Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Overview Discrete Random Variables • Random Variables – variables that assume numerical values associated with random outcomes from an experiment • Random variables can be: • • • Dr Tom Ilvento • • • Continuous We will look at the probability Distribution for Random Variables • Department of Food and Resource Economics Discrete Expected Values (mean) and Variance of Random Variables Special Distributions Discrete and then the Binomial Distribution Nitrous Oxide Example Example of a Random Variable • Suppose we were recording the number of dentists that use nitrous oxide (laughing gas) in their practice • We know that 60% of dentists use the gas (a priori for this problem, but most likely based on a survey result) • • • • • Let X = number of dentists in a random sample of five dentists that use use laughing gas. • 0, 1, 2, 3, 4, 5 and then list the probabilities associated with each value P(X) P(No) = .4 X is a random variable that can take on the following values: We can list the values of this Random Variable X P(Yes) = .6 • 2 Continuous: Normal Distribution 3 0 1 2 3 4 5 0.0102 0.0768 0.2304 0.3456 0.2592 0.0778 • The table shows the probability distribution for the discrete random variable X • • The table is also referred as probability distribution table Properties of the probability distribution • • Each probability is between 0 and 1 The sum of the probabilities for all values of x is equal to 1 4 Nitrous Oxide Example How did I assign the probabilities? • • P(X) • P(1) = (.6)(.4)(.4)(.4)(.4) x 5 = .0768 5 is the the number of combinations with only 1 dentist using the gas: • • • • • Or use the nCr formula Yes No No No No 5! 120 '5$ = =5 % "= & 1 # 1!(5 ! 1)! 24 No Yes No No No No No Yes No No • • No No No Yes No 1 2 3 4 5 0.0102 0.0768 0.2304 0.3456 0.2592 0.0778 Note: we are assuming independence P(X) 0 1 5 2 3 4 • The table provides the probability distribution • The graph provides a visual picture of the probability distribution of our discrete random variable I said the expectation is 3: this reflects the center of the distribution • 5 0.4 0.3 0.2 • • 0.1 1 2 P(X<2) = P(X=0) + P(X=1) P(X<2) = .0102+.0768 = .0870 If I randomly selected 5 dentists, how many would I expect to use laughing gas? Expectation = 3 We don’t have a way to solve this yet, but we will 6 The variable in the dentist example is called a Discrete Random Variable 3 4 5 7 Finite countable number of distinct possible values We can assume that the values can be listed or counted • Random Variables that fall along points on an interval, and can’t be fully counted, are call Continuous Random Variables • How can we tell that it is countable? • • 0 Just read it from the table: P(X=4) = .2592 What is the probability of less than 2 of 5 using laughing gas? • • Probability Distribution of X 0 • Types of Random Variables 0.0102 0.0768 0.2304 0.3456 0.2592 0.0778 P(X) X What is the probability of 4 of 5 dentists selected randomly Using laughing gas? • • No No No No Yes Nitrous Oxide Example • 0 P(0) = (.4)(.4)(.4)(.4)(.4) = .45 = .01024 • • X Usually is described as “the number of…” Tends to be whole numbers • • • • Number of students applying to a university Number of errors on a test Number of bacteria per cubic centimeter of water Number of heart beats of a patient 8 Specifying a discrete random variable – Tossing two coins and noting the number of Heads To describe a Discrete Random Variable Specify all the possible values it can assume • • • Assign corresponding probabilities to each value Example: Toss a coin twice, note the number of Heads • Let X=number of Heads observed • • • • • Head Head Tails Tails Head Tails Head Tails HH HT TH TT 0 TT 1 HT or TH 2 HH • • # Heads Sample Points P(X) 0.500 0 TT .5 * .5 = .25 0.375 1 HT TH (.5 * .5)* 2 = .50 2 HH .5 * .5 = .25 • Graph Table Formula – will come later Requirements: P(X) ! 0 and P(X) " 1 for all values of x #P(X) = 1 11 • Notation Specifies the probability associated with each value • • 0.250 0.125 0 0 1 2 10 Probability Distribution: P(x) for Flipping Two Coins and Noting the Number of Heads The Probability Distribution can be shown by a • • • Connecting probabilities to the values results in the probability distribution 9 Probability Distribution: P(X) • This completely defines the discrete random variable X Probability Distribution of X So X takes on the following values of Heads • • • We can assign probabilities based on a priori knowledge P(X) • • • Probability P(x = 0) = .25 P(x = 1) = .50 P(x = 2) = .25 P(x < 2) = .75 P(X=1) + P(X=0) = .50 + .25 = .75 P(x > 1) = .25 P(X=2) = .25 # Heads Sample Points P(X) 0 TT .5 * .5 = .25 1 HT TH (.5 * .5)* 2 = .50 2 HH .5 * .5 = .25 12 The manager of a large computer network has developed the following probability distribution of the number of interruptions per day • The data and assigned probabilities based on past experience • You try it: Open up Excel and enter these numbers in a worksheet • Graph it using Inset, Graph, Bar Chart X 0 1 2 3 4 5 6 Mean and Variance of a Discrete Random Variable P(X) 0.32 0.35 0.18 0.08 0.04 0.02 0.01 • We can think of our observed probability distribution for x as having a mean and variance • An expected value is another term for the mean when dealing with a probability distribution • The expected value of a discrete random variable is given by the following formula 1.000 Probability Distribution for Network Interuptions • • 0.40 0.35 0.30 P(X) 0.25 0.20 The sum of each value times the probability of that value It is a weighted average, where the weights are the probabilities n 0.15 E ( x) = " xi ! p ( xi ) = µ 0.10 0.05 0.00 0 1 2 3 4 5 Interuptions per Day 13 The Expectation of a Discrete Random Variable 2. Multiply by the probability associated with the value • X Look at the results from Excel n E ( x) = " xi ! p ( xi ) = µ i =1 0 1 2 3 4 5 6 P(X) 0.32 0.35 0.18 0.08 0.04 0.02 0.01 X*P(X) 0.00 0.35 0.36 0.24 0.16 0.10 0.06 Sum 1.000 1.270 14 Variance of a Discrete Random Variable 1. I take each value of the discrete variable 3. Sum the calculations i =1 6 • The variance of a discrete random variable is given by the following formula • From this formula, we can see that the Variance is the Expectation of the squared deviations about the population mean • Sum of the squared deviation of each value about the mean, multiplied times the probability of the value 2 We expect 1.27 interuptions per day 15 n E[( x " µ ) ] = # ( xi " µ ) 2 p ( xi ) = ! 2 i =1 16 The Variance of a Discrete Random Variable Computational Formula for the Variance 1. I take each value of the discrete variable 2. Subtract the mean 1. Square each X value 3. Square the result 4. Multiply by the probability associated with the value X 5. Then sum each of these calculations Look at the results from Excel n 0 1 2 3 4 5 6 P(X) 0.32 0.35 0.18 0.08 0.04 0.02 0.01 X*P(X) 0.00 0.35 0.36 0.24 0.16 0.10 0.06 Var Calc 0.516 0.026 0.096 0.239 0.298 0.278 0.224 Sum 1.000 1.270 1.677 E[( x " µ ) 2 ] = # ( xi " µ ) 2 p ( xi ) = ! 2 i =1 3. Add them all together 4. Then subtract µ2 n " = $ ( x i2 # P(x)) % µ 2 2 i=1 The Variance is 1.677 17 18 ! Probability Distribution of the number of interruptions per day The Standard Deviation • Remember, we said the variance is the mean squared deviation about the mean • • • The standard deviation is the square root of the variance • • • In our example, !2 = 1.677 X 0 1 2 3 4 5 6 µ = 1.27 $2 = 1.677 $ = 1.295 Probability Distribution for Network Interuptions $ = (1.677).5 = 1.295 0.40 0.35 0.30 P(X) 0.32 0.35 0.18 0.08 0.04 0.02 0.01 0.25 P(X) • 2. Then multiply by P(X) 0.20 0.15 0.10 0.05 0.00 19 0 1 2 3 Interuptions per Day 4 5 6 20 Probability Distribution for the Nitrous Oxide Example X 0 P(X) • 1 2 3 4 5 X 0.0102 0.0768 0.2304 0.3456 0.2592 0.0778 P(X) If I randomly select 5 dentists, how many can I expect to use nitrous oxide? E(X)= 0(.0102) + 1(.0768) + 2(.2304) + 3(.3456) + 4(.2592) + 5(.0778) E(X) = 3.0002 • If I randomly selected 5 dentists, what is the variance ($2)? ! = 1.0954 21 We discussed what we mean by a discrete random variable • • • variables that assume numerical values associated with random outcomes from an experiment by discrete we mean the outcomes are not continuous they are countable • We showed the probability distribution by a table and by graphing it • We solved for the mean and variance of a discrete random variable • Next we will look at a special class of discrete random variables - binomial random variables 2 3 4 5 0.0102 0.0768 0.2304 0.3456 0.2592 0.0778 Probability Distribution of X µ = 3.00 0.4 $2 = 1.20 0.3 $ = 1.01 0.2 0 Summary • 1 0.1 !2 = (0-3)2(.0102) + (1-3)2(.0768) + (2-3)2(.2304) + (3-3)2(.3456) + (4-3)2(.2592) + (5-3)2(.0778) !2 = 1.1988 • • • 0 P(X) Nitrous Oxide Example 23 0 1 2 3 4 5 22