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Chapter 4 & 5 – Continuous & Discrete Distributions Assignment #3 Your answers must be clear, concise (one or two sentences) and supported by attached output from R (include code and outputs). 1. A technical support call center measures the amount of time that each customer waits on hold. After analyzing the data it has been determined that the wait times follow a Uniform distribution with a minimum wait time of 0 minutes and a maximum wait time of 22 minutes. a) Sketch the distribution, calculate the mean and include this on your sketch. b) What is the probability that a customer waits on hold for less than 5 minutes? c) What is the probability that a customer waits on hold for more than 10 minutes? d) What is the probability that a customer waits on hold for between 5 and 15 minutes? e) The call center wants to publish a service standard on their website i.e. “90% of calls will be answered within ……… minutes.” Calculate what the service standard should be. f) Simulate the wait time of 100 callers by generating random values from your Uniform distribution. How many of the 100 callers waited for longer than the service standard computed in part e? 2. The final grades for a business statistics class follow a Normal distribution with a mean of 78.3 and a standard deviation of 12.1. a) Sketch the distribution, include the mean on your sketch. b) What is the probability that a student scored less than 65? c) What is the probability that a student scored more than 90? d) What is the probability that a student scored between 70 and 90? e) The professor wishes to award an ‘A’ letter grade to the top 25% of students. Calculate what the cutoff-grade for an ‘A’ should be. f) Simulate the grades of 50 students by generating random values from your Normal distribution. How many of the 50 students scored above the ‘A’ cut-off computed in part e? 3. Suppose the number of customers using a bank’s ATM during the lunch period between 12noon and 2pm can be modelled with a Poisson distribution with a mean of 18. a) Calculate the standard deviation of the distribution. b) What is the probability that exactly 18 customers use the ATM between 12noon and 2pm? c) What is the probability that 15 or less customers use the ATM between 12noon and 2pm (i.e. X=25)? e) Simulate 100 lunch periods by generating random values from your Poisson distribution. In how many of these lunch periods were there more than 25 customers? The Practice of Statistics for Business and Economics Fourth Edition David S. Moore George P. McCabe Layth C. Alwan Bruce A. Craig © 2016 W.H. Freeman and Company Distribution for Counts and Proportions The Binomial Distributions PSBE Chapter 5.1 © 2016 W. H. Freeman and Company Objectives (PSBE Chapter 5.1) The Binomial Distributions ▪ The Binomial setting ▪ The Binomial distribution ▪ Binomial probabilities ▪ Binomial mean and standard deviation ▪ Sample proportions ▪ The Normal approximation for counts and proportions Binomial setting Binomial distributions are models for some categorical variables, typically representing the number of successes in a series of n trials. The observations must meet these requirements:  The total number of observations n is fixed in advance.  The outcomes of all n observations are independent.  Each observation falls into just one of two categories: success or failure.  All n observations have the same probability of “success,” p. We record the next 50 vehicles sold at a dealership. Each vehicle sold is either an SUV (success) or not (failure). Binomial distribution The distribution of the count X of successes in the binomial setting is the binomial distribution with parameters n and p: B(n,p). The parameter n is the total number of observations.  The parameter p is the probability of success on any one observation.  The count of successes X can be any whole number between 0 and n.  A coin is flipped 10 times. Each outcome is either a head or a tail. The variable X is the number of heads among those 10 flips, our count of “successes.” On each flip, the probability of success, “head,” is 0.5. The number X of heads among 10 flips has the binomial distribution B(n = 10, p = 0.5). Applications for binomial distributions Binomial distributions describe the possible number of times that a particular event will occur in a sequence of observations. They are used when we want to know about the occurrence of an event, not its magnitude.  In a clinical trial, a patient’s condition may improve or not. We study the number of patients who improved, not how much better they feel.  Was a sales transaction considered pleasant? The binomial distribution describes the number of pleasant transactions, not how pleasant they are.  In quality control we assess the number of defective items in a lot of goods, irrespective of the type of defect. Binomial coefficient The number of ways of arranging k successes in a series of n observations (with constant probability p of success) is the number of possible combinations (unordered sequences). This can be calculated with the binomial coefficient: n!  n    k  k!(n  k )!    Where k = 0, 1, 2, ..., or n. n , “n choose k” uses the factorial notation “!”. k The factorial n! for any strictly positive whole number n is: n! = n× (n – 1) × (n – 2) × … × 3× 2× 1. The binomial coefficient For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. Note 0! = 1. Calculations for binomial probabilities The binomial coefficient counts the number of ways in which k successes can be arranged among n observations. The binomial probability P(X = k) is this count multiplied by the probability of any specific arrangement of the k successes: P( X  k )   n  p k (1  p) n k k The probability that a binomial random variable takes any range of values is the sum of each probability for getting exactly that many successes in n observations. P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) Finding binomial probabilities: tables  You can also look up the probabilities for some values of n and p in Table C in the back of the book.  The entries in the table are the probabilities P(X = k) of individual outcomes.  The values of p that appear in Table C are all 0.5 or smaller. When the probability of a success is greater than 0.5, restate the problem in terms of the number of failures. Customer satisfaction Each consumer has probability of 0.25 of preferring your product over a competitor’s product. If we question five consumers, what is the probability that exactly two of them prefer your product? P(X= 2)= P(X= 2)= n! 5! pk(1 p)n-k = ×0.252×0.753 2! × 3! k!(n  k)! 5×4 × 0.252×0.753 2×1 P(X= 2)= 10 ×0.0625 ×0.421875 = 0.2637 Binomial mean and standard deviation 0.3 a) distribution for a count X are defined by P(X=x) The center and spread of the binomial 0.25 0.2 0.15 0.1 0.05 the mean μ and standard deviation σ: 0 0   np(1  p) 2 0.3 3 4 5 6 7 8 9 10 8 9 10 8 9 10 Number of successes 0.25 b) P(X=x)   np 1 0.2 0.15 0.1 0.05 0 Effect of changing p when n is fixed. a) n = 10, p = 0.25 0 1 2 3 4 5 6 7 Number of successes 0.3 b) n = 10, p = 0.5 c) n = 10, p = 0.75 For small samples, binomial distributions are skewed when p is different from 0.5. c) P(X=x) 0.25 0.2 0.15 0.1 0.05 0 0 1 2 3 4 5 6 7 Number of successes Sample proportions Let X be the count of successes in n samples. Then X pො = . n In an SRS of size n drawn from a large population with population proportion p of successes, the sample proportion pො has mean and standard deviation 𝜇p = p 𝜎p = p 1−p . n Use when the population is at least 20 times as large as the sample. Normal approximation If n is large, the binomial distribution can be approximated by the Normal distribution. X is approximately N np, pො is approximately N p, np(1−p) p(1−p) n The Normal approximation to the binomial distribution can be used when by np ≥ 10 and n(1-p) ≥ 10. Normal approximation If a quality control engineer determines that 8% of switches fail to meet specifications, what is the probability that no more than 10 of 150 switches will be nonconforming? µ= np = (150)(0.08) = 12 σ= np(1  p)= 150(0.08)(0.92)= 3.3226 So Xis approximately N(12, 3.3226); Note: np≥10 and n(1 p) ≥10. P(X ≤ 10) = P X – 12 10 – 12 ≤ 3.3226 3.3226 = P(Z≤ -0.60) = 0.2743 The continuity correction When using the Normal approximation to the binomial distribution, a continuity correction is needed since the binomial distribution puts probability only on whole numbers. To estimate P(X ≤ k) for a binomial distribution, find P(X ≤ k + 0.5) in the Normal distribution. P(X ≤ 10) = P(X ≤ 10.5) in the binomial distribution as shown in the probability histogram. Looking at the Normal curve, P(X ≤ 10.5) will give a better approximation of the binomial probability than P(X ≤ 10). Distribution for Counts and Proportions The Poisson Distributions PSBE Chapter 5.2 © 2016 W. H. Freeman and Company Objectives (PSBE Chapter 5.2) The Poisson Distributions ▪ The Poisson setting ▪ The Poisson distribution ▪ Approximations to the Poisson ▪ Assessing Poisson assumptions with data The Poisson setting  A count X of successes has a Poisson distribution in the following Poisson setting: ➢ The number of successes that occur in two non-overlapping units unit of measure are independent. ➢ The probability that a success will occur in a unit of measure is the same for all units of equal size and is proportional to the size of the unit. ➢ The probability that more than one event occurs in a unit of measure is negligible for very small-sized units. (Events occur one at a time.) Poisson distribution  Poisson distribution  Let 0.9 be the average number of Wi-Fi interruptions on your home network per day. The count of X flaws per day is modeled by the Poisson distribution with μ= 0.9.  The probability of two or less flaws per square yard is P(X ≤ 2).  P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) e−0.9(0.9)0 e−0.9(0.9)1 e−0.9(0.9)2 = + + 0! 1! 2! = 0.4066 + 0.3659 + 0.1647 = 0.9372 Poisson approximations Using the Normal to approximate the Poisson:  Use N(μ, μ) to approximate Poisson probabilities when μ is large.  Americans send an average of approximately 616 text messages per week. What is the probability that you would send more than 650 text messages over a week? P(X> 650) =P X−616 650−616 > 616 616 = P(Z>1.37) = 1 - 0.9147 = 0.0853  The actual Poisson probability is 0.083176, so this approximation is quite accurate. Poisson approximations Using the Poisson approximation to the binomial distribution: Use when n is large and p is small such that np 0.8) = 1 – P(0.5 < X < 0.8) = 0.7 Continuous random variable and population distribution % individuals with X such that x1 < X < x2 The shaded area under a density curve shows the proportion, or %, of individuals in a population with values of X between x1 and x2. Because the probability of drawing one individual at random depends on the frequency of this type of individual in the population, the probability is also the shaded area under the curve. Normal probability distributions The probability distribution of many random variables is a normal distribution, N(μ, σ). Example: Probability distribution of women’s heights. Here, because we chose a woman randomly, her height, X, is a random variable. To calculate probabilities with the normal distribution, we will standardize the random variable (z score) and use Table A. What is the probability, if we pick one woman at random, that her height will be some value X? For instance, between 68 and 70 inches P(68 < X < 70)? Because the woman is selected at random, X is a random variable. z (x  ) N(µ, ) = N(64.5, 2.5)  As before, we calculate the zscores for 68 and 70. For x1 = 68", z (68  64.5)  1.4 2.5 For x2 = 70", z (70  64.5)  2.2 2.5 0.9192 0.9861  The area under the curve for the interval [68" to 70"] is 0.9861 − 0.9192 = 0.0669. Thus, the probability that a randomly chosen woman falls into this range is 6.69%. P(68 < X < 70) = 6.69% Probability: The Study of Randomness Means and Variances of Random Variables PSBE Chapter 4.5 © 2016 W.H. Freeman and Company Objectives (PSBE Chapter 4.5) Means and Variances of Random Variables ▪ Mean of a random variable ▪ Law of large numbers ▪ Rule for means ▪ Variance of a random variable ▪ Rule for variances Mean of a random variable The mean of a set of observations is their arithmetic average. The mean µ of a random variable X is a weighted average of the possible values of X, reflecting the fact that all outcomes might not be equally likely. A basketball player shoots three free throws. The random variable X is the number of baskets successfully made (“H”). MMM HMM MHM MMH HHM HMH MHH HHH Value of X 0 1 2 3 Probability 1/8 3/8 3/8 1/8 The mean of a random variable X is also called expected value of X. We would be interested in the µX, the expected number of baskets successfully made in three shots. Mean of a discrete random variable For a discrete random variable X with probability distribution  the mean µ of X is found by multiplying each possible value of X by its probability, and then adding the products. A basketball player shoots three free throws. The random variable X is the number of baskets successfully made. Value of X 0 1 2 3 Probability 1/8 3/8 3/8 1/8 The mean µ of X is µX = (0*1/8) + (1*3/8) + (2*3/8) + (3*1/8) = 12/8 = 3/2 = 1.5 Law of large numbers As the number of randomly drawn independent observations in a sample x increases, the mean of the sample gets closer and closer to the population mean . This is the law of large numbers. It is valid for any population. Note: We often intuitively expect predictability over a few random observations, but this is wrong. The law of large numbers only applies to really large numbers. Rules for means Rule 1: If X is a random variable and a and b are fixed numbers, then µa+bX = a + bµX Rule 2: If X and Y are random variables, then µX+Y = µX + µY Rule 3: If X and Y are random variables, then µX-Y = µX - µY Investment You invest 20% of your funds in Treasury bills and 80% in an “index fund” that represents all U.S. common stocks. Your rate of return over time is proportional to that of the T-bills (X) and of the index fund (Y), such that R = 0.2X + 0.8Y. Based on annual returns between 1950 and 2003:  Annual return on T-bills µX = 5.0% σX = 2.9%  Annual return on stocks µY = 13.2% σY = 17.6%  Correlation between X and Y ρ = −0.11 µR = 0.2µX + 0.8µY = (0.2*5) + (0.8*13.2) = 11.56% This portfolio has a smaller mean return than an all-stock portfolio. Variance of a random variable The variance and the standard deviation are the measures of spread that accompany the choice of the mean to measure center. The variance σ2X of a random variable is a weighted average of the squared deviations (X − µX)2 of the variable X from its mean µX. Each outcome is weighted by its probability in order to take into account outcomes that are not equally likely. The larger the variance of X, the more scattered the values of X on average. The positive square root of the variance gives the standard deviation σ of X. Variance of a discrete random variable For a discrete random variable X with probability distribution and mean µX, the variance σ2 of X is A basketball player shoots three free throws. The random variable X is the number of baskets successfully made. µX = 1.5. Value of X 0 1 2 3 Probability 1/8 3/8 3/8 1/8 The variance σ2 of X is σ2 = 1/8*(0−1.5)2 + 3/8*(1−1.5)2 + 3/8*(2−1.5)2 + 1/8*(3−1.5)2 = 2*(1/8*9/4) + 2*(3/8*1/4) = 24/32 = 3/4 = 0.75 Rules for variances Rule 1: If X is a random variable and a and b are fixed numbers, then σ2a+bX = b2σ2X Rule 2: If X and Y are independent random variables, then σ2X+Y = σ2X + σ2Y σ2X-Y = σ2X + σ2Y Rule 3” If X and Y are NOT independent but have correlation ρ, then σ2X+Y = σ2X + σ2Y + 2ρσXσY σ2X-Y = σ2X + σ2Y - 2ρσXσY Investment You invest 20% of your funds in Treasury bills and 80% in an “index fund” that represents all U.S. common stocks. Your rate of return over time is proportional to that of the T-bills (X) and of the index fund (Y), such that R = 0.2X + 0.8Y. Based on annual returns between 1950 and 2003:  Annual return on T-bills µX = 5.0% σX = 2.9%  Annual return on stocks µY = 13.2% σY = 17.6%  Correlation between X and Y ρ = −0.11
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Hi, I am done your assignment :). See attachment :). I am just want to say that 2 or 3 of the data or Code is not fully coded in R, this is because it was depend on the given data of the calculation okay :). Everything is there..

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Chapter 4 & 5 – Continuous & Discrete Distributi...


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