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Continuous Random Variables and the Normal Distribution What are Continuous Random Variables? n n FRE C408 Fall 1999 Dr. Ilvento When dealing with a PDF n It is not particularly useful to think of a probability when our variable takes on a particular value n n n n P(a # x # b) = some proportion of the curve For every distribution with a mean (:) and a standard deviation (F) there is a different normal curve Thus, there are an infinite number of normal curves If x is distributed as a normal variable then it is designated as: n x~N PDF f(x) Normal Distribution n n P(x=a) = 0 Normal Distribution n n One bell shaped, symmetrical distribution is the normal distribution It is defined by two parameters n But, we can think of areas under the curve as reflecting a probability n n n Unlike Discrete Random Variables, Continuous Random Variables take on any point in the interval Thus the probability function is continuous It is referred to as a Probability Density Function n : F f ( x) = 2 1 e − (1/ 2)[( x − µ ) / σ ] σ 2π Properties of the Normal Distribution n n n n n Symmetrical, Bell-shaped curve Defined by the mean and standard deviation Mean = Median = Mode Since its properties are defined by a formula, we can a priori define probabilities associated with the curve If we convert our variable to a z-score, we make it possible to use one table for all normal pdf n n mean = 0 std dev = 1 1 Standard Normal Distribution n n n :=0 F=1 If x ~ N, then n Finding Areas under the Curve n n Use the Table on page 779 Steps n z = (x - :)/F ~ N n n n Draw the curve and the area we are interested in Convert the values to zscores Read the proportions in the table Do any calculations necessary Look at the table on page 779 n n Only ½ of the curve since the distribution is symmetrical Allows for two decimal places n n n n Vertical axis is the ones and first decimal place Horizontal axis is the second decimal place The probabilities in the table represent the probability up to that half of the curve Thus, a z-score of n n 3.00 corresponds to .4987 of one half of the curve 3.09 corresponds to .4990 of one half of the curve Problem n n Find the area under the standard normal curve for a z-score between 0 and 2.00. Answer: n n n n A z-score of zero is at the mean, with a probability of zero A z-score of 2 is two standard deviations above the mean, which corresponds to a probability of .4772 We want the area from the mean to 2 standard deviations from the mean Equal to .4772 Problems in Class n Suppose a variable is distributed normally with a mean = 300 and a standard deviation of 30 n n X~N : = 300 F = 30 What is the probability that x is more than 2 standard deviations from the mean? n n n n Draw it out Calculate z-score Check the table Do any final calculations 2 Problem n X~N n n n n n Problem n X~N n n n n Z=2 In table when z = 2.00 we have a probability up to that point on one side of the curve of .4772 .5 - .4772 = .0228 one side of curve 2 x .0228 = .0456 both sides of curve Problem : = 300 F = 30 More than 3 std deviations n : = 300 F = 30 More than 2 std deviations n n Z = 3.00 In table when z = 3.00 we have a probability up to that point on one side of the curve of .4987 .5 - .4987 = .0013 one side of curve 2 x .0013 = .0026 both sides of curve X~N : = 300 F = 30 Probability that x is between 260 and 360? n n n n Draw it Calculate z-scores Look up in the table Do any calculations Problem n n X~N : = 300 F = 30 Probability that x is between 260 and 360? n n n n n X =300 z = (260 – 300)/30 = -1.33 X = 340 z = (360 – 300)/30 = 2.00 Z for 1.33 = .4082 Z for 2.00 = .4772 .4082 + .4772 = .8854 3 What is the value at the 80th percentile? n n Given X ~ N : = 300 F = 30 What am I looking for? n n n n n The 80th percentile reflects everything up to the mean (50th percentile) Plus .30 more Look in the table for .30 It is between .84 (p=.2995) and .85 (p=.3023) I could extrapolate, but I know it is a lot closer to .84 n The normal distribution can be used as an approximation for the Binomial distribution of x successes in n trials n n n n n n n n n .841 = (x – 300)/30 30 A .841 = x – 300 300 + (30 A .841) = x 325.23 = x The 80th percentile is at .841 Example of test calculation n n Yes, because : = np = .06(200) = 12 F = (200 A.06 A.94).5 = (11.28).5 = 3.36 and 12.0 " 3(3.36) = 1.92 to 22.08 To solve the problem Example on page 214 Quality control for pocket calculators n n Calculate : "3F = np " 3 (npq).5 If this interval lies in the range of 0 to n, we can go forward And we use a correction for continuity of .5 Can we use Binomial Approximation? n n n This save a lot of calculations Calculator problem n Solve for x n Provided we can assume the following n n n .841 is a good approximation Normal Distribution as an approximation of the Binomial Distribution n Next - n Defective rate is .06 or 6% Sample is 200 (I.e., n= 200 trials) What is the probability of 20 or more defects are observed? Calculator Problem n Next we look at what we want to solve for n n n n Probability that 20 or more defects are observed Or we can look at the complement, which is 1- p(19 or less) We use the correction for continuity of 19 + .5 = 19.5 And solve for the probability of 19.5 or less. 4 Calculator Problem n n n n n n n n Z = (19.5 – 12)/3.36 Z = 2.23 Look it up in the table = .4871 The probability of less than 19.5 = .5 + .4871 = .9871 The probability of the complement is 1 - .9871 = .0129 5