Definition 1. A standard (one-dimensional) Wiener process (also called Brownian motion) is a stochastic process {Wt }t≥0+ indexed by nonnegative real numbers t with the following properties: (1) (2) (3) (4) W0 = 0. With probability 1‚ the function t → Wt is continuous in t. The process {Wt }t≥0 has stationary‚ independent increments. The increment Wt+s − Ws has the N ORMAL(0‚ t) distribution. A Wiener process with initial value W0 = x is gotten by adding x to a standard Wiener process
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selected by lecturer. c. Solutions from each group must be submitted by 19 April 2013. SPECIAL DISTRIBUTIONS I. Concept of probability (3%) 1. Explain why the distribution B(n‚p) can be approximated by Poisson distribution with parameter if n tends to infinity‚ p 0‚ and = np can be considered constant. 2. Show that – and + are the turning points in the graph of the p.d.f. of normal distribution with mean and standard deviation . 3. What is the relationship
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Statistics Math 1342 Final Exam Review Name___________________________________ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Provide an appropriate response. 1) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a heart. 17 7 3 4 A) B) C) D) 52 52 13 13 Answer: D 2) The events A and B are mutually exclusive. If P(A) = 0.7 and P(B) = 0.2‚ what is P(A or B)? A) 0.5 B) 0.9 C) 0.14 D)
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Examining Stock Returns for Normal Distributions July11‚ 2012 Part A. A1 (CRSP 2000-2008) | VW Daily | EW Daily | VW Monthly | EW Monthly | Mean | 0.00% | 0.05% | -0.12% | 0.50% | σ | 1.35% | 1.12% | 4.66% | 6.14% | Table A1 shows return means and standard deviations for the CRSP market portfolio from 2000-2008. In comparing daily vs monthly returns in both cases‚ equally weighted (EW) and value weighted (VW)‚ Table A1 shows the mean and standard deviation are bigger
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paper. 1) The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0. Suppose 36 golfers played the course today. Find the probability that the average score of the 36 golfers exceeded 71. 2) At a computer manufacturing company‚ the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the probability that
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Marks Q.No 1 Distinguish between Classification and Tabulation. Explain the structure and components of a Table with an example. Meaning of Classification and Tabulation Differences between Classification and Tabulation 2 Structure and Components of a Table with an example 2 2 10 6 a) Describe the characteristics of Normal probability distribution. b) In a sample of 120 workers in a factory‚ the mean and standard deviation of wages were Rs. 11.35 and Rs.3.03 respectively. Find
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above deal with joint probability? C‚b. Which part(s) deal with conditional probability? B‚D The Standard Normal Curve 3. In a recent year‚ about two-thirds of U.S. households purchased ground coffee. Consider the annual ground coffee expenditures for households who purchase coffee‚ assuming that these expenditures are approximately normally distributed with a mean of $45.16 and a standard deviation of $10.00. a. Find the probability that a household spent less than $25.00. Z=(25-45
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-0.40573 0.01292 -31.404 < 2e-16 *** my -0.25862 0.01292 -20.018 < 2e-16 *** mz -0.36557 0.01292 -28.296 < 2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 1.809 on 2043 degrees of freedom Multiple R-squared: 0.8874‚ Adjusted R-squared: 0.8872 F-statistic: 4026 on 4 and 2043 DF‚ p-value: < 2.2e-16 > data=read.table("d:/111113/2.txt"‚header=T) Call: lm(formula = S ~ u_direction
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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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STA1101 Normal Distribution and Continuous random variables CONTINUOUS RANDOM VARIABLES A random variable whose values are not countable is called a _CONTINUOUS RANDOM VARIABLE._ THE NORMAL DISTRIBUTION The _NORMAL PROBABILITY DISTRIBUTION_ is given by a bell-shaped(symmetric) curve. THE STANDARD NORMAL DISTRIBUTION The normal distribution with and is called the _STANDARD NORMAL DISTRIBUTION._ Example 1: Find the area under the standard normal curve between z = 0 and z = 1.95 from z = -2
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