m16819 Continuous Random Variables: The Uniform Distribution 1.14 2008/06/05 09:38:02 GMT-5 2009/02/20 19:18:32.684 US/Central Susan Dean Susan Dean deansusan@deanza.edu Barbara Illowsky Barbara Illowsky, Ph.D. illowskybarbara@deanza.edu Susan Dean Susan Dean deansusan@deanza.edu Barbara Illowsky Barbara Illowsky, Ph.D. illowskybarbara@deanza.edu Connexions Connexions cnx@cnx.org Maxfield Foundation Maxfield Foundation cnx@cnx.org continuous distribution elementary function probability random statistics uniform variable Mathematics and Statistics This module describes the properties of the Uniform Distribution which describes a set of data for which all values have an equal probability. en The previous problem is an example of the uniform probability distribution.Illustrate the uniform distribution. The data that follows are 55 smiling times, in seconds, of an eight-week old baby. 10.4 19.6 18.8 13.9 17.8 16.8 21.6 17.9 12.5 11.1 4.9 12.8 14.8 22.8 20.0 15.9 16.3 13.4 17.1 14.5 19.0 22.8 1.3 0.7 8.9 11.9 10.9 7.3 5.9 3.7 17.9 19.2 9.8 5.8 6.9 2.6 5.8 21.7 11.8 3.4 2.1 4.5 6.3 10.7 8.9 9.4 9.4 7.6 10.0 3.3 6.7 7.8 11.6 13.8 18.6 sample mean = 11.49 and sample standard deviation = 6.23We will assume that the smiling times, in seconds, follow a uniform distribution between 0 and 23 seconds, inclusive. This means that any smiling time from 0 to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution. Let X = length, in seconds, of an eight-week old baby's smile.The notation for the uniform distribution isX ~ U(a, b) where a = the lowest value of X and b = the highest value of X.The probability density function is fX = 1b-a for a X b.For this example, X ~ U(0, 23) and fX = 123-0 for 0 X 23.Formulas for the theoretical mean and standard deviation are μ a+b 2 and σ (b-a)2 12 For this problem, the theoretical mean and standard deviation are μ 0+23 2 11.50 seconds and σ (23-0)2 12 6.64 secondsNotice that the theoretical mean and standard deviation are close to the sample mean and standard deviation. What is the probability that a randomly chosen eight-week old baby smiles between 2 and 18 seconds? Find P(2 X 18) . P(2 X 18) (base)(height) (18-2)⋅123 1623 . Find the 90th percentile for an eight week old baby's smiling time. Ninety percent of the smiling times fall below the 90th percentile, k, so P(X k) 0.90 P(X k) 0.90 (base)(height)=0.90 (k-0)⋅123=0.90 k 23⋅0.90 20.7 Find the probability that a random eight week old baby smiles more than 12 seconds KNOWING that the baby smiles MORE THAN 8 SECONDS. Find P(X 12|X 8) There are two ways to do the problem. For the first way, use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than 8 seconds. Write a new fX: fX 1 23-8 115 for 8 X 23 P(X 12|X 8) (23-12)⋅115 1115 For the second way, use the conditional formula from Probability Topics with the original distribution X ~ U(0,23): P ( A | B ) = P ( A AND B ) P ( B ) For this problem, A is ( X 12 ) and B is ( X 8 ) . So, P(X 12|X 8) (X>12 AND X >8) P ( X 8 ) P ( X 12 ) P ( X 8 ) 11 23 15 23 0.733 Uniform: The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between 0 and 15 minutes, inclusive. What is the probability that a person waits fewer than 12.5 minutes? Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. X~U(0,15). Write the probability density function. fX = 115-0 =115 for 0 X 15. Find P(X 12.5). Draw a graph. P(X k) (base)(height) =(12.5-0) ⋅115 =0.8333 The probability a person waits less than 12.5 minutes is 0.8333. On the average, how long must a person wait? Find the mean, μ, and the standard deviation, σ. μ= a+b 2 = 15+0 2 = 7.5. On the average, a person must wait 7.5 minutes.σ= (b-a)2 12 = (15-0)2 12 = 4.3. The Standard deviation is 4.3 minutes. Ninety percent of the time, the time a person must wait falls below what value? NoteThis asks for the 90th percentile. Find the 90th percentile. Draw a graph. Let k = the 90th percentile. P(X k) (base)(height) =(k-0) ⋅(115) 0.90= k⋅115 k = (0.90)(15) = 13.5k is sometimes called a critical value.The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes. Uniform: The average number of donuts a nine-year old child eats per month is uniformly distributed from 0.5 to 4 donuts, inclusive. Let X = the average number of donuts a nine-year old child eats per month. Then X ~ U(0.5, 4). The probability that a randomly selected nine-year old child eats an average of more than two donuts is _______. 0.5714 Find the probability that a different nine-year old child eats an average of more than two donuts given that his or her amount is more than 1.5 donuts. The second probability question has a conditional (refer to "Probability Topics"). You are asked to find the probability that a nine-year old eats an average of more than two donuts given that his/her amount is more than 1.5 donuts. Solve the problem two different ways (see the first example). You must reduce the sample space. First way: Since you already know the child eats more than 1.5 donuts, you are no longer starting at a = 0.5 donut. Your starting point is 1.5 donuts.Write a new f(X):fX =14-1.5 =25 for 1.5 X 4.Find P(X 2|X 1.5) . Draw a graph. P(X 2|X 1.5) =(base)(new height) = (4-2) (2/5) =? 45 The probability that a nine-year old child eats an average of more than 2 donuts when he/she has already eaten more than 1.5 donuts is 45.Second way: Draw the original graph for X ~ U(0.5, 4). Use the conditional formula P(X 2|X 1.5) P(X> 2AND X>1.5) P ( X 1.5 ) P ( X 2 ) P ( X 1.5 ) 2 3.5 2.5 3.5 0.8 =45 See "Summary of the Uniform and Exponential Probability Distributions" for a full summary. Conditional Probability The likelihood that an event will occur given that another event has already occurred. Uniform Distribution A continuous random variable (RV) that has equally likely outcomes over the domain, a<x<b size 12{a<x<b} {}. Often referred as the Rectangular distribution because the graph of the pdf has the form of a rectangle. Notation: X~U(a,b) size 12{X "~" U \( a,b \) } {}. The mean is μ=a+b2 size 12{μ= { {a+b} over {2} } } {} and the standard deviation is σ (b-a)2 12 The probability density function is fX = 1b-a for a X b. The cumulative distribution is P(X≤x)=x−ab−a size 12{P \( X <= x \) = { {x-a} over {b-a} } } {}.