Probability Distributions

Learning Objectives Probability Distributions Section 1: How Can We Summarize Possible Outcomes and Their Probabilities? 1. Random variable 2. Probability distributions for discrete random variables 3. Mean of a probability distribution 4. Summarizing the spread of a probability distribution 5. Probability distribution for continuous random variables 1 2 Learning Objective 1: Randomness! Suppose that the numerical values that a variable assumes are the result of some random phenomenon, e.g.,! Selecting a random sample for a population or! Performing a randomized experiment Learning Objective 1: Random Variable! A random variable is a numerical measurement of the outcome of a random phenomenon. 3 4 1

Learning Objective 1: Random Variable! Use letters near the end of the alphabet, such as x, to symbolize a particular value of the random variable! Use a capital letter, such as X, to refer to the random variable itself. Example: Flip a coin three times! X=number of heads in the 3 flips; defines the random variable! x=2; represents a possible value of the random variable Learning Objective 2: Probability Distribution! The probability distribution of a random variable specifies its possible values and their probabilities. Note: In spite of the randomness of the variable, there is a pattern of randomness that allows us to specify probabilities for the outcomes 5 6 Learning Objective 2: Probability Distribution of a Discrete Random Variable! A discrete random variable X has separate values (such as 0,1,2, ) as its possible outcomes! Its probability distribution assigns a probability P(x) to each possible value x:! For each x, the probability P(x) falls between 0 and 1! The sum of the probabilities for all the possible x values equals 1 Learning Objective 2: Example! What is the estimated probability of at least three home runs? P(3)+P(4)+P(5)=0.13+0.03+0.01=0.17 7 8 2

Learning Objective 3: The Mean of a Discrete Probability Distribution! The mean of a probability distribution for a discrete random variable is µ = x p(x) where the sum is taken over all possible values of x.! The mean of a probability distribution is denoted by the parameter, µ.! The mean is a weighted average; values of x that are more likely receive greater weight P(x) 9 Learning Objective 3: Expected Value of X! The mean of a probability distribution of a random variable X is also called the expected value of X.! The expected value reflects not what we ll observe in a single observation, but rather that we expect for the average in a long run of observations.! It is not unusual for the expected value of a random variable to equal a number that is NOT a possible outcome. 10 Learning Objective 3: Example! Find the mean of this probability distribution. The mean: µ = x p(x) = 0(0.23) + 1(0.38) + 2(0.22) + 3(0.13) + 4(0.03) + 5(0.01) = 1.38 11 Learning Objective 4: The Standard Deviation of a Probability Distribution The standard deviation of a probability distribution, denoted by the parameter, σ, measures its spread.! Larger values of σ correspond to greater spread.! Roughly, σ describes how far the random variable falls, on the average, from the mean of its distribution 12 3

Learning Objective 5: Continuous Random Variable! A continuous random variable has an infinite continuum of possible values in an interval.! Examples are: time, age and size measures such as height and weight.! Continuous variables are measured in a discrete manner because of rounding. Learning Objective 5: Probability Distribution of a Continuous Random Variable! A random variable is continuous if its set of possible values forms an interval.! Its probability distribution is specified by a density curve: the probability of any interval is the area under the curve and above that interval.! Each interval has probability between 0 and 1.! The interval containing all possible values has probability equal to 1. 13 14 Learning Objectives Section 2: How Can We Find Probabilities for Bell-Shaped Distributions? 1. Normal Distribution 2. 68-95-99.7 Rule for normal distributions 3. Z-Scores and the Standard Normal Distribution 4. The Standard Normal Table: Finding Probabilities 5. Using the TI-calculator: find probabilities 15 16 4

Learning Objectives 6. Using the Standard Normal Table in Reverse 7. Using the TI-calculator: find z-scores 8. Probabilities for Normally Distributed Random Variables 9. Percentiles for Normally Distributed Random Variables 10. Using Z-scores to Compare Distributions Learning Objective 1: Normal Distribution The normal distribution is symmetric, bellshaped and characterized by its mean µ and standard deviation σ.! The normal distribution is the most important distribution in statistics! Many distributions have an approximate normal distribution! Approximates many discrete distributions well when there are a large number of possible outcomes! Many statistical methods use it even when the data are not bell shaped 17 18 Learning Objective 1: Normal Distribution! Normal distributions are! Bell shaped! Symmetric around the mean! The mean (µ) and the standard deviation (σ) completely describe the density curve! Increasing/decreasing µ moves the curve along the horizontal axis! Increasing/decreasing σ controls the spread of the curve Learning Objective 1: Normal Distribution! Within what interval do almost all of the men s heights fall? Women s height? 19 20 5

Learning Objective 2: 68-95-99.7 Rule for Any Normal Curve! 68% of the observations fall within one standard deviation of the mean! 95% of the observations fall within two standard deviations of the mean! 99.7% of the observations fall within three standard deviations of the mean Learning Objective 2: Example : 68-95-99.7% Rule! Heights of adult women! can be approximated by a normal distribution! µ = 65 inches; σ = 3.5 inches! 68-95-99.7 Rule for women s heights " 68% are between 61.5 and 68.5 inches [ µ - σ = 65-3.5 ] " 95% are between 58 and 72 inches [ µ + 2σ = 65 + 2(3.5) = 65 + 7 ] " 99.7% are between 54.5 and 75.5 inches [ µ - 3σ = 65-3(3.5) = 65-10.5 ] 21 22 Learning Objective 2: Example : 68-95-99.7% Rule! What proportion of women are less than 69 inches tall?? = 84% 16% -1 68% (by 68-95-99.7 Rule)? +1 23 65 68.5 (height values) Learning Objective 3: Z-Scores and the Standard Normal Distribution! The z-score for a value x of a random variable is the number of standard deviations that x falls from the mean z = x µ σ! A negative (positive) z-score indicates that the value is below (above) the mean! z-scores can be used to calculate the probabilities of a normal random variable using the normal tables in the back of the book 24 6

Learning Objective 3: Z-Scores and the Standard Normal Distribution! A standard normal distribution has mean µ=0 and standard deviation σ=1! When a random variable X has a normal distribution and its values are converted to z- scores (by subtracting the mean and dividing by the standard deviation), the new random variable Z whose values are these z-scores has the standard normal distribution. Learning Objective 4: Table A: Standard Normal Probabilities Table A enables us to find normal probabilities! It tabulates the normal cumulative probabilities falling below the point µ+zσ To use the table:! Find the corresponding z-score! Look up the closest standardized score (z) in the table.! First column gives z to the first decimal place! First row gives the second decimal place of z! The corresponding probability found in the body of the table gives the probability of falling below the z-score 25 26 Learning Objective 4: Example: Using Table A! Find the probability that a normal random variable takes a value less than 1.43 standard deviations above µ; P(z<1.43)=.9236 Learning Objective 4: Example: Using Table A! Find the probability that a normal random variable takes a value greater than 1.43 standard deviations above µ: P(z>1.43)=1-.9236=.0764 27 28 TI Calculator = Normcdf(-1e99,1.43,0,1)=.9236 TI Calculator = Normcdf(1.43,1e99,0,1)= 0.0764 7

Learning Objective 4: Example:! Find the probability that a normal random variable assumes a value within 1.43 standard deviations of µ! Probability below 1.43σ =.9236! Probability below -1.43σ =.0764 (1-.9236)! P(-1.43<z<1.43) =.9236-.0764=.8472 Learning Objective 5: Using the TI Calculator To calculate the cumulative probability! 2 nd DISTR; 2:normalcdf(lower bound, upper bound,mean,sd)! Use 1E99 for negative infinity and 1E99 for positive infinity TI Calculator = Normcdf(-1.43,1.43,0,1)=.8472 29 30 Learning Objective 5: Find Probabilities Using TI Calculator! Find probability to the left of -1.64! P(z<-1.64)=normcdf(-1e99,-1.64,0,1)=.0505! Find probability to the right of 1.56! P(z>1.56)=normcdf(1.56,1e99,0,1)=.0594! Find probability between -.50 and 2.25! P(-.5<z<2.25)=normcdf(-.5,2.25,0,1)=.6793 Learning Objective 6: How Can We Find the Value of z for a Certain Cumulative Probability?! To solve some of our problems, we will need to find the value of z that corresponds to a certain normal cumulative probability! To do so, we use Table A in reverse! Rather than finding z using the first column (value of z up to one decimal) and the first row (second decimal of z)! Find the probability in the body of the table! The z-score is given by the corresponding values in the first column and row 31 32 8

Learning Objective 6: How Can We Find the Value of z for a Certain Cumulative Probability?! Example: Find the value of z for a cumulative probability of 0.025.! Look up the cumulative probability of 0.025 in the body of Table A.! A cumulative probability of 0.025 corresponds to z = -1.96.! Thus, the probability that a normal random variable falls at least 1.96 standard deviations below the mean is 0.025. Learning Objective 6: How Can We Find the Value of z for a Certain Cumulative Probability?! Example: Find the value of z for a cumulative probability of 0.975.! Look up the cumulative probability of 0.975 in the body of Table A.! A cumulative probability of 0.975 corresponds to z = 1.96.! Thus, the probability that a normal random variable takes a value no more than 1.96 standard deviations above the mean is 0.975. 33 34 Learning Objective 7: Using the TI Calculator to Find Z-Scores for a Given Probability! 2 nd DISTR 3:invNorm; Enter! invnorm(percentile,mean,sd)! Percentile is the probability under the curve from negative infinity to the z-score! Enter Learning Objective 7: Examples! The probability that a standard normal random variable assumes a value that is z is 0.975. What is z? Invnorm(.975,0,1)=1.96! The probability that a standard normal random variable assumes a value that is > z is 0.0275. What is z? Invnorm(.975,0,1)=1.96! The probability that a standard normal random variable assumes a value that is z is 0.881. What is z? Invnorm(1-.881,0,1)=-1.18! The probability that a standard normal random variable assumes a value that is < z is 0.119. What is z? Invnorm(.119,0,1)= -1.18 35 36 9

Learning Objective 7: Example! Find the z-score z such that the probability within z standard deviations of the mean is 0.50.! Invnorm(.75,0,1)=.67! Invnorm(.25,0,1)= -.67! Probability = P(-.67<Z<.67)=.5 Learning Objective 8: Finding Probabilities for Normally Distributed Random Variables 1. State the problem in terms of the observed random variable X, i.e., P(X<x) 2. Standardize X to restate the problem in terms of a standard normal variable Z P(X < x) = P Z < z = x µ σ 3. Draw a picture to show the desired probability under the standard normal curve 4. Find the area under the standard normal curve using Table A 37 38 Learning Objective 8: P(X<x)! Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure less than 100? 100 120! P(X<100) = P Z < ( ) = P(z < 1.00) =.1587 20! Normcdf(-1E99,100,120,20)=.1587! 15.9% of adults have systolic blood pressure less than 100 Learning Objective 8: P(X>x)! Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure greater than 100?! P(X>100) = 1 P(X<100) 100 120 P Z < ( ) = P(Z < 1.00) =.1587 20! P(X>100)= 1-.1587=.8413! Normcdf(100,1e99,120,20)=.8413! 84.1% of adults have systolic blood pressure greater than 100 39 40 10

Learning Objective 8: P(X>x)! Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure greater than 133?! P(X>133) = 1 P(X<133) 133 120 P Z < ( ) = P(Z <.65) =.7422 20! P(X>133)= 1-.7422=.2578! Normcdf(133,1E99,120,20)=.2578! 25.8% of adults have systolic blood pressure greater than 133 41 Learning Objective 8: P(a<X<b)! Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure between 100 and 133?! P(100<X<133) = P(X<133)-P(X<100) 133 120 P Z < ( ) P Z < ( 100 120 ) = 20 20 P(Z <.65) P(Z < 1.00) =.7422.1587 =.5835! Normcdf(100,133,120,20)=.5835! 58% of adults have systolic blood pressure between 100 and 133 42 Learning Objective 9: Find X Value Given Area to Left! Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What is the 1 st quartile?! P(X<x)=.25, find x:! Look up.25 in the body of Table A to find z= -0.67! Solve equation to find x:! Check: x = µ + zσ =120 + ( 0.67) * 20 =106.6! P(X<106.6) P(Z<-0.67)=0.25! TI Calculator = Invnorm(.25,120,20)=106.6 Learning Objective 9: Find X Value Given Area to Right! Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. 10% of adults have systolic blood pressure above what level?! P(X>x)=.10, find x.! P(X>x)=1-P(X<x)! Look up 1-0.1=0.9 in the body of Table A to find z=1.28! Solve equation to find x:! Check: x = µ + zσ =120 + (1.28) *20 =145.6! P(X>145.6) =P(Z>1.28)=0.10! TI Calculator = Invnorm(.9,120,20)=145.6 43 44 11

Learning Objective 10: Using Z-scores to Compare Distributions Z-scores can be used to compare observations from different normal distributions! Example:! You score 650 on the SAT which has µ=500 and σ=100 and 30 on the ACT which has µ=21.0 and σ=4.7. On which test did you perform better?! Compare z-scores SAT: ACT: 650 500 30 21 z = =1.5 z = =1.91 100 4.7! Since your z-score is greater for the ACT, you performed better on this exam 45 Section 3: How Can We Find Probabilities When Each Observation Has Two Possible Outcomes? 46 Learning Objectives 1. The Binomial Distribution 2. Conditions for a Binomial Distribution 3. Probabilities for a Binomial Distribution 4. Factorials 5. Examples using Binomial Distribution 6. Do the Binomial Conditions Apply? 7. Mean and Standard Deviation of the Binomial Distribution 8. Normal Approximation to the Binomial Learning Objective 1: The Binomial Distribution! Each observation is binary: it has one of two possible outcomes.! Examples:! Accept, or decline an offer from a bank for a credit card.! Have, or do not have, health insurance.! Vote yes or no on a referendum. 47 48 12

Learning Objective 2: Conditions for the Binomial Distribution! Each of n trials has two possible outcomes: success or failure.! Each trial has the same probability of success, denoted by p.! The n trials are independent. Learning Objective 3: Probabilities for a Binomial Distribution! Denote the probability of success on a trial by p.! For n independent trials, the probability of x successes equals:! The binomial random variable X is the number of successes in the n trials. P(x) = n! x!(n - x)! px (1 p) n x, x = 0,1,2,...,n 49 50 Learning Objective 4: Factorials Rules for factorials:! n!=n*(n-1)*(n-2) 2*1! 1!=1! 0!=1 For example,! 4!=4*3*2*1=24 Learning Objective 5: Example: Finding Binomial Probabilities! John Doe claims to possess ESP.! An experiment is conducted:! A person in one room picks one of the integers 1, 2, 3, 4, 5 at random.! In another room, John Doe identifies the number he believes was picked.! Three trials are performed for the experiment.! Doe got the correct answer twice. 51 52 13

Learning Objective 5: Example 1 Learning Objective 5: Example 1 If John Doe does not actually have ESP and is actually guessing the number, what is the probability that he d make a correct guess on two of the three trials? # The three ways John Doe could make two correct guesses in three trials are: SSF, SFS, and FSS. # Each of these has probability: (0.2)2(0.8)=0.032. # The total probability of two correct guesses is 3(0.032)=0.096. 53 The probability of exactly 2 correct guesses is the binomial probability with n = 3 trials, x = 2 correct guesses and p = 0.2 probability of a correct guess. P(2) = 3! 2!1! (0.2)2 (0.8) 1 = 3(0.04)(0.8) = 0.096 2nd Vars 0:binampdf(n,p,x) Binampdf(3,.2,2)=0.096 54 Learning Objective 5: Binomial Example 2! 1000 employees, 50% Female! None of the 10 employees chosen for management training were female.! The probability that no females are chosen is: P(0) = 10! 0!10! (0.50)0 (0.50) 10 = 0.001 Learning Objective 6: Do the Binomial Conditions Apply?! Before using the binomial distribution, check that its three conditions apply:! Binary data (success or failure).! The same probability of success for each trial (denoted by p).! Independent trials.! Binompdf(10,.5,0)=9.765625E-4! It is very unlikely (one chance in a thousand) that none of the 10 selected for management training would be female if the employees were chosen randomly 55 56 14

Learning Objective 6: Do the Binomial Conditions Apply to Example 2?! The data are binary (male, female).! If employees are selected randomly, the probability of selecting a female on a given trial is 0.50.! With random sampling of 10 employees from a large population, outcomes for one trial does not depend on the outcome of another trial Learning Objective 7: Binomial Mean and Standard Deviation! The binomial probability distribution for n trials with probability p of success on each trial has mean µ and standard deviation σ given by: µ = np, σ = np(1- p) 57 58 Learning Objective 7: Example: Racial Profiling? Learning Objective 7: Example: Racial Profiling?! Data:! 262 police car stops in Philadelphia in 1997.! 207 of the drivers stopped were African-American.! In 1997, Philadelphia s population was 42.2% African-American.! Does the number of African-Americans stopped suggest possible bias, being higher than we would expect (other things being equal, such as the rate of violating traffic laws)? 59! Assume:! 262 car stops represent n = 262 trials.! Successive police car stops are independent.! P(driver is African-American) is p = 0.422.! Calculate the mean and standard deviation of this binomial distribution: µ = 262(0.422) = 111 σ = 262(0.422)(0.578) = 8 60 15

Learning Objective 7: Example: Racial Profiling? Learning Objective 7: Example: Racial Profiling?! Recall: Empirical Rule! When a distribution is bell-shaped, close to 100% of the observations fall within 3 standard deviations of the mean. u - 3σ = 111-3(8) = 87 µ + 3σ = 111+ 3(8) = 135 61! If there is no racial profiling, we would not be surprised if between about 87 and 135 of the 262 drivers stopped were African-American.! The actual number stopped (207) is well above these values.! The number of African-Americans stopped is too high, even taking into account random variation. Limitation of the analysis: Different people do different amounts of driving, so we don t really know that 42.2% of the potential stops were African- American. 62 Learning Objective 8: Approximating the Binomial Distribution with the Normal Distribution! The binomial distribution can be well approximated by the normal distribution when the expected number of successes, np, and the expected number of failures, n(1-p) are both at least 15. 63 16