Chapter_6_Formulas - CHAPTER FORMULAS Uniform Probability...

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Unformatted text preview: CHAPTER FORMULAS Uniform Probability Density Function for a Random Variable x —— for a S x S b b f (x) = (6.1) 0 elsewhere Mean and Variance of a Uniform Continuous Probability Distribution _ a + b p 2 2 0,2 = (b ‘ a) 12 Normal Probability Density Function f(x) = ;e'(”)2’r2 “2 for — co 5 x s 00 (6.2) c 27: where p. = mean of the random variable x 0'2 = variance of the random variable x it =3.14159... e = 2.71828... The transformation of any Random Variable x with Mean 14 and Standard Deviation a to the Standard Normal Distribution Z = (X - H) (6, 3 ) o where z = the number of standard deviations Exponential Probability Density function f(x) = ie—x’“ (6.4) p Exponential Distribution Probabilities p(xS yin,)-—-1—e"‘*""M (6.5) EXERCISES *1. The driving time for an individual from his home to his work is uniformly distributed between 300 to 480 seconds. (a) Give a mathematical expression for the probability density function. Answer: The probability density function of a uniform probability distribution is given in equation 6.1 as 1 — foranSb b-a f(X) = 0 elsewhere In our example, b = 480 and a = 300. Therefore, the probability density function is — 1 _i m— 480-300 180 This figure indicates that for any value of x (from 300 to 480), the f(x) is constant and its value is equal to U180. (b) Compute the probability that the driving time will be less than or equal to 435 seconds. Answer: Inthis part, we are interested in determining the p(300 S x S 435) . This probability is the area under a uniform distribution with a height of 1:180 and a width of 135 seconds (i.e., 435 - 300 = 135). Hence, the area is (135)(1!180)= 0.75. This indicates that p(300 S x s 435) = 0.75. The shaded area in Figure 6.1 represents this probability. fix) 12’180 300 435 480 Driving Time in Seconds Figure 6.1 (c) Determine the expected driving time and its standard deviation. Answer: The expected value of a uniform continuous random variable is a+b E(X) = Hence, in our example, the expected value of x is Em = Eli—300 = 390 Furthermore, the variance is given by 2 Variance (x)= —(b - a) 12 Therefore 2 Variance (x) = Egoizfl = 2?00 Hence, the standard deviation is O = a} 2700 =51.96 ...
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Chapter_6_Formulas - CHAPTER FORMULAS Uniform Probability...

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