TEM1116 Probability and Statistics Tri1 2013/14 Chapter 1 Chapter 1: Discrete and Continuous Probability Distributions Section 1: Probability Contents: 1.1 1.2 1.3 1.4 1.5 Some basics of probability theory Axioms‚ Interpretations‚ and Properties of Probability Counting Techniques and Probability Conditional Probability Independence TEM1116 1 TEM1116 Probability and Statistics Tri1 2013/14 Chapter 1 1.1 Basics of Probability Theory Probability refers to the study
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Statistics Chapter 5 Some Important Discrete Probability Distributions 5-1 Chapter Goals After completing this chapter‚ you should be able to: Interpret the mean and standard deviation for a discrete probability distribution Explain covariance and its application in finance Use the binomial probability distribution to find probabilities Describe when to apply the binomial distribution Use Poisson discrete probability distributions to find probabilities 5-2 Definitions Random Variables A
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Discrete and Continuous Probability All probability distributions can be categorized as discrete probability distributions or as continuous probability distributions (stattrek.com). A random variable is represented by “x” and it is the result of the discrete or continuous probability. A discrete probability is a random variable that can either be a finite or infinite of countable numbers. For example‚ the number of people who are online at the same time taking a statistics class at CTU on
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Subject CT3 Probability and Mathematical Statistics Core Technical Syllabus for the 2014 exams 1 June 2013 Subject CT3 – Probability and Mathematical Statistics Core Technical Aim The aim of the Probability and Mathematical Statistics subject is to provide a grounding in the aspects of statistics and in particular statistical modelling that are of relevance to actuarial work. Links to other subjects Subjects CT4 – Models and CT6 – Statistical Methods: use the statistical concepts
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Karan negi 12.2 12.3 We use equation 2 to find out probability: F(t)=1 – e^-Lt 1-e^-(0.4167)(10) = 0.98 almost certainty. This shows that probability of another arrival in the next 10 minutes. Now we figure out how many customers actually arrive within those 10 minutes. If the mean is 0.4167‚ then 0.4167*10=4.2‚ and we can round that to 4. X-axis represents minutes (0-10) Y-axis represents number of people. We can conclude from this chart that the highest point with the most visitors
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True/False Questions 1. The standard deviation of any normal random variable is always equal to one. Answer: False Type: Concept Difficulty: Easy 2. For any normal random variable‚ the probability that the random variable will equal one is always zero. Answer: True Type: Concept Difficulty: Medium 3. The graph of a standard normal random variable is always symmetric. Answer: True Type: Concept Difficulty: Easy 4. The formula will convert any normal
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Tutorial on Discrete Probability Distributions Tutorial on discrete probability distributions with examples and detailed solutions. ------------------------------------------------- Top of Form | Web | www.analyzemath.com | | Bottom of Form | | Let X be a random variable that takes the numerical values X1‚ X2‚ ...‚ Xn with probablities p(X1)‚ p(X2)‚ ...‚ p(Xn) respectively. A discrete probability distribution consists of the values of the random variable X and their corresponding
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Chapter 3 Probability Distributions 1. Based on recent records‚ the manager of a car painting center has determined the following probability distribution for the number of customers per day. Suppose the center has the capacity to serve two customers per day. |x |P(X = x) | |0 |0.05 | |1 |0.20 | |2 |0.30 | |3 |0.25 | |4 |0.15 | |5 |0.05 | a. What is the probability that one
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EXERCISES (Discrete Probability Distribution) EXERCISES (Discrete Probability Distribution) P X x n C x p 1 p x BINOMIAL DISTRIBUTION n x P X x n C x p 1 p x BINOMIAL DISTRIBUTION n x 1. 2. 3. The probability that a certain kind of component will survive a given shock test is ¾. Find the probability that exactly 2 of the next 4 components tested survive. The probability that a log-on to the network is successful is 0.87. Ten users
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consecutive points falling on one side of the centerline When the process is in statistical control‚ find the false alarm probability (Type-I error) for each case. The corresponding probability measures are obtained from the Normal table as P(3 " Z) = 0.00135 P(2 " Z) = 0.02275 P(1 " Z) = 0.1587 Solution: ! i) Use the Binomial distribution to ! calculate the probability measures. ! 3! 3! P(Y ! 2 n = 3‚ p = 0.02275) = (0.02275)2 (1" 0.02275) + (0.02275)3 = 0.00153 2!1! 3!0! Type-1
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