Study Set for Midterm II‚ Chapters 7 & 8 ESSAY. Write your answer in the space provided or on a separate sheet of 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
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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. As is customary in the land of Markov processes‚ the initial value x is indicated (when appropriate) by putting a superscript x on the probability
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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 between exponential distribution and Poisson distribution? II. Computation
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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 the percentage of workers getting wages between Rs.9 and Rs.17 in the whole factory assuming that the wages are normally distributed. Characteristics of Normal probability distribution Formula/Computation/Solution to the problem 3 4 10 6 a) The
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> data=read.table("d:/111113/1.txt"‚header=T) > model1=lm(S~u_direction+mx+my+mz‚data) > summary(model1) Call: lm(formula = S ~ u_direction + mx + my + mz‚ data = data) Residuals: Min 1Q Median 3Q Max -11.8430 -0.3962 0.3252 0.7887 18.3963 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -0.50372 0.12738 -3.955 7.93e-05 *** u_direction -0.40368 0.07996 -5.048 4.85e-07 *** mx -0.40573 0
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60/500=0.12=12% d. A randomly selected man would choose either the camera or the bike? 177/239=0.74=74% 2. a. Which part(s) of question 1 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
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Frequencies A sound wave is created as a result of a vibrating object. The object that is vibrating‚ is the source of the disturbance that moves throughout the medium. The object creating the disturbance could be the vocal cords of a person‚ the vibrating strings and soundboard on a string instrument‚ or the vibrating diaphragm of a radio speaker. If an object has the ability to vibrate‚ then it will produce sound. Almost every
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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
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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 Risk considering both sides: ! = 0.00306
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Relative frequency of a category = Frequency of that category Sum of all frequencies Class width = Upper boundary– Lower boundary Class midpoint or mark = Lower limit+ Upper limit/2 Approximate class width=Largest value -mallest value / Number of classes Relative frequency of a class=Frequency of that class/Sum of all frequencies Cumulative relative frequency=Cumulative frequency of a class/Total observations in the data set Mean = Average ‚ Median ‚ Mode Range = Largest value–
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