...If an entire world can be expressed in ones and zero, then why not our living world? In seeing a random sequence of numbers such as 413270, it brings a slew of thoughts to mind. 314 are the first three digits of pi. 4, 3, 2, 1 speaks to me of the applied math and cause of history. 2.71 is my favorite number though an outdated concept for a mature individual. A random sequence of numbers can provide many memories. One of the saddest things lost in childhood is the interest in a favorite number. Nobody over the age of 10 asks another what their favorite number is. Perhaps as we mature we become less attached to certain digit sequences and closer to the meanings of such digits. A favorite number may end up being any numeral that represents a large sum of money. Favorite colors, shapes, and numbers lose importance as we grow and become more aware of our own wants and needs. My favorite number when I was 12 years old was the number 7. I had no particular love of 6 or 8 or any even number for that matter. My birth month, July, is the seventh month and 7 seemed to me to be a significant piece of my personality. It may seem absurd that a number may be of note for someone, but to a 12 year old girl or to a regular lottery ticket purchaser a number is the difference between happiness and misery. As I aged, my favorite number...
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...Random numbers in C++ and The Pythagorean Theorem Literature Review Name Course Date Literature Review The increase in technological advancements has seen a similar increase in the number of computer programs which are designed to command a computer to carry out a given specified task. The number of languages that are available which are used in this creation and design include Java Script, C++, Java and Sage. It is worth noting that while these are the most notable ones, the number of languages in computer programming design might be higher. However, computer programmers argue that the rest of the languages, despite being of equal capabilities, have not met the required usage to warrant widespread literature review. Hiscotta is particularly critical of this in 10 programming languages you should learn in 2014 by asserting that The field of computer programming is particularly important with regards to the increasing use and adoption of the internet use. This has seen the field carve out a distinct field of study which is purely dedicated to the understanding of how the programs work. The first step in the design of the computer programs is the basic understanding of the dynamics that are involved in the working of computers. This forms the initial step which will eventually be accompanied by software writing involving random numbers with the sole undertaking of coming up with a particular outcome. Of critical importance is the adherence to source code representation...
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...COPIES CASE STUDY James C. Kessler Strayer University Introduction to Management Science Math 540 Dr. Yaw Kyei January 3, 2013 Jet Copies Case Study 2 Days-to-Repair I typed the name jet copies into the A1 slot. Then I made the letters bold and used the cells section to increase the row height and the column width. I also increased the font size. I then typed in the probability of weekly demand section. Increasing the font size and making it bold. I set up three columns, P(x), cumulative, and repair time. This was taken from the table that was given. Because we need 52 weeks of information I set numbers for 60 weeks, starting with column E6. I then went to G6 and typed in =RAND() to start a random number sequence and hit enter. I then clicked on G6 and right clicked and held down to set up 60 random numbers. I then went to the clipboard section and clicked copy. Then we clicked on F6, then right clicked, then clicked on past special, then clicked on values, then clicked ok to set the values we would use. Then I left clicked on G6, then clicked on delete, then clicked entire column to remove. To develop the probability of breakdowns I had to assume values for P(y). We were given six operating times and given .33 as the highest probability, we then assumed the others with the total equaling 1.00. I then set up the cumulative column, and the operating time to breakdown column. Now I can do the calculation to get the times between...
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...Frequencytest.m r = rand(1000,1); x = zeros(20,1); for i=1:1000 for j=0:19 if(r(i)>=j*0.05 && r(i)<(j+1)*0.05) x(j+1)=x(j+1)+1; end end end a=0; for k = 1:20 a = a + (x(k)-50)*(x(k)-50)/50; end a Pokers_test.m r = randi([0,999],1000,1); a=0; b=0; c=0; for i = 1:1000 if r(i)==0 b=b+1; elseif (r(i)>0)&&(r(i)<=10) c=c+1; elseif (r(i)>10)&&(r(i)<100) x=mod(r(i),10); y=(r(i)-x)/10; if (x==y || x==0) c=c+1; else a=a+1; end else x=mod(r(i),10); y=mod(((r(i)-x)/10),10); z=(((r(i)-x)/10)-y)/10; if (x==y)&&(y==z) b=b+1; elseif (x~=y)&&(x~=z) a=a+1; else c=c+1; end end end a b c chisq = (a-720)*(a-720)/720 + (b-10)*(b-10)/10 + (c-270)*(c-270)/270 normal.m r=rand(200,1); r=sort(r); x=zeros(100,1); a=zeros(100,2); for n=1:100 i=2*n-1 x(n,1)= sqrt(-2*log(r(i,1)))*cos(2*pi*r(i+1,1)); a(n,2)=x(n,1); a(n,1)=exp(-x(n,1)*x(n,1))/(2*pi); end plot(a(:,1),a(:,2)) Exp_dist.m r=rand(100,1); r=sort(r); x=zeros(100,1); a=zeros(100,2); for n=1:100 x(n,1)=-log(1-r(n,1)); a(n,2)=x(n,1); a(n,1)=exp(-x(n,1)); end plot(a(:,1),a(:,2)) r = rand(1000,1); x = zeros(20,1); for i=1:1000 for j=0:19 ...
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...between breakdowns; lost revenue; putting it together). Days to repair With regards to the days to repair component, it was calculated by using the probability distribution of repair times that was given. This was used along with a set of random numbers based on 100 breakdowns a year. Then, a vlookup was used and the probability distribution per day to come up with the days to repair, which varies based on the random number that excel generates. The random number represents the probability of how many days it would take to repair the copier. Interval between breakdowns The formula is x=4√r1 as espoused by Taylor III (2011)where x equals the weeks between machine breakdowns and r1 equals the random number was used . Here the time between breakdowns component was implemented by taking the formula for elapsed time between the various times the copier would break down. Excel automatically calculated the formula once it was put into it and the breakdowns were calculated based on that information. Lost revenue The lost revenue was calculated by taking the median revenue to be earned in a given day and multiplying this by the calculated days to repair. Since the repair time can vary from 1 to 4 days, we generate four random numbers r1, r2, r3 and r4 between 2000 and 8000. Therefore, it was calculated annually by taking the sum of the lost revenue column and dividing it by the cumulative time divided by 52 for 52 weeks in a year. This calculates the annual loss of revenue...
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...Jamie Banks, Ernie Moore and Terri Jones were students in State University. They always have to go far to make copies in Klecko’s copy center as there was no copying service nearby where they live by the south gate. One day while James was standing in line at Klecko’s copy center waiting for his turn, he realizes how much time they are wasting just by waiting as most students use the same machine to get copies. James got an idea from other student who was waiting for his turn as James, if he can have a copy center by the Southgate where most of the students live, he easily can make lots of money. James shares his idea with friends Ernie and Terri and they liked the idea. Three of them decided to start the copying business and they would call it JET Copies named after the first letter of their name. They bought a copier similar to the one used at the university for $18,000.00.They enquire about the reliability of the copier and came to know that the university copier broke down frequently and when it did it often require between 1 and 4 days to get it repaired. As Repair time include the time portion during which technicians are working on a machine to effect repair starting from preparation, fault location, fault correction and final check time, it needs time between 1 and 4 days to get it repaired. During this repair time they will always lose Revenue and they become worried. So they decided they might need back up copier to use in these repair time in order...
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...will be able to distribute the envelopes back to the volunteers with the according signature. Second Step: a) Identify the component to be repeated. The component to be repeated is to simulate randomly matching people with their signature. b) Explain how you will model the component’s outcome. I will use the numbers 0 through 9 to represent the envelopes. Using a random number generator I will list out 10 random numbers in the order they were received. I’ll then count how many of the random numbers are still in their numerical place. For example, if 0 was the first random number and 9 was the last random number picked then that would be counted as two correct matches. c) Explain how you will combine the components to model a trial. By using the numbers 0-9 to model the envelopes, I will use random number generator to simulate matching the envelopes to the volunteers. I will not count any numbers that are repeated because each number is representative of one specific envelope and cannot be given to two volunteers. After counting how many numbers are in their regular numerical place one trial is complete. After running 20 trials I will take the average number of people matched to their signatures by chance. Anything above this average will be considered...
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...frustrated with what I now know is called RNG. I would drive myself crazy as a child asking myself “how is it that he got a shiny Pokémon when I am playing the same game as him?” pseudo-RNG was the answer, but at the time I couldn’t even have fathomed a complex numbering system/algorithm that takes things like, frame rate, date and time, and battery life, and positioning to determine rather or not I would encounter a rare Pokémon or not. Which, in essence, is why I believe that pseudo-RNG should be done away with in game development in substitute for a more decipherable algorithm that rewards those who strategize instead of the luck of the draw. RNG is an acronym for Random Number Generator, which is essentially any algorithm used in computing when an outcome needs to be as random as possible. But it can never truly be random. There will always...
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...average number of days that it took for the company to repair the copier if it broke down. The average was between 1 and 4 days to have the machine repaired. For this data, a probability distribution table was developed and programmed into excel. Once the probability table is loaded into excel, they are averaged into a random number range that generates numbers from 0.00 to 1.0. The system then generated an average number days it would take to repair the machine base to the random number generated. 2. Write a brief description of how you implemented the intervals between break downs. When the company looked at the number of intervals between breakdowns, there was no exact numbers to go by to create a probability distribution. What they did instead was to talk to the college who already owned one and go by their data. It was then determined that the average repair days would be between 0 and 6 days. This time the formula took the maximum number of and averaged it with the square root of the random numbers generated again from 0.00 to 1.0. Excel would then come up with the average number of days it would be between machine break down based on these numbers. 3. Write a brief description of how you implemented lost revenue. When looking at the lost revenue, the first thing I did was look at the average revenue that was generated per day. JET averaged between 2000 and 8000 copies per day at $0.10 per copy. What I did in this instance is took the median number which...
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...Jet Copies Case Study 1. In Excel, use a suitable method for generating the number of days needed to repair the copier, when it is out of service, according to the discrete distribution shown. 2. In Excel, use a suitable method for simulating the interval between successive breakdowns, according to the continuous distribution shown. 3. In Excel, use a suitable method for simulating the lost revenue for each day the copier is out of service. 4. Put all of this together to simulate the lost revenue due to copier breakdowns over 1 year to answer the question asked in the case study. 5. In a word processing program, write a brief description/explanation of how you implemented each component of the model. Write 1-2 paragraphs for each component of the model (days-to-repair; interval between breakdowns; lost revenue; putting it together). 6. Answer the question posed in the case study. How confident are you that this answer is a good one? What are the limits of the study? Write at least one paragraph. Answers 1. # of days P(x) Cumulative 1 0.2 0 2 0.45 0.2 3 0.25 0.65 4 0.1 0.9 Q: 2-4. Break Random times b/w Random Repair Random Lost cumulative down # 1 Break (weeks) # 2 Time #3 Revenue time 1 0.78468 5.314929 0.88991 3 2237 $6,711 5.314929 2 0.512227 4.294201 0.831365 2 3244 $6,488 9.60913 3 0.389251 3.743399 0.912647 2 5874 $11,748 13.35253 4 0.998082 5.994243 0.216353 1 3330 $3,330 19.34677 5 0.963834 5.890502 0.415313 4 5487 $21,948...
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...law of large numbers 2.slide * Statistical concept that larger the sample population (or the number of observations) used in a test, the more accurate the predictions of the behavior of that sample, and smaller the expected deviation in comparisons of outcomes. * As a general principle it means that, in the long run, the average (mean) of a long series of observations may be taken as the best estimate of the 'true value' of a variable. 3.slide * In other words, what is unpredictable and chancy in case of an individual is predictable and uniform in the case of a large group. * This law forms the basis for the expectation of probable-loss upon which insurance premium rates are computed. Also called law of averages. Law of Large Numbers Observe a random variable X very many times. In the long run, the proportion of outcomes taking any value gets close to the probability of that value. The Law of Large Numbers says that the average of the observed values gets close to the mean μ X of X. 4.slide ; Law of Large Numbers for Discrete Random Variables * The Law of Large Numbers, which is a theorem proved about the mathematical model of probability, shows that this model is consistent with the frequency interpretation of probability. 5.slide ; Chebyshev Inequality * To discuss the Law of Large Numbers, we first need an important inequality called the Chebyshev Inequality. * Chebyshev’s Inequality is a formula in probability theory that relates...
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...CHAPTER 6 RANDOM VARIABLES PART 1 – Discrete and Continuous Random Variables OBJECTIVE(S): • Students will learn how to use a probability distribution to answer questions about possible values of a random variable. • Students will learn how to calculate the mean and standard deviation of a discrete random variable. • Students will learn how to interpret the mean and standard deviation of a random variable. Random Variable – Probability Distribution - Discrete Random Variable - The probabilities of a probability distribution must satisfy two requirements: a. b. Mean (expected value) of a discrete random variable [pic]= E(X) = = 1. In 2010, there were 1319 games played in the National Hockey League’s regular season. Imagine selecting one of these games at random and then randomly selecting one of the two teams that played in the game. Define the random variable X = number of goals scored by a randomly selected team in a randomly selected game. The table below gives the probability distribution of X: Goals: 0 1 2 3 4 5 6 7 8 9 Probability: 0.061 0.154 0.228 0.229 0.173 0.094 0.041 0.015 0.004 0.001 a. Show that the probability distribution for X is legitimate. b. Make a histogram of the probability distribution. Describe what you see. 0.25 0.20 0.15 0.10 ...
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...Objectives 1. Understand the concepts of a random variable and a probability distribution. 2. Be able to distinguish between discrete and continuous random variables. 3. Be able to compute and interpret the expected value, variance, and standard deviation for a discrete random variable. 4. Be able to compute and work with probabilities involving a binomial probability distribution. 5. Be able to compute and work with probabilities involving a Poisson probability distribution. A random variable is a numerical description of the outcome of an experiment. A discrete random variable may assume either a finite number of values or an infinite sequence of values. A continuous random variable may assume any numerical value in an interval or collection of intervals. Discrete Probability Distributions n Random Variables n Discrete Probability Distributions n Expected Value and Variance n Binomial Distribution n Poisson Distribution [pic] A random variable is a numerical description of the outcome of an experiment. A discrete random variable may assume either a finite number of values or an infinite sequence of values. A continuous random variable may assume any numerical value in an interval or collection of intervals. Example: JSL Appliances n Discrete random variable with a finite number of values n Let x = number of TVs sold at the store in one day, where...
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...Expected Value Category name Value Frequency Probability Cumulative Probability Random No interval Category 1 $ 22.00 0.02 0.02 0.02 2 Category 2 $ 23.00 0.35 0.35 0.37 37 Category 3 $ 24.00 0.16 0.16 0.53 53 Category 4 $ 25.00 0.44 0.44 0.97 97 Category 5 $ 28.00 0.03 0.03 1 100 Category 6 0 1 100 Total 1 The probability for each cost is shown. The total of probability is 1. We find the cumulative probability to assign range to each category. Each category is represented by a range of random numbers between 1 to 100. To match the probability with random numbers we multiply cumulative probability by 100 So Labor Cost per Unit Random numbers 1 to 2 represents $ 22.00 Random numbers 3 to 37 represents $ 23.00 Random numbers 38 to 53 represents $ 24.00 Random numbers 54 to 97 represents $ 25.00 Random numbers 98 to 100 represents $ 28.00 so if random number is 75 then it is in range 54--97 so labor cost is $ 25 Similarly for Utilities cost Expected Value Category name Value Frequency Probability Cumulative Probability Random No interval Category 1 $ 3.00 0.36 0.36 0.36 36 Category 2 ...
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...costs (between $5,040 and $6,860) U = rent, utilities, other unavoidable costs = $3,995 What are the monthly variable costs? F = food costs M = number of meals served in month F = $11 x M What are the monthly total costs? L + U + F = L + 3,995 + 11 x M 2 Gentle Lentil’s Monthly Earnings,cont. What are the monthly revenues? R = monthly revenues P = price of meal R = PxM What are the monthly earnings? X = monthly earnings = revenues – costs = P x M – ( L + 3,995 + 11 x M ) = (P – 11 ) x M – L – 3,995 Which of these quantities are random variables? P M L X = = = = price of prix fixe meal number of meals sold labor cost monthly earnings 3 (X is a function of random variables, so it is a random variable) Assumptions Regarding the Behavior of the Random Variables M P = number of meals sold per month We assume that M obeys a Normal distribution with µ = 3,000 and σ = 1,000 = price of the prix fixe meal We assume that P obeys the following discrete probability distribution Scenario Very healthy market Healthy market Not so healthy market Unhealthy market Price of Prix Fixe Meal $20.00 $18.50 $16.50 $15.00 Probability 0.25 0.35 0.30 0.10 L = labor costs per month We assume that L obeys a uniform distribution with a minimum of $5,040 and maximum of $6,860 4 The Behavior of the Random Variables, cont. X = earnings per month We do not know the distribution of X . We assume, however, that X = (P – 11 ) x M – L – 3,995 Always...
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