriting.net ...

Ilab Week 6 Math 221 Essay
Elementary Statistics iLab Week 6
Statistical Concepts: * Data Simulation * Discrete Probability Distribution * Confidence Intervals
Calculations for a set of variables
Mean Median
3.2 3.5
4.5 5.0
3.7 4.0
3.7 3.0
3.1 3.5
3.6 3.5
3.1 3.0
3.6 3.0
3.8 4.0
2.6 2.0
4.3 4.0
3.5 3.5
3.3 3.5
4.1 4.5
4.2 5.0
2.9 2.5
3.5 4.0
3.7 3.5
3.5 3.0
3.3 4.0
Calculating Descriptive Statistics
Descriptive Statistics: Mean, Median
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
Mean 20 0 3.560 0.106 0.476 2.600 3.225 3.550 3.775 4.500
Median 20 0 3.600 0.169 0.754 2.000 3.000 3.500 4.000 5.000

Calculating Confidence Intervals for ... Show more content on Helpwriting.net ...
The mean for the column "mean" is 3.56. It is very close to the parameter of interest but is not equal
to it. You can calculate a confidence interval for the mean of the mean column, but a specific
confidence interval would need to be provided. In that case, the confidence interval would be
centered on 3.56, not 3.5. |
4. Give the mean for the median column of the Worksheet. Is this estimate centered about the
parameter of interest (the parameter of interest is the answer for the mean in question 2)
The mean for the median column is 3.6, which is close to the mean in question 2 but not as close as
the answer in question 3. |
5. Give the standard deviation for the mean and median column. Compare these and be sure to
identify which has the least variability?
Standard Deviation of Mean= 0.4762Standard Deviation of Median= 0.7539The standard deviation
of the Mean is smaller, which means all of the data points will tend to be very close to the Mean.
The Median with a larger Standard Deviation will tend to have data points spread out over a large
range of values. Since the Mean has the smaller value of the Standard Deviations, it has the least
variability. |
6. Based on questions 3, 4, and 5 is the mean or median a better estimate for the parameter of
interest? Explain your reasoning.
The Mean seems to be the better estimate as

Questions On Faults And Degradation
section{Simple POdtSHS System Example} label{sec:shsexample} Since we are interested in faults
and degradation, the system will have three modes and one continuous state, i.e.: egin{itemize}
item $mathcal{Q}={q_1,q_2,q_3}={mathrm{OK},mathrm{FAULTY},mathrm{BROKEN}}$ item
$n(q_1)=n(q_2)=n(q_3)=1$ end{itemize} The modes signalize the faults of the system, with $q_1$
being the faultless mode, $q_2$ denoting a fault in the system that compromises its operation and
finally $q_3$ denoting completely broken system that is incapable of operating at all. The
continuous state $x$ denotes the degradation level of the equipment and can have any value from
interval $[0,100]$. The outputs will be observed degradation level and observed mode. Thus the
observation vector will be a two–element vector $y=(y_1,y_2)$ with $y_1in,[0,100]$ meaning an
observed (estimated) level of degradation, e.g. from computing some key performance indicator
(KPI) from sensor measurements. The
$y_2in{mathrm{OK},mathrm{FAULTY},mathrm{BROKEN}}$ is again observed (or estimated)
mode, e.g. from some fault detection engine in place which itself aggregates and transforms sensor
readings into this one discrete signal. The outputs are subject to sensor noise etc., so they do not
perfectly correlate with the true system state. The actions will depict the system maintenance and
thus the action space $mathcal{U}$ will be a three–element set ${u_1,u_2,u_3}$ with the first
element meaning doing no maintenance action, the

Queuing Theory : Queuing Systems
Queuing Theory: Queuing theory is described as the study of waiting lines (Render et al, 2015).
Believe it or not it is a theory we use daily. Some instances you may encounter applying this theory,
could be when deciding on which line to wait on when making a purchase or when initiating a
phone call for service to be placed on hold. Sometimes when being placed on hold in queue the
company, for example Comcast, may tell that you are the fifth person on hold or in queue; they may
even give you an estimated wait time for your call to be answered by a representative, such as your
estimated wait time is 8 minutes, you are the third person in queue. You then decide will you wait
the 8 minutes or try again later. Well just how you may use this theory in this example and other
more simplistic daily decisions, managers utilize this same process to help evaluate the cost and
effectiveness of service system (Render et al, 2015). Queuing Characteristics: Queuing system has
three characteristics they are: arrivals of inputs to the system, the waiting list and the service facility
(Render et al, 2015). A simulation modeling process is based mainly on feeding the quantitative data
into a model to produce quantitative results in a structured sequential process (Eldabi, 2002). This
method assists managers by allowing them the opportunity to create a simulation model to see the
various advantages and disadvantages of any changes they would like to integrate into their

Essay about Ilab Week 6 Devry
Statistics – Lab #6
Name:__________
Statistical Concepts: * Data Simulation * Discrete Probability Distribution * Confidence Intervals
Answer:
Calculating Descriptive Statistics
Answer:
Mean 20 0 3.560 0.106 0.476 2.600 3.225 3.550 3.775 4.500
Median 20 0 3.600 0.169 0.754 2.000 3.000 3.500 4.000 5.000
Calculating Confidence Intervals for one Variable
Answer:
Mean 20 0 3.560 0.106 0.476 2.600 3.225 3.550 3.775 4.500
Median 20 0 3.600 ... Show more content on Helpwriting.net ...
Is this estimate centered about the parameter of interest (the parameter of interest is the answer for
the mean in question 2)?
The mean for the median column of the worksheet is 3.6Yes, the estimate is centered about the
parameter of interest. |
5. Give the standard deviation for the mean and median column. Compare these and be sure to
Standard Deviation for the mean column is 0.476Standard Deviation for the median column is

0.754Standard deviation for the mean column has least variability |
6. Based on questions 3, 4, and 5 is the mean or median a better estimate for the parameter of
interest? Explain your reasoning.
Mean is the better estimate for the parameter of interest. The mean is more centered with least
variability. |
7. Give and interpret the 95% confidence interval for the hours of sleep a student gets.
One–Sample T: Sleep Variable N Mean StDev SE Mean 95% CISleep 20 6.950 1.572 0.352 (6.214,
7.686)95% of students get between 6.2 hour to 7.69 hour of sleep. |
8. Give and interpret the 99% confidence interval for the hours of sleep a student gets.
One–Sample T: Sleep Variable N Mean StDev SE Mean 99% CISleep 20 6.950 1.572 0.352 (5.944,
7.956) 99% of students get between 5.9 hour to 7.95 hour of sleep. |
9. Compare the 95% and 99% confidence intervals for the hours of sleep a student gets. Explain

PracticeS8 Essay
A) mean, median, SD, correlation, histograms 1) Find the mean, median, SD, IQR for the following
data set: 5 5 8 10 12 15 15 19 20 21 What are the values of the SD IQR ? 2) Find the mean,
median, SD IQR for the data in (1) after it has been transformed as follows: new value = 2.8(old
value) – 7.2 Which statement is true ? a) all four measures change so that new value = 2.8(old
value) – 7.2 b) the mean, median IQR change so that new value = 2.8(old value) – 7.2, but new
SD = 2.8(old SD). c) the mean median change so that new value = 2.8(old value) – 7.2, but the
SD IQR change so that new value = 2.8(old value) 3) Find the height of the density ... Show more
content on Helpwriting.net ...
What was the average tip charge? a) $8.75 b) $9.00 c) $9.25 d) $9.75 e) $10.40 (19) In problem (18)
suppose the standard deviation of bill amounts was $18. What was the standard deviation of tip
sizes? a) 1.18 b) 1.36 c) 1.54 d) 1.88 e) 2.70 (20) What can you say about the correlation r between
heights of husbands wives? Bob (height = avr. husband height) Bev (height way above avr. wife
height) 0 Joe (much taller than avr. husband) Jen (much shorter than avr. wife) + Lou (height = avr.
husband height) Ann (height = avr. wife height) 0 Ted (shorter than avr. husband) Deb (slightly
taller than avr. wife) + Gus (much shorter than avr. husb.) Sue (height = avr. wife height) 0 a) r 0
b) r = 0 c) r 0 d) any of these could be true. (21) Which one of the following statements is true
about the correlation coefficient, r ? a) r does not depend on the units of measurement (inches, feet,
meters, etc.) b) r is not affected by the presence of outliers c) r is always between 0 and 1 d) a high (
+ or – ) r is proof of a cause effect relationship between x y e) none of the above statements are
true (22) The mean age of the 30 members of the TC Jets is 28. After the current season, five
members whose mean age is 35 will retire. They will be replaced by five new members whose mean
age is 24. What will the new mean age of the Jets be ? a) 25.88 b) 26.17 c) 26.52 d) 27.12 e) 27.38
B) normal

Math Midterm Essay
Deterministic techniques assume that no uncertain exists in model parameters.
A: True
An inspector correctly identifies 90% of the time. For the next 10 products, the probability that he
makes fewer than 2 incorrect inspections is .736.
A: Use Binomial table to discover , add 3 probabilities for 0,1,2
A continuous random variable may assume only integer values within a given interval.
A: False
A decision tree is a diagram consisting of circles decision nodes, square probability nodes and
branches.
A: False
A table of random numbers must be normally distributed and efficiently generated
A: False
Simulation results will always equal analytical results if 30 trials of the simulation have been
conducted.
A: False
Data cannot ... Show more content on Helpwriting.net ...
A: .01
Coefficient of determination is the percentage of the variation in the _ variable that results from the
_ variable.
A: Dependent/independent
In exponential smoothing the closer alpha is to _ the greater the reaction to the most recent demand.
A:
_ is a linear regression model relating demand to time.
A: linear trend
_ is a measure of the strength of the relationship between independent and dependent variables.
A: Double check definition possibly coefficient of determination or correlation
Consider the following graph of sales: which of the following characteristics is exhibited by the
data:
A: trend plus seasonal
Which of the follow characteristics is exhibited by the data
A: None of the above
_ is the difference between the forecast and actual demand
A: forecast error
Given the following data on number of pints of ice cream sold at a local ice cream store for a 6
month period time frame:

If the forecast for period 5 is equal to 275 use exponential smoothing a to compute a forecast period
7.
THe Drying rate in an industrial process is dependent on many factors and varies according to the
following distribution.
Compute the drying time. Use two places after the decimal.
A: takes values of variable and multiply with the relative frequency then add them up to get answer.
Life insurance company .... With a mean of 68 years and a standard deviation of 4 years. What
proportion of the plan recipients would receive payments beyond age 75?

Noise in Electronic Communications Systems
An additive noise is characteristic of almost all communication systems. This additive noise
typically arises from thermal noise generated by random motion of electrons in the conductors
comprising the receiver. In a communication system the thermal noise having the greatest effect on
system performance is generated at and before the first stage of amplification. This point in a
communication system is where the desired signal takes the lowest power level and consequently
the thermal noise has the greatest impact on the performance. This characteristic is discussed in
more detail in Chapter 10. This chapter's goal is to introduce the mathematical techniques used by
communication system engineers to characterize and predict the ... Show more content on
Helpwriting.net ...
It turns out that W(t) is accurately characterized as a stationary, Gaussian, and white random
process. Consequently, our first task is to define a random process (Section 9.1). The exposition of
the characteristics of a Gaussian random process (Section 9.2) and a stationary Gaussian random
process (Section 9.3) then will follow. A brief discussion of the characteristics of thermal noise is
then followed by an analysis of stationary random processes From this point forward in the text the
experimental outcome index will be dropped and random processes will be represented as N(t).
EXAMPLE 9.1 A particular random process is defined as N(t) = U exp[−|t|] + V (9.1) where U and
V are independent random variables. It is clear that with each sample value of the random variables
U(ω) and V (ω) there will be a time function N(t, ω). This example of a random process is not
typical of a noise process produced in real communication systems but it is an example process that
proves insightful as we develop tools to characterize noise in communications. EXAMPLE 9.2 A
noise generator and a lowpass filter are implemented in Matlab with a sample rate of 22,050 kHz.
Recall each time Matlab is run this is equivalent to a different experiment outcome, i.e., a different
ω. A sample path of the input noise to the filter, W(t), and a sample path at the output of the filter,
N(t), is shown in Figure 9.3 for a filter with a bandwidth of 2.5 kHz. It is

Quantitative Techniques
DESCRIPTIVE STATISTICS PROBABILITY THEORY
1. Consider the following data: 1, 7, 3, 3, 6, 4 the mean and median for this data are a. 4 and 3 b. 4.8
and 3 c. 4.8 and 3 1/2 d. 4 and 3 1/2 e. 4 and 3 1/3
2. A distribution of 6 scores has a median of 21. If the highest score increases 3 points, the median
will become __. a. 21 b. 21.5 c. 24 d. Cannot be determined without additional information. e. none
of these
3. If you are told a population has a mean of 25 and a variance of 0, what must you conclude? a.
Someone has made a mistake. b. There is only one element in the population. c. There are ... Show
more content on Helpwriting.net ...
All of the other passengers are sober, and will go to their proper seats unless it is already occupied;
In that case, they will randomly choose a free seat. You 're person number 100. What is the
probability that you end up in your seat (i.e., seat #100)?. Solution: Let's consider seats #1 and #100.
There are two possible outcomes: E1: Seat #1 is taken before #100; E2: Seat #100 is taken before
#1. If any passenger takes seat #100 before #1 is taken, surely you will not end up in you own seat.
But if any passenger takes #1 before #100 is taken, you will definitely end up in you own seat. By
symmetry, either outcome has a probability of 0.5. So the probability that you end up in your seat is
50%. Explanation: If the drunk passenger takes #1 by chance, then it's clear all the rest of the
passengers will have the correct seats. If he takes #100, then you will not get your seat. The
probabilities that he takes #1 or #100 are equal. Otherwise assume that he takes the n–th seat, where
n is a number between 2 and 99. Everyone between 2 and (n–1) will get his own seat. That means
the n–th passenger essentially becomes the new drunk guy with designated seat #1. If he chooses
#1, all the rest of the passengers will have the correct seats. If he takes #100, then you will not get
your seat. (The probabilities that he takes #1 or #100 are again

The Multiplicative Laws Of Probability
QUESTION 1 Probability Show all calculations/reasoning
Question: 1(a)
Basic Laws of Probability
The additive law of probabilities
Probability is known as mutually exclusive events, The sum of Separate probabilities likely to be
one event occur or another.
Example:
Place 100 marbles in a box; 35 blue, 45 red, and 20 yellow.
P(blue)=.35 P(red)=.45 P(yellow)=.20
What is the probability of choosing either a red or a yellow marble from the box?
P(red or yellow) = P(red)+ P(yellow)
= .45+.20
= .65
The multiplicative law of probabilities
The multiplicative law of probability is defined as the probability of the joint occurrence of two or
more of the events which are independent events or the product of ... Show more content on
Helpwriting.net ...
3. Calculate from the table, showing your calculation methods:
The marginal probability that any person selected at random from the population is a male.
The marginal probability that any person selected at random from the population is aged between 25
and 54.
The joint probability that any person selected at random from the population is a female and aged
between 55 and 64.
The conditional probability that any person selected at random from the population is 25 or over
given that the person is a male.
QUESTION 2 Statistical Decision Making and Quality Control Show all calculations/reason
Question: 2(a)
1. If management wishes to establish x ̅ control limits covering the 95% confidence interval,
calculate the appropriate UCL and LCL. UCL = 46.63 LCL = 43.36 2. If management wishes to use
smaller samples of 16 observations calculate the control limits covering the 95% confidence
interval. (Round calculations to 2 decimal

Tutorial on Probablity
Mathematics Department
Tutorial Sheet No. 3
MAL250(Probability and Stochastic Processes)
1. The percentage of alcohol (100X ) in a certain compound may be considered as a random
variable, where
X (0 X 1) has pdf fX (x) = 20x3 (1 − x), 0 x 1.
Suppose that the selling price of the above compound depends on the alcohol contents. Specifically,
if
1/3 X 2/3, the compound sells for c1 dollars/gallon otherwise it sells for c2 dollars/gallon. If the
cost is c3 dollars/gallon, find the probability distribution of the net profit per gallon.
2. The pdf of a random variable X is given by fX (x) = 6x(1 − x), 0 x 1.
Find the distribution of
(i) Y = X/X+1

 2X, −∞ X 1/4
1
, 1/4 ≤ X 3/4
(ii) Y =
2
2
, 3/4 ≤ X ∞. ... Show more content on Helpwriting.net ...
Also find the variance of the profit P.
17. Suppose that an electronic device has a life length X (in units of 100 hours) which is considered
as a continuous random variable with the pdf fX (x) = e−x , x 0. Suppose that the cost of
manufacturing such item is $2.00. The manufacturer sells the item for $5.00, but guarantees a total
refund if X 0.9. What is the manufacturer's expected profit per item?. 18. Suppose that X is a
continuous r.v. having the following pdf: f (x)
=
=
ex if x ≤ 0
2
e−x if x 0.
2

Let Y = |X |. Obtain E(Y ) and Var(Y ).
19. A company rents out time on a computer for periods of t hours,for which it receives $400 an
hour. The number of times the computer breaks down during t hours is a r.v. having the Poisson
distribution with λ = (0.8)t, and if the computer breaks down x times it costs 50x dollars to ﬁx it.
How should the company select t in order to maximize its expected proﬁt?

Advantages And Disadvantages Of Stochastic Model
3.1 Deterministic models There are two types of model that we are going to look at, firstly the
deterministic model and then the stochastic model. [23]A deterministic model is used in a situation
where the result can be established straightforwardly from a series of conditions. It has no stochastic
elements and both the input and the outputs are determined conclusively. On the other hand a
stochastic model is one where the cause and effect relationship is stochastically or randomly
determined. Therefore the system having stochastic element is generally not solved analytically and
hence there are several cases for which it is difficult to build an intuitive perspective. When
simulating a stochastic model a random number is usually generated ... Show more content on
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This is illustrated in figure 3 below. The chain ladder method explicitly relies on the assumption that
the expected cumulative losses settled up to and including the development year divided by the
expected cumulative claims losses settled up to and including the previous development year hold
for all claim occurrence years. 3.1.4. A Loss development data Let us consider a range of risks and
assume that each claim of the portfolio is settled either in the accident year or in the following n
development years. The data can be modelled by cumulative losses and incremental losses. 3.1.4. B
Incremental losses Let CI,J where i, j Ɛ{1.2...n} (a) represent incremental losses of accident year i
which is settled with a delay of j years and therefore in development year j. Let us also assume that
incremental losses C I, j are observable for calendar years i + J ≤ n and are non–observable for
calendar years i + J ≥ n + 1. The runoff triangle below shows the incremental losses for accident
years 2000 developing over 10 years. In this case the incremental loss for 2000 development year 5
(C2000,5) is given by 89837.06 1 2 3 4 5 6 7 8 9 10 2000 24698 58384 112485 61605 89837 36174
22525 48206 19747

An Integrated Energy And Environmental Management System...
The manager of an integrated energy and environmental management system is responsible for
allocating carbon–emission treatment amount to each carbon emitter within a multi–period horizon.
In a regional energy system, there are some limitations for total emission amounts set up by the
emission constraints in certain periods. If carbon emission is larger than the emission target, the
over–limit part should be treated by carbon mitigation facilities. In period k, carbon emission
amounts should be no more than the summation of carbon emission permit and the surplus credits
from previous period (k −1), in which surplus of emission credits are to be transferable to a
consecutive period (Li et al. 2012). Moreover, carbon emission inventory of different emitters may
vary from each period under the different operating condition, which can be expressed as a random
variable, the study system can be formulated through a MSP approach. Thus, in a MSP model for
carbon emission management system, the objective is to maximize system benefits while satisfying
the total carbon emission requirement. Specifically, the economic objective can be formulated as
follows: （1a） where are decision variables that are reflected in the objective function and
determined by the capacity of carbon capture measures.
During a certain period, there is a maximum limit for total carbon emission amount (1b) (1c)

Monte Carlo Simulation : A Computerized Mathematical...
I. Introduction The Monte Carlo Simulation is a computerized mathematical technique that allows
people to account for risk in quantitative analysis and decision. It furnishes the decision–maker with
a range of possible outcomes and probabilities that they will occur for any chance of action. It
shows the extreme possibilities of things as well. The system calculates results over and over, each
time using a different set of random values from the probability functions. The simulation could
involve tens and thousands of recalculations before its complete. There will be many different
probability distributions. In this exploration I will find how to use the Monte Carlo Simulation in
order to find future stock prices for the Gold Share Market (GLD). II. Exploration First, we need to
go in depth and see what we need to find in order to start the simulation. Each day, the the price of
an asset, such as a stock is: Today's Stock Price= Yesterday's Stock Price x er r=periodic daily
return, the rate that the asset increased or decreased that day. Because the rate of return on an asset is
a random number, to model the movement to determine possible future values, a formula is needed
that model random movements. This was first done about 100 years ago by Louis Bachelier who
first applied the Brownian Motion; a formula used to model random movements and physics to the
movement of the price of an asset. His work expanded on the Black Scholes Finance formula, and
some of

Search Theory, Linear And Random Walk Moving Target On The...
Abstract. This paper presents the cooperation between two searchers at the origin to seek for a
random walk moving target on the line. Any information of the target position is not available to the
searchers all the time. Rather than finding the conditions that make the expected value of the first
meeting time between one of the searchers and the target is finite, we show the existence of the
optimal search strategy which minimizes this first meeting time. The effectiveness of this model is
illustrated using numerical example.
Mathematics Subject Classification: 37A50, 60K30, 90B40.
Keywords: Search theory, Linear search, Random walk. 1.Introduction The prime focus of searching
for a randomly moving particle on the real line is to ... Show more content on Helpwriting.net ...
Recently, Mohamed et al. [15] studied this problem when the target moves on one of n–intersected
real lines in which any information of the target position is not available to the searchers all the
time. Mohamed et al. formulate a search model and find the conditions under which the expected
value of the first meeting time between one of the searchers and the target is finite. Furthermore,
they showed the existence of the optimal search plan that minimizes the expected value of the first
meeting time and found it. More recently, El–Hadidy [16] considered a search problem for a d–
dimensional Brownian target that moves randomly on d–space. He found the conditions that make
the expected value of the first meeting time between one of the searchers and the target is finite. In
addition, he showed the existence of the optimal search plan that minimizes the expected value of
this first meeting time. A comprehensive discussion of many aspects of search problem is found in
El–Hadidy et al. [18–30]. Finding a Random walk moving target inside a cylinder has numerous
applications in physics. One of the famous search methods, is the coordinated linear search method
which consider two unit–speed searchers starting at the origin point seek for a random walk target. A
target is assumed to move randomly on a line according to a

Case Study Of Monte Carlo Simulation
Mental disorder can be classified into 4 categories: depression, bipolar affective disorder,
schizophrenia and other psychoses and dementia. According to World Health Organization,
depression is one of the most common mental disorders worldwide. Globally, about 400 million
people of all ages suffer from depression. More women are affected than men. (WHO, Mental
disorders, 2014) In Malaysia, some national surveys were conducted in community households by
trained medical professionals. According to the survey, a mental health problem in year 1996 was
10.7%. (The Second National Health and Morbidity Survey (NHMS II) 1996, 1996) It had increased
from to 11.2% in year 2006. (The Third National Health and Morbidity Survey NHMS III 2006,
2008) In ... Show more content on Helpwriting.net ...
It shows all possible outcome and of decisions and assess the impact of risk and allows for better
decision making under uncertainty. The technique is used by professionals in such widely disparate
fields including operational research, finance, insurance, medicine and engineering. Monte Carlo
simulation is used to solve both probabilistic and deterministic problems. In the case of a
probabilistic problem a simple Monte Carlo approach can be used to observe the random numbers,
which is chosen in such a way that they directly simulate the physical random processes of the
original problem, and to assume the preferred solution from the behavior of these random numbers.
Monte Carlo simulation has wide application in performing risk analysis by building models of
possible results by substituting a probability distribution for any factor that has inherent uncertainty.
It then calculates results by using a different set of random values over and over from the probability
functions. Monte Carlo simulation produces probability distributions of possible outcome values.
Different outcome occurred can lead to different probabilities of variables. Probability distributions
are a realistic way of describing uncertainty in variables of a risk analysis. (James, T., Reeve,T.,
Nasiri,

Managing Customer Value
Decision Sciences and Management Information Systems
MBA 608 Statistical Models for Business Decisions
Review for Class test #1
1. A summary measure that is computed to describe a characteristic from only a sample of the
population is called
a) a parameter. b) a census. c) a statistic. d) the scientific method.
2. The British Airways Internet site provides a questionnaire instrument that can be answered
electronically. Which of the 4 methods of data collection is involved when people complete the
questionnaire? a) published sources b) experimentation c) surveying d) observation
3. Which of the following is a continuous quantitative ... Show more content on Helpwriting.net ...
If X is the weight of school children sampled in a nationwide study, then X is an example of a) a
categorical random variable. b) a discrete random variable. c) a continuous random variable. d) a
parameter.
14. The manager of the customer service division of a major consumer electronics company is
interested in determining whether the customers who have purchased a videocassette recorder made
by the company over the past 12 months are satisfied with their products. The population of interest
is a) all the customers who have bought a videocassette recorder made by the company over the past
12 months. b) all the customers who have bought a videocassette recorder made by the company
and brought it in for repair over the past 12 months. c) all the customers who have used a
videocassette recorder over the past 12 months. d) all the customers who have ever bought a
videocassette recorder made by the company.
15. True or False: A continuous variable may take on any value within its relevant range even
though the measurement device may not be precise enough to record it.
16. True or False: The Coefficient of variation is a measure of dispersion
17. True or False: A statistic is usually used to provide an estimate for a usually unobserved
parameter.

18. True or False: The type of TV one owns is an example of an ordinal

Project Completion Method: Stochastic Project Scheduling...
Manuscript ID: CO/2003/022870 Specialty Area: Cost Schedule Audience: Researchers
PROBABILITY OF PROJECT COMPLETION USING STOCHASTIC PROJECT SCHEDULING
SIMULATION (SPSS) Dong–Eun Lee1 ABSTRACT This paper introduces a software, Stochastic
Project Scheduling Simulation (SPSS), developed to measure the probability to complete a project
in a certain time specified by the user. To deliver a project by a completion date committed to in a
contract, a number of activities need to be carried out. The time that an entire project takes to
complete and the activities that determine total project duration are always questionable because of
the randomness and stochastic nature of the activities' durations. Predicting a project completion
probability ... Show more content on Helpwriting.net ...
It means PERT assumes that the duration of each activity is represented by a random variable with a
known probability density function. PERT extends CPM by introducing the concept of uncertainty
in estimating activity durations. PERT uses expected mean time (te) with standard deviation or
variance. The expected mean time (te) of an individual activity is an estimate having an approximate
chance of 50 percent success. Three time estimates, i.e., the most likely (m), optimistic (a), and
pessimistic (b) durations, are required for each activity. (Khisty and Mohammadi; 2001). But PERT
has also been criticized for systematically underestimating the total project duration. That is why a
new methodology is required to increase the accuracy of scheduling project activities. As stated by
Crandall (1977), the most reliable method of predicting the total behavior or a network comprised of
probabilistic activities is simulation. To complement the PERT system, simulation can be applied to
run a network a certain number of times. Complementing the existing approaches with simulation
can reduce the errors that might be introduced by the PERT assumptions as studied by MacCrimmon
and Ryavec (1962) and Van Slyke (1963). The Probability Density Function (PDF) of the duration
of a construction activity is unknown and needs to be selected depending on the type of project. In
general, the variability of the time estimates of an activity can be assumed to follow the Beta
distribution.

MAT 540 MIDTERM EXAM Essay
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MAT 540 MIDTERM EXAM
1. Deterministic techniques assume that no uncertainty exists in model parameters.
2. A continuous random variable may assume only integer values within a given interval.
3. A joint probability is the probability that two or more events that are mutually exclusive can occur
simultaneously.
4. A decision tree is a diagram consisting of circles decision nodes, square probability nodes, and
branches.
5. A table of random numbers must be normally distributed and efficiently generated.
6. Starting conditions have no impact on the validity of a simulation model.
7. The Delphi develops a consensus forecast about what will occur in the future.
8. Qualitative methods are ... Show more content on Helpwriting.net ...
27. The drying rate in an industrial process is dependent on many factors and varies according to the
following distribution. Compute the mean drying time. Use two places after the decimal.
28. A loaf of bread is normally distributed with a mean of 22 oz and a standard deviation of 0.5 oz.
What is the probability that a loaf is larger than 21 oz? Round your answer to four places after the
decimal.
29. An investor is considering 4 different opportunities, A, B, C, or D. The payoff for each
opportunity will depend on the economic conditions, represented in the payoff table below.
Economic Condition
Poor Average Good Excellent
Investment (S1) (S2) (S3) (S4)
A 50 75 20 30
B 80 15 40 50

C –100 300 –50 10
D 25 25 25 25
If the probabilities of each economic condition are 0.5, 0.1, 0.35, and 0.05 respectively, what is the
highest expected payoff?
30. The local operations manager for the IRS must decide whether to hire 1, 2, or 3 temporary
workers. He estimates that net revenues will vary with how well taxpayers comply with the new tax
code. The following payoff table is given in thousands of dollars (e.g. 50 = $50,000).
If he uses the maximin criterion, how many new workers will he hire?
31. Consider the following distribution and random numbers: If a simulation begins with the first
random number, what would the first simulation value would be __________.
32. Given the

Essay about Math 540 Midterm
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User | | Course | Quantitative Methods | Test | Midterm Exam | Started | 2/9/13 10:35 PM | Submitted
| 2/11/13 5:07 PM | Status | Completed | Score | 150 out of 200 points | Time Elapsed | No data |
Instructions | |
Question 1
5 out of 5 points | | | Deterministic techniques assume that no uncertainty exists in model ... Show
The average number of breakdowns from the simulation trials was 1.93 with a standard deviation of

0.20. No. of breakdowns per week | Probability | Cumulative probability | 0 | .10 | .10 | 1 | .25 | .35 |
2 | .36 | .71 | 3 | .22 | .93 | 4 | .07 | 1.00 |
What is the probability of 2 or fewer breakdowns?Answer | | | | | Selected Answer: | .71 | Correct
Answer: | .71 | | | | |
Question 16
5 out of 5 points | | | rob
14, and 15)estion worth 2 points, 1 hour time limit (chapters 1,ue units EXCEPT:The U.S.
Department of Agriculture estimates that the yearly yield of limes per acre is distributed as follows:
Yield, bushels per acre | Probability | 350 | .10 | 400 | .18 | 450 | .50 | 500 | .22 |
The estimated average price per bushel is $16.80.
What is the expected yield of the crop?Answer | | | | | Selected Answer: | 442 | Correct Answer: | 442
| | | | |
Question 17
5 out of 5 points | | | __________ is a linear regression model relating demand to time.Answer | | | | |
Selected Answer: | Linear trend | Correct Answer: | Linear trend | | | | |
Question 18
0 out of 5 points | | | Consider the following graph of

Statistics
STATISTICS – Lab #6
Statistical Concepts: Data Simulation Discrete Probability Distribution Confidence Intervals
Open the class survey results that were entered into the MINITAB worksheet.
We want to calculate the mean for the 10 rolls of the die for each student in the class. Label the
column next to die10 in the Worksheet with the word mean. Pull up Calc Row Statistics and select
the radio–button corresponding to Mean. For Input variables: enter all 10 rows of the die data. Go to
the Store result in: and select the mean column. Click OK and the mean for each observation will
show up in the Worksheet.
We also want to calculate the median for the 10 rolls of the die. Label the ... Show more content on
Helpwriting.net ...
Either show work or explain how your answer was calculated.
Mean: Summation xP(x) = 1(1⁄6) +2(1⁄6) + 3(1⁄6) + 4(1⁄6) + 5(1⁄6) +6(1⁄6) = 21⁄6= μ 3.5
Standard deviation: sq. root ((1–3.5)^2 (1⁄6) + (2–3.5)^2(1⁄6) + (3–3.5)^2(1⁄6) + (4–3.5)^2(1⁄6) + (5–
3.5)^2(1⁄6) + (6–3.5)^2(1⁄6))= sq. root2.916=σ 1.707
3.) Give the mean for the mean column of the Worksheet. Is this estimate centered about the
parameter of interest (the parameter of interest is the answer for the mean in question 2)?
μ of Mean: 3.560. Yes, this is very closely centered around the parameter of interest (3.5)
4.) Give the mean for the median column of the Worksheet. Is this estimate centered about the
parameter of interest (the parameter of interest is the answer for the mean in question 2)?
μ of Median: 3.600. Yes, this too is also centered around the parameter of interest (3.5).
5.) Give the standard deviation for the mean and median column. Compare these and be sure to
σ of Mean: .0476 σ of Median: .0754

The standard deviation of the Means is smaller, thus having less variability than the Median,
meaning the data for the Means is grouped closer

Applied Statistics in Business and Economics Quiz 2 With...
University of Phoenix OnlineCourse: RES / 341QUIZ # 2(Chapter 5+6+7 from Applied Statistics in
Business and Economics )45 Questions [Each Question = 1 Point]SOLUTIONPlease mark one
answer for all multiple choice questions with RED!Chapter 51. Events A and B are mutually
exclusive whenA) the joint probability of the two events is zero.
B) they are independent events.
C) P(A)P(B) = 0D) P(A)P(B) = P(A | B)Answer: A2. Independent events A and B would be
consistent with which of the following statements:A) P(A) = .3, P(B) = .5, P(A B) = .4B) P(A) = .4,
P(B) = .5, P(A B) = .2C) P(A) = .5, P(B) = .4, P(A B) = .3D) P(A) = .4, P(B) = .3, P(A B) =
.5Answer: B3. The probability of event A occurring given event B has occurred is an example ofA)
a ... Show more content on Helpwriting.net ...
D) has none of the above properties.
Answer: B20. Which of the following distributions is not discrete?A) Binomial.
B) Geometric.
C) PoissonD) Normal.
Answer: D21. In a randomly–chosen month, which probability model would you use to describe the
number of accidents at the intersection of two streets?A) BinomialB) PoissonC) HypergeometricD)
GeometricAnswer: B22. A carnival has a game of chance: a fair coin is tossed. If it lands heads you
win $1.00 and if it lands tails you lose 50 cents. How much should a ticket cost if the carnival wants
to breakeven?A) $0.50B) $0.25C) $ 0.75D) $1.00Answer: BAt Break evenE(X) = 0X 1 –.50P (X =
x) . 5 .5E ( X) = 1 * .5 – .5 * .5 = .5 – .25 = 0.2523. A die is rolled. If it rolls to 1, 2 you win $2. If it
rolls to a 3, 4, 5, 6 you lose $1. Find the expected winnings.
A) $1B) $2C) $0.50D) $0.25Answer: AUse the following to answer 24–27The discrete random
variable X is the number of students that show up for Professor Smith's office hours on Monday
afternoons. The table below shows the probability distribution for X24. What is the E(X) for this
distribution?A) 0B) 1C) 1.5D) 2Answer: BE ( X ) = 0 * .40 + 1 *.30 + 2 *.20 + 3*.10= .30 +.40
+.30= 125. What is the probability that at least 1 student comes to office hours on any given
Monday?A) .30B) .40C) .50D) .60Answer: DP ( X ≥ 1 ) = P ( X = 1) + P ( X = 2 ) + P ( X = 3)= .30

+.20 +.10= 0.6026. What is the probability that fewer than 2 students come to office hours on any
given

Introduction for Expected Value of Sample Information...
Introduction for expected value of sample information tutor:
Expected value is the main thought in probability, in an intellect more general than probability itself.
The expected value of a real–valued selection variable offers a compute of the center of the
distribution of the variable. More considerably, by taking the expected value of various functions of
a common random variable, we can calculate a lot of interesting features of its distribution,
including spread and correlation. Tutor is a personality working in the education of others, either
separately or in group.
Formula for expected value of sample information tutor:
The following formula for expected value of sample information tutor which is used to compute
expected ... Show more content on Helpwriting.net ...
Solution: Expected value is recognized for the discrete possibility variable by utilize the formula,
E(x) = sum xi P (xi) E(x) = 0 (1/13) + 1 (1/13) + 2 (1/13) + 3(1/13) + 4(1/13) E(x) = 0 + 0.0769 +
0.1538 + 0.2307 + 0.3076 E(x) = 0.769 The Expected value is: 0.769
Expected value of sample information tutor – Example 2:
2) Evaluate the expected value to the discrete chance variable 1/6 from 2 to 7.
Solution:
The formula for finding the expected value for a discrete chance variable is E(x) = sum xi P (xi)
Here, i = 2 to 7 E(x) = 2 (1/6) + 3 (1/6) + 4(1/6) + 5(1/6) + 6(1/6) + 7(1/6) = 0.3333 + 0.5 + 0.6666
+ 0.8333 + 1 + 1.1666 = 4.4998 Answer value is: 4.4998
Additional problems for expected value of sample information tutor:
Expected value of sample information tutor – Example 3:
3) Evaluate the expected value for the discrete possibility variable. (1/18). Where x value begins
from 1 to 6.
Solution:
Expected value formula for discrete random variables: E(x) = sum xi P(xi) E(x) = 1 (1/18) + 2
(1/18) + 3 (1/18) + 4(1/18) + 5 (1/18) + 6 (1/18) E(x) = 0.0555 + 0.1111 + 0.1666 + 0.2222 + 0.2777
+ 0.3333 E(x) = 1.1664 The Expected value is: 1.1664
Expected value of sample

Monte Carlo Simulation Analysis And Decision
I. Introduction Have you ever wondered what the next Stock Prices were going to be? Did you ever
know that you could calculate these future prices? Have you heard of the Monte Carlo Simulation?
The Monte Carlo Simulation is a computerized mathematical technique that allows people to
account for risk in quantitative analysis and decision. It furnishes the decision–maker with a range
of possible outcomes and probabilities that they will occur for any chance of action. It shows the
extreme possibilities of different situations as well. The system calculates results over and over,
each time using a different set of random values from the probability functions. The simulation
could involve tens and thousands of recalculations before its ... Show more content on
Helpwriting.net ...
His work expanded on the Black Scholes Finance formula which leads to the Brownian Motion
theory made. Example Graphs of Brownian Motion Brownian Motion expresses that there are two
parts around a movement. The first is an overall cost of driving force (Drift). The second is a
random component. This component is defined as a random number to vary the results. Hence, the
rate that the asset changes in value each day, the r value that the e is raised to can be broken down
into two parts; an overall drift and a random stochastic component. Amount change in the stock
price=the expected growth over time + the effect of the constant volatility of people randomly
buying and selling the stock over time has on that expected growth. Amount change in the stock
price= fixed drift rate + random stochastic variable To create a Monte Carlo Simulator to model the
possible future outcomes, it's necessary to find the two parts around a movement. periodic return
(continuous compounding) ln (St /St–1)=α+ztσ ←random shock ^constant drift For the drift, the
expected rate of return is used. In other words, we use the rate that we expect the price to change
each day. Drift: Expected Periodic Daily Rate of Return. The expected rate is a way to change with
the greatest odds of occurring. There are different theories of what this rate

Probability Modeling And Statistics Essay
Project On Probability Modeling Statistics.
Topic : Binomial Poisson and Normal.
Please mention The Measures of central tendency
The use of these distributions (in which cases these distributions are used) with illustrations.
Binomial approximation to the normal distribution.
What is Skewness and Kurtosis? How it is used and interpreted?
Binomial Distribution :
This kind of distribution is applied to single variable discrete data where results are the number of
successful outcomes in a given scenario.
E.g. :
no. of times the lights are red in 20 sets of traffic lights,
No of students with green eyes in class of 40,
No. of plants with diseased leaves from a sample of 50 plants.
Binomial distribution is used to calculate the ... Show more content on Helpwriting.net ...
Normal Distribution:
The normal distribution is a very common continuous probability distribution. Normal distribution
is important in statistics and is often used in the natural and social sciences to represent real–valued
random variables whose distributions are not known.
This distribution is useful because of the central limit theory. In its most general form, under some
conditions which include finite variance), it states that average of random variables independently
drawn from independent distributions converge in distribution to the normal, that is, become
normally distributed when the number of random variables is sufficiently large.
The normal distribution is sometimes informally called the bell curve. However, many other
distributors are bell–shaped( such as the Cauchy, students and logistic distribution).
Measure of Central Tendency :
The Mean, Median and Mode are all valid measures of central tendency, but under different
conditions, some measures of central tendency become more appropriate to use than others.
Mean
The Mean is essentially a model of your data set. It is the value that is the most common. However,
Mean is not often one of the actual values that you have observed in your data set, but it has one of
its important properties is that it maximizes error in the prediction of any one value in your data set.
That is, it is the value that produces the lowest amount of error

Random Variable and Previous Work Experience
(ISOM2500)[2012](f)midterm1~=0zvopee^_78631.pdf downloaded by mhwongag from
http://petergao.net/ustpastpaper/down.php?course=ISOM2500id=0 at 2013–12–16 02:44:12.
Academic use within HKUST only.
Business Statistics, ISOM2500 (L3, L4 L5)
Practice Quiz I
1. The following bar chart describes the results of a survey concerning the relevance of study to
present job by school.
Focus on the School of Business and Management. What are the mode and the median respectively?
(a) Relevant, Neutral
(b) Relevant, Relevant
(c) Neutral, Relevant
(d) Neutral, Neutral
2–4. The manager of a specialty outdoor store has gathered the following data concerning sales (in
hundreds of dollars) of all the items sold by the store in the previous 2 months:
Item ... Show more content on Helpwriting.net ...
(a) (1.87, 1.1)
(b) (1.87, 8.9)
(c) (8.13, 1.1)
(d) (8.13, 8.9)
11. Two events with positive probabilities are impossible to have the following relationship:
(a) Independent and mutually exclusive.
(b) Dependent and mutually exclusive.
(c) Independent and Not mutually exclusive.
(d) Dependent and Not mutually exclusive
12. If you roll two fair dice, what is the probability that the sum on the two dice equals 6?
(a) 3/36
(b) 4/36

(c) 5/36
(d) 6/36
13. Based on the following mosaic plot, which of the following statements is false?
(a) P (B) = 0.5
(b) P (B|A) = 0.25
(d) Events A and B are independent.
3
(c) P (B|Not A) = 0.75
(ISOM2500)[2012](f)midterm1~=0zvopee^_78631.pdf downloaded by mhwongag from
http://petergao.net/ustpastpaper/down.php?course=ISOM2500id=0 at 2013–12–16 02:44:12.
Academic use within HKUST only.
14. The human resources manager at a company has classiﬁed applicants according to whether or
not they have any computer skills (Yes or No), and whether or not they have previous work
experience
(Yes or No). The results are summarized below:
Work Experience
Yes No
Yes 80
40 120
No 60
30 90
140 70 210
Computer skills
Which of the following statements is not correct?
(a) If an applicant is randomly selected and the applicant has previous work experience, then the
probability that the applicant has computer skills is about 0.57
(b) If an applicant is randomly selected, the

An Introduction Of Martingale Theory
Introduction to Martingale Theory
Martingale Theory is a simple mathematical model which models sequence of a fair game [1]. It is a
stochastic process on some probability space {Ω, F, P}. Originally it was used as a betting strategy
during 18th century in France. In 1934, Paul Lévy introduced concept of Martingale in probability
theory. It is named after 'La Grande Martingale,' which means strategy for even odd bets where bets
are doubled every time we lose. Idea behind this betting strategy is tht one cannot expect gains
without taking risks.
To understand Martingale, we need to understand Filtration first.
Filtration
Available information is modelled by a sub–σ–algebra F. A sequence of σ–algebras Fn such that F0
⊂ F1 ⊂...⊂ F. This ... Show more content on Helpwriting.net ...
Every time the person lost he had to double his bet so that the first win would help him regain from
the previous losses plus win profit equal to the original stake [2]. If X1, X2,..., is a sequence of
independent and identically distributed random variables with P(Xn = 1) = 1/2 , P(Xn = –1)= ½.
Filtration (F_n )_n Fn = σ(X1,...,Xn). Then sequence (s_n )_(n=1)^∞(simple random variable walk
on Z) is martingale w.t.r. (F_n )_n as
E(Sn | Fn–1) = E(Sn–1 + Xn | Fn–1) = Sn–1 + E(Xn | Fn–1) = Sn–1 + E(Xn) = Sn–1 [1] Poyla's Urn
– A container has balls of two color, say red and blue (r and b). We pick out one ball, observe its
color and place it back in the container along with another ball of the same color. Draw another ball
from the same container. Probability of drawing red ball in first try is r/(r+b) and that of blue ball is
b/(r+b). In second draw, probability of red ball being drawn if a red ball was drawn first time will be
(r+1)/(r+b+1). Probability of red ball being drawn if a blue ball was drawn first time will be
r/(r+b+x).
Probability of a red ball being drawn on second draw
P(Red : Draw =2) = ((r+1)/(r+b+1))*(r/(r+b))+((r/r+b+1)*(b/r+b))
= r/(r+b) [3] Example of Submartingale, Supermartingale : In a coin toss we have three coins. One
unbiased and two biased. In the two biased coins, one has P(HEADS) = ¾ and the other has
P(HEADS) = ¼. After tossing the coin if heads come, we win, tails, we lose. If we

Jet Copies Essay
JET Copies Problem The simulation of Jet Copies can be done by generating random numbers from
given probability distributions. The different steps of this simulation and assumption made are
explained below. 1. Simulation for the repair time. It is given that the repair time follows Repair
Time (days) Probability 1. .20 2. .45 3. .25 4. .10 ––––– 1.00 To generate a random number from the
above distribution, we use the following procedure. Generate a random number denoted by r2 from
between 0 and 1. If this generated random number is less than or equal to 0.2 take repair time = 1. ...
Show more content on Helpwriting.net ...
Since the break down time is given in weeks, we stop the simulation when total number of weeks
greater than 52 5. Summary Generate random number from the break down calculate breakdown
time. Generate random number from the repair distribution and calculate repair time require to get
the machine repaired. Generate random numbers between 2000 to 8000 and estimate lost revenue
due to this particular break down. Calculate the lost revenue by multiplying this number with repair
time and revenue from 1 copy ($0.1). Repeat this process until sum of all the break down time
exceeds 52 weeks and calculate the total loos. The excel output under two different assumptions are
given below. |Repair Distribution | | |P(y) |Cumulative |Repair Time | |0.2 |0 |1 | |0.45 |0.2 |2 | |0.25
|0.65 |3 | |0.1 |0.9 |4 | Break |Random |Time between Break |Random |Repair |Random |Lost
|Cumulative | |Down |r1 |downs x (weeks) |r2 |Time |r3 |Rvenue |Time Σx | |1 |0.129 |2.158 |0.520 |2
|2252 |450.4 |2.158 | |2 |0.331 |3.450 |0.092 |1 |5563 |556.3 |5.608 | |3 |0.413 |3.854 |0.642 |2 |5225
|1045 |9.462 | |4 |0.358 |3.588 |0.889 |3 |7515 |2254.5 |13.050 | |5 |0.571 |4.536 |0.606 |2 |3637

The And Of A Bernoulli Random
Abstract
Abstract here
Humans, and animals, often perceive events as being random. Many lab oratory and controlled
studies have been conducted over the last half–century to determine human ability to detect whether
a particular string of events is random. Often, these studies consist of detecting whether some binary
sequence has the properties of a bernoulli random variable. From the re sults of these studies, we
can conclude experimental subjects hold the Law of
Small Numbers (Tversky Kahneman, 1971) to be true. Particularly, many studies have found
humans have a skewed perception of and researchers do not understand randomness. In this paper,
we seek to answer three focused questions: 1. Can randomness be defined, and if so, how can
humans determine whether a sequence is random?
2. What have previous studies concluded with regard to decision makers skill in determining
whether a sequence is random?
3. Are we able to develop a rigorous, mathematical argument regarding randomness? Before we
begin to answer these questions, we will present definitions from the literature to facilitate the
discussion.
1
1 Definitions
Two ubiquitous concept used in studies of randomness are the gambler's fal lacy and the hot
hand. Using the definitions from Oskarsson, van Boven,
McClelland, and Hastie (2009), we will define the gambler's fallacy as the judgement the streak
will end. The gambler's fallacy is also known as neg ative recency. The hot hand is the judgement
that a streak will

Binomial Distribution
Binomial Distribution This is a discrete random variable, where the process of obtaining the
Binomial distribution is called Bernoulli process. An experiment that often consists of repeated
trials, each with two possible outcomes, which could be labeled as success or failure. This
experiment is known as binomial experiment. A binomial experiment is one that possesses the
following properties: 1. The experiment consists of n repeated trials. 2. Each trial has only 2
possible outcomes that can be classified as Success or Failure. 3. The probability of a success
and failure , denoted by pand q, remains constant from trial to trial. 4. The repeated, trials are
independent. Formula for ... Show more content on Helpwriting.net ...
of heads from 5 toss of a coin. a) Find the value of X. b) Calculate the value of P(X). c) What is the
probability for at least 3 heads in a try? 2. A die is thrown 5 times. Calculate the probability when
the lands at 4, if a) twice b) 3 times. 3. 25% of a local university who registered for the first year
needs additional class for mathematics. If 6 students are chosen at random, find the probability a) 1
student needs the additional class b) 2 students need the additional class c) 3 students who need the
additional class. 4. If X represents the number of broken pencils from 5 pencils chosen at random in
a box of 100 pencils, where 10 are broken. Find the probability a) P(X = 3) b) P(X ≥ 3) c) P(X ≤ 2)
5. In an examination of 10 objective questions, every one contains 5 answers but only one that is
correct. If one student that has never study, sit for examination could only answer by guessing the
right answer. Find the probability a) none of the answer are correct b) only 3 answer are correct. 6.
Probability of a man between 20–24 years married is 0.2. 20 men are chosen from the age group,
find a) the probability that 9 has gotten married b) the probability that less than 3 are already
married c) µ , the number of men from the group who has already married. 7. 40% from Kedah's
population visit Langkawi

Statistics Assignment
Week 4 Exercises
Chapter 5 – Section 1. Question 5
To perform a certain type of blood analysis, lab technicians must perform two procedures. The first
procedure requires either one or two separate steps, and the second procedure requires either one,
two, or three steps.
a. List the experimental outcomes associated with performing the blood analysis.
Answer: There are two procedures that a lab technician must perform. The first procedure requires
either one or two separate steps, which could be named as x1 and x2. The second procedure requires
either one, two or three steps, which could be named as y1, y2 and y3. The experimental outcomes
associated with performing the blood analysis from 1st procedure (x1,x2) and 2nd procedure:
(y1,y2,y3): ... Show more content on Helpwriting.net ...
What is the probability that it takes at least two sessions to gain the patient's trust?
Answer: The probability that is takes at least two sessions to gain the patient's trust is 0.83, because
Probability x≥2=f2+f3=26+36=0.8333=0.83
Chapter 5 – Section 3. Question 17
a. Let x be a random variable indicating the number of times a student takes the SAT. Show the
probability distribution for this random variable.
Answer: The probability distribution: Number of Times (x) | Number of Students | f(x) | 1 | 721,769 |
0.4752 | 2 | 601,325 | 0.3959 | 3 | 166,736 | 0.1098 | 4 | 22,299 | 0.0147 | 5 | 6,730 | 0.0044 | |
1,518,859 | 1.0000 |
b. What is the probability that a student takes the SAT more than one time?
Answer: The probability that a student takes the SAT more then one time is 0.5248, because
Probabilityxgt;1=f2+f3+f4+f5=0.3959+0.1098+0.0147+0.0044=0.5248
c. What is the probability that a student takes the SAT three or more times?
Answer: The probability that a student takes the SAT three or more times is 0.1289, because
Probabilityx≥3=f3+f4+f5=0.1098+0.0147+0.0044=0.1289
d. What is the expected value of the number of times the SAT is taken? What is your interpretation
of the expected value?
Answer: The expected value of the number of times the SAT is taken is 1.6772, because
Ex=μ=Σxfx
x | f(x) | xf(x) | 1 | 0.4752 | 0.4752 | 2 | 0.3959 | 0.7918 | 3 | 0.1098 | 0.3293 | 4 | 0.0147 |

The Classical Expected Utility Theory And The Dual Theory...
INTRODUCTION:
Identifying, defining risks (market risks as well as non–market risks), presenting and justifying a
unified framework for the analysis, construction and implementation of risk measures are important
components of insurance pricing.
According to the Oxford's advanced learners dictionary, risk can be defined as the possibility that
something uncertain (not predictable) and unpleasant will happen. Both financial and insurance
organisations are therefore faced with this concept of risk in their everyday activities. Financial risk
can be said to be the possibility that the return achieved on an investment will be different from that
expected, and also takes into account the size of the difference. Whereas insurers will define risk as
a chance of harm, damage or loss against something which is insured.
Several literatures reveal a good number of different approaches and theories to the price of risk.
The two main competing economic theories we shall consider are the classical expected utility
theory, and the dual theory of risk which was developed by Yaari(1987).
They defined the price of an insurance risk excluding other expenses as the risk adjusted premium.
The rest of this paper is structured as follows: section 2 presents the class of distortion operators
used in insurance pricing, their properties and an application of pricing by distortion. Section 3
incorporates a new pricing principle by Wang (2002) and its relevance to natural hedging.
Section 4

Sampling Techniques
Sampling Techniques Worksheet For each description of sampling, decide if the sampling technique
is A. Simple Random B. Stratified C. Cluster D. Systematic E. Convenience 1. In order to estimate
the percentage of defects in a recent manufacturing batch, a quality control manager at Intel selected
every 8th chip that comes off the assembly line starting with the 3rd, until she obtains a sample of
140 chips. 2. In order to determine the average IQ of ninth–grade students, a school psychologist
obtains a list of all high schools in the local public school system. She randomly selects five of these
schools and administers an IQ test to all ninth–grade students at the selected schools. 3. In an effort
to determine customer satisfaction, ... Show more content on Helpwriting.net ...
the average b. the standard deviation c. the first quartile d. the median e. the mode f. the third
quartile g. the 90th percentile h. the 40th percentile 1. This data is a list of three–year rates of return
of 40 small–capitalization growth mutual funds. 27.4 16.7 10.8 24.1 25.9 12.7 28.5 22.2 18.4 17.4
22.6 29.6 11.6 45.9 16.6 32.1 47.7 10.9 18.4 23.3 18.2 32 25.5 23.7 38.1 23.7 14.7 12.8 31.1 21.9
18.4 21.3 27 19.6 15.8 14.7 37 19.2 18.5 29.1 2. This data is a list of the percent of persons living in
poverty in each of the 50 US states in 1997. 14.8 9.4 13 10.4 11.2 16.3 23.4 15.2 14.1 12.5 8.5 10.1
13.3 16.4 10.7 10.0 16.6 11.7 15.1 10.5 18.8 9.1 11.6 18.4 9.7 9.6 11.8 11.4 16.7 17.5 18.4 14.3 8.2
10.7 18.6 7.7 12.3 11.9 8.3 8.5 16.8 14.7 9.6 9.3 10.6 9.2 11.8 13.1 10.9 12.7 3. The following table
gives the number of children under the age of five in 50 households. Number of children under five
Number of households 0 16 1 18 2 12 3 3 4 1 Chapter 2 – Histogram The following is a frequency
table of the number of customers who enter a Wendy's restaurant in a 15–minute interval. Complete
the Cumulative Frequency column. Number of customers Frequency Relative Frequency
Cumulative Frequency 1 1 0.025 0.025 2 6 0.15 3 1 0.025 4 4 0.1 5 7 0.175 6 11 0.275 7 5 0.125 8 2
0.05 9 2 0.05 10 0 0.0 11 1 0.025 Construct a histogram below. Make sure to label and scale both
axis. Chapter 2 – Worksheet The following is a

Reflection Paper On Inequality
The lecture presented in class (Inequality in Education and Employment) was a particularly salient
topic for me. As part of my Bachelor's degree, in Philosophy, I was required to present and defend a
thesis. My thesis was a defense of democracy from the problems of distributive justice. Over the
course of two years, I spent countless hours reading, writing, and debating the fine points on
inequality. Therefore, the subject matter is not something new for me, nor do I think there is a
simple solution.
Reflecting on the information from the lecture, there were various aspects of inequality that
coincided with my research; however, there were two areas that I thought required further
consideration and contemplation. First, nowhere in the slides, in the lecture, nor in the discussions,
was justice mentioned. I find that particularly odd as the preliminary point of any inquiry into
inequality should start from the structural foundations of equality viz., justice.
Second, I was slightly confused by the slide (23) titled Workers with Disabilities Earn Less Than
Their Peers. The slide indicated that people with disabilities (PWDs) earned 37% less pay than
people without disabilities (PWODs). While I looked at the slide's bar graph, I could visually see
that the 37% figure did not match what was on the graph. This forced me to do some quick math on
what was represented. According to the values on the slide, the actual wage gap was 21%. While
this is still significant, it is not

Advantages And Disadvantages Of Distribution-Free Control...
We present an overview of literature on nonparametric or distribution–free control charts for uni–
variate variable data. We highlight various advantages of these charts while pointing out some of the
disadvantages of the more traditional, distribution based control charts. Specific observations are
made in the course of review of articles and constructive criticism is offered so that opportunities for
further research can be identified. Connections to some areas of active research are made, such as
sequential analysis, which are relevant to process control. We hope that this article leads to a wider
acceptance of distribution– free control charts among practitioners and serves as an impetus to
future research and development in this area. ... Show more content on Helpwriting.net ...
Clearly, the quicker the detection and the signal, the more efficient the chart is. The number of
samples or subgroups that need to be collected before the first out of–control signal is given by a
chart is a random variable called the ran length. The efficiency of a CC depends on the probability
distribution of the run length. The most common efficiency criterion is to consider the average run
length (ARL), which is the expected value of the run length distribution. It is desirable (often
stipulated) that the ARL of a chart be large when the process is in–control and small when the
process is out–of control. The false alarm rate is the probability that a chart signals a process change
when in fact there is no change, that is, when the process is in–control. This is similar to the
probability of a Type I error in the context of hypothesis testing. In practice, the stability of a
process is usually determined relative to one or more of its output characteristics such as the mean
and/or the variance. Two control charts are often compared on the basis of out–of–control ARL,
such that their respective in–control ARL's are roughly the same. This parallels comparing two
statistical tests on the basis of power against some alternative hypothesis when they are roughly of
the same

Quantitative Analysis
Linear Programming D.V. – Decision Variables O.F. – Objective Funtion S.T. or CONST –
Constraints Constrained Mathematical Model – a model with an objective and one or more
constraints EXAMPLE: 50D + 30C + 6M is the total profit for a production run($50 profit for Desk,
$30 profit for Chair and $6 per pound for steel) Functional Constraints – ≤ ≥ or = ––Restrictions that
involve expressions with 1 or more variables EXAMPLE: 7d+3c+1.5M = 2000 (constraint on raw
steel) Variable Constraints – Involve only 1 variable – Nonnegativity Constraint – X≥0 Lower
Bound Constraint – X≥L(a number other than 0) Upper Bound Constraint – X≤U Interger
Constraint – X=integer Binary Constraint – X=0 or 1 EXAMPLE FROM HOMEWORK 1 of
Constraints ... Show more content on Helpwriting.net ...
Slope: the amount X2 increases given one unit increase of X1 Intercept: the point where the line
intersects with X2 axis. Use above for Slope – great formula Redundant Constraints – if removed
will not affect the feasible region Feasible Region – The set of all points that satisfy all constraints
of the model HW EXAMPLE Apply Graphical analysis on the Golden Electronics

Safety and Reliability Engineering Past Paper
EG40JQ/12
UNIVERSITY OF ABERDEEN
SESSION 2011 – 2012
Degree Examination in EG40JQ SAFETY AND RELIABILITY ENGINEERING
Friday 20 January 2012
Notes: (i)
(ii)
2.00 p.m. – 5.00 p.m.
Candidates ARE permitted to use an approved calculator
Data sheets are attached to the paper.
Candidates should attempt all FIVE questions.
REGULATIONS:
(i)
You must not have in your possession any material other than that expressly permitted in the rules
appropriate to this examination. Where this is permitted, such material must not be amended,
annotated or modified in any way.
(ii)
You must not have in your possession any material that could be determined as giving you an
advantage in the examination.
(iii)
You must ... Show more content on Helpwriting.net ...
It is given that t = 6.25mm.
Basic

variable σy D
Mean
Coefficient of variation 240 MPa 0.11
225 mm 0.004
[6 marks]
Qu. 3 continued overleaf/
3/7
EG40JQ/12
Qu. 3 continued/
c)
A projectile of mass M is flying at a velocity V following a minor gas explosion o n an offshore
installation. Write down the safety margin corresponding to an event that the Kinetic Energy (KE =
0.5MV2 ) of the projectile is greater than 22500 J. Using an appropriate FORM approach, determine
the reliability index corresponding to this safety margin and comment on the sensitivity of the
variables (limit your calculations to 2 iterations). Assume that the variables are statistically
independent and normally distributed with the following parameters.
Basic variable
M
V
Mean
12 kg
59 m/s
Standard deviation
0.12 kg
2.95 m/s
[11 marks]
Question 4.
A tank containing gasoline is surrounded by a circular dike of diameter 10 m. The gasoline leak s
and occupies the area bounded by the dike. If there is ignition, calculate:
(a) The total radiative flux from the flame. Use the following correlation for the geometry of the
flame: H/D = 42{m'/[ a(gD)]} 0.61 where H = height of the flame (m), d = diameter of the pool
of liquid (m), m' = mass transfer rate from pool to flame (kg s–1), a = density of air = 1.17 kg m–3
and g = acceleration due to gravity =
9.8 m s–2. Make

Questions: The Multiplicative Laws Of Probability
Question: 1(a)
Basic Laws of Probability
The additive law of probabilities
Probability is known as mutually exclusive events, The sum of Separate probabilities likely to be
one event occur or another.
Example:
Place 100 marbles in a box; 35 blue, 45 red, and 20 yellow.
P(blue)=.35 P(red)=.45 P(yellow)=.20
What is the probability of choosing either a red or a yellow marble from the box?
P(red or yellow) = P(red)+ P(yellow)
= .45+.20
= .65
The multiplicative law of probabilities
The multiplicative law of probability is defined as the probability of the joint occurrence of two or
more of the events which are independent events or the product of their individual probabilities.
_________________________________________________________________________________
... Show more content on Helpwriting.net ...
3. Calculate from the table, showing your calculation methods:
The marginal probability that any person selected at random from the population is a male.
The marginal probability that any person selected at random from the population is aged between 25
and 54.
The joint probability that any person selected at random from the population is a female and aged
between 55 and 64.
The conditional probability that any person selected at random from the population is 25 or over
given that the person is a male.
QUESTION 2 Statistical Decision Making and Quality Control Show all calculations/reason
Question: 2(a)
1. If management wishes to establish x ̅ control limits covering the 95% confidence interval,
calculate the appropriate UCL and LCL. UCL = 46.63 LCL = 43.36 2. If management wishes to use
smaller samples of 16 observations calculate the control limits covering the 95% confidence
interval. (Round calculations to 2 decimal

Essay about Dai Park Textbook
Stochastic Manufacturing Service Systems Jim Dai and Hyunwoo Park School of Industrial and
Systems Engineering Georgia Institute of Technology October 19, 2011 2 Contents 1
Newsvendor Problem 1.1 Proﬁt Maximization 1.2 Cost Minimization . 1.3 Initial Inventory . . 1.4
Simulation . . . . . . 1.5 Exercise . . . . . . . 5 5 12 15 17 19 25 25 27 29 29 31 32 33 34 39 39 40 40 42
44 46 47 48 49 51 51 51 52 54 55 57 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... Show
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Exercise Answers 6.1 Newsvendor Problem . . . . . . . 6.2 Queueing Theory . . . . . . . . . 6.3 Discrete
Time Markov Chain . . 6.4 Poisson Process . . . . . . . . . . 6.5 Continuous Time Markov Process . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . Chapter 1 Newsvendor Problem In this course, we will learn how to design, analyze, and
manage a manufacturing or service system with uncertainty. Our ﬁrst step

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