Chapter 05

Chapter 05 - Chapter 5 Continuous Random Variables...

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Unformatted text preview: Chapter 5 Continuous Random Variables Continuous Probability Distributions Continuous Probability Distribution: areas under curve correspond to probabilities for x Area A corresponds to the probability that x lies between a and b. The Uniform Distribution Uniform Probability Distribution: distribution resulting when a continuous random variable is evenly distributed over a particular interval 1 d -c Probability Distribution for a Uniform Random Variable x Probability density function: f ( x ) = Mean: = 2 c+d Standard Deviation: = d -c 12 P( a < x < b ) = ( b - a ) / ( d - c ) , c a < b d The Uniform Distribution A bus is scheduled to stop at a certain bus stop every half hour on the hour and the half hour. At the end of the day, busses still stop after every 30 minutes, but because delays often occur earlier in the day, the bus is never early and likely to be late. The director of the bus line claims that the length of time a bus is late is uniformly distributed and the maximum time that a bus is late is 20 minutes. If the director's claim is true, what is the expected number of minutes a bus will be late? If the director's claim is true, what is the probability that the last bus on a given day will be more than 19 minutes late? 10 .05 The Normal Distribution Bellshaped curve Symmetrical about its mean Spread determined by the value of it's standard deviation. 1 f ( x) = e 2 1 - x - 2 2 The Normal Distribution The mean and standard deviation affect the flatness and center of the curve, but not the basic shape. The Normal Distribution Probabilities associated with values or ranges of a random variable correspond to areas under the normal curve. Calculating probabilities can be simplified by working with a Standard Normal Distribution. A Standard Normal Distribution is a Normal distribution with = 0 and = 1. The standard normal random variable is denoted by the letter Z. The Normal Distribution What is P(1.33 < Z < 1.33)? Table gives us area A1 Symmetry about the mean tell us that A2 = A1 P(1.33 < Z < 1.33) = P(1.33 < Z < 0) +P(0 < Z < 1.33)= A2 + A1 = 0.4082 + 0.4082 = 0.8164 The Normal Distribution What is P(Z > 1.64)? Table gives us area A2 Symmetry about the mean tell us that A2 + A1 = 0.5 P(Z > 1.64) = A1 = 0.5 A2 = 0.5 0.4495 = 0.0505 The Normal Distribution What is P(Z < 0.67)? Table gives us area A1 Symmetry about the mean tell us that A2 = 0.5 P(Z < 0.67) = A1 + A2 = 0.2486 + 0.5 = 0.7486 The Normal Distribution What is P(|Z| > 1.96)? Table gives us area 0.5 A2 = 0.4750, so A2 = 0.0250 Symmetry about the mean tell us that A2 = A1 P(|Z| > 1.96) = A1 + A2 = 0.0250 + 0.0250 = 0.05 The Normal Distribution What if values of interest were not normalized? Find p(8 < X < 12), where X is normally distributed with = 10 and = 1.5 Convert to standard normal using x- z= P(8<x<12) = P(1.33<z<1.33) = 2(0.4082) = 0.8164 The Normal Distribution You can also use the table in reverse to find a zvalue that corresponds to a particular probability What is the value of z that will be exceeded only 10% of the time? Look in the body of the table for the value closest to 0.4, and read the corresponding z value Z = 1.28 The Normal Distribution Which values of z enclose the middle 95% of the standard normal z values? Using the symmetry property, z0 must correspond with a probability of 0.475. From the table, we find that z0 and z0 are 1.96 and 1.96 respectively. The Normal Distribution Given a normally distributed variable X with mean 550 and standard deviation of 100, what value of x identifies the top 10% of the distribution? The z value corresponding with 0.40 is 1.28. Solve for X. X = 550 + 1.28(100) = 550 + 128 = 678 ...
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