Understanding Random Variables in Statistics

Statistics Session 12 Random Variables part 1 Ezra Halleck, City Tech (CUNY), Fall 2021

Opening Example Now that you know a little about probability, do you feel lucky enough to play the lottery? If you have $20 to spend on lunch , how much of it are you willing to spend on $5 lottery tickets? Do you think you will profit, on average, if you continue buying lottery tickets over time? When you purchase a lottery ticket for $5 and you don t win anything, then your earnings are 5. If you get a $20 prize, then your earnings are 20 5 = 15. Playing the lottery and determining your earnings is an experiment with a numeric outcome and hence is an example of a random variable. 2

5.1 Random Variables A random variable (RV) is a variable whose numerical value is determined by the outcome of a random experiment. There are two types of RV s depending on the nature of the numeric outcome: A discrete random variable (DRV) assumes countable values. A continuous Random Variable (CRV) assumes any value in one or more intervals. Playing the lottery is discrete. 3

2nd example of DRV: # of Vehicles Owned by Families Number of Vehicles Owned Frequency 0 1 2 3 4 Blank N = 2000 Note how the DRV is given as a table, with outcomes in one column and their probabilities (relative frequencies) in another. Relative Frequency .015 .235 .425 .245 .080 30 470 850 490 160 Sum = 1.000 Vehicles here would mean cars and light trucks. Would this table be valid for NYC? Where do you think the data for this table were collected? 4

Examples of Discrete Random Variables The number of: 1. vehicles owned by a family in a suburban community 2. cars sold at a dealership during a given month 3. houses in a certain block 4. fish caught on a fishing trip 5. complaints received at the office of an airline on a given day 6. customers who visit a bank during any given hour 7. heads obtained in three tosses of a coin Earnings (prize minus cost of ticket) for lottery 5

Continuous Random Variable Definition A random variable that can assume any value in one or more intervals is called a continuous random variable. 6

Examples of Continuous Random Variables 1. life of a battery in hours (depicted on previous slide) 2. length of a room in cm 3. time taken to commute from home to work in minutes 4. amount of milk in a gallon jug in gallons (note that we do not expect a gallon to contain exactly 1 gallon of milk but either slightly more or slightly less than one gallon) 5. weight of a letter in oz (for determining postage) 6. value (worth) of a house in dollars Note that worth may be treated as a continuous random variable, although often it is rounded to the nearest $ or . 7

The Hypergeometric Probability Distribution We now turn our attention to one specific DRV. In the next session, we will go into how to find the mean and standard deviation for a DRV. For a Hypergeometric Probability Distribution, the probability of x successes in n trials is given by C C ( ) = n r P x r x N r C N n These are the binomial coefficients or combinations which were introduced in the previous session. 8

The Hypergeometric Probability Distribution For a Hypergeometric Probability Distribution, the probability of x successes in n trials is given by C P x = C ( ) n r r x N r C N n We would like to develop an intuition behind the formula. The lower-case r and x refer to the sizes of the success pool and successes selected, respectively. The upper-case N and lower-case n refer to the sizes of the entire pool which includes both successes and failures . ? ? and ? ? are the sizes of the failure pool and failures selected, respectively. 9

Exercise: fill in the #s successes failures C C ( ) = n r P x r x N r C N overall n symbol value 2 x 6 = 12 3 36 9 r x N n N-r 4 x 6 = 24 6 n-x Now fill in formula 10 Diagram courtesy of https://www.wizeprep.com/

symbol value 2 x 6 = 12 3 36 9 r x N n n-r 4 x 6 = 24 6 N-x C C ( ) = n r P x r x N r C N n ? 3 12?3 24?6 36?9 = As check, sum of left # s in top = left # in bottom Ditto for right: sum of right # s in top = right # in bottom 11

C C ( ) = n r P x r x N r C N n ? 3 12?3 24?6 36?9 = These # s are too big to do by hand Instead, we use Excel (combination is COMBIN): = COMBIN(12,3)* COMBIN(24,6) /COMBIN(36,9) For clarification, parentheses would not hurt. However, they are not necessary. Why? 12

Example 5-16 Brown Manufacturing makes auto parts. Last week the company shipped 25 auto parts to a dealer. Later, it found out that 5 of those parts were defective. By the time the company manager contacted the dealer, 4 auto parts from that shipment had already been sold. What is probability that 3 of those 4 parts were good and 1 was defective? N 25 4 ( ) 3 P = symbol value 20 3 r x 20! 5! - ( ) 25! ( ) 3! 20 3 ! 1! 5 1 ! - C C C C n N-r = = - n x - 20 3 5 1 r x N r C C 5 1 25 4 N n ( ) - 4! 25 4 ! These # s are big, so again we use Excel: = COMBIN(20,3)* COMBIN(5,1) /COMBIN(25,4)=.4506 n-x ( )( ) 1140 5 12,650 = = Thus, the probability that three of the four parts sold are good and one is defective is .4506. .4506 13

Example 5-16: Defective parts Brown Manufacturing makes auto parts. Last week the company shipped 25 auto parts to a dealer. Later, it found out that 5 of those parts were defective. By the time the company manager contacted the dealer, 4 auto parts from that shipment had already been sold. What is the probability that 3 of those 4 parts were good and 1 was defective? ( ) 3 N n C These # s are a little big to do by hand, so again we use Excel: = COMBIN(20,3)* COMBIN(5,1) /COMBIN(25,4)=.4506 20! 5! - ( ) 25! ( ) 3! 20 3 ! 1! 5 1 ! - C C C C = = = N r - n x - 20 3 5 1 P r x C 25 4 ( ) - 4! 25 4 ! ( )( ) 1140 5 12,650 = = .4506 14

Example 5-17: employee selection by gender Dawn Corp. has 12 employees who are managers. Of them, 7 are female and 5 are male. The company will send 3 of the managers to a conference. If the 3 managers are randomly selected, find probability that (a) all 3 of them are female; (b) at most 1 of them is female. 15

Example 5-17: employee selection by gender symbol value 12 employees: 7 female and 5 male. Selection of 3: (a) chance all 3 of them are female? 7 3 r x 12 3 N n N-r 5 0 N-x )( ) 220 ( 35 1 C C C C C ( ) 3 = = = = n x .1591 P 7 3 5 0 r x N r C 12 3 N n Thus, chance that all 3 managers selected are female is .1591. 16

Example 5-17: employee selection by gender symbol value 12 employees: 7 female and 5 male. Selection of 3: (b) chance at most 1 of them is female? r 7 0, 1 x 12 3 5 3,2 N ( )( 1 10 220 7 10 220 = = ) ( )( 1 10 ) ( )( 1 10 220 ( )( = ) x r C 0 C - C 12 C C C 5 C C C r C = C C 7 C = - C C n x - n x C N r C ( ) 0 ( ) 0 P = = ( ) 0 n N-r = = = = = = = .0455 .0455 = 220 7 10 220 .3637 = - n x - N r - .0455 7 0 5 3 P r x N r C N C x C 0 7 3 5 3 P x 12 1 7 3 C 5 C C N n N 12 3 C n 3 n ( )( ( )( 7 10 220 ) ) ) x r - 2 C r C = C C 7 C = 5 1 - C n x - n x C N r C N-x ( ) 1 ( ) 1 P = = ( ) 1 = = = = = .3182 .3182 = = - n x - N r - .3182 7 1 P r x N r C N C P 2 5 2 P C C 12 12 .0455 .3182 + ( ) 1 .0455 .3182 = 3 12 N n N 3 n 3 n ( ) ( ) 0 ( ) 0 P = ( ) 1 ( ) 1 P + ( P ) + + ( ) 0 ( ) = = 1 = .0455 .3182 + = + 1 1 .3637 .3637 P x P x P P x P Thus, chance that at most 1 of 3 managers selected is female is .3637. 17

Example 5-17: employee selection by gender Dawn Corp. has 12 employees who are managers. Of them, 7 are female and 5 are male. The company will send 3 of the managers to a conference. If the 3 managers are randomly selected, find probability that (a) all 3 of them are female; (b) at most 1 of them is female. ? ? = ???????.???? ?,?,?,?,????? = ???????.???? 3,7,3,12,????? ? ? = ???????.???? ?,?,?,?,???? = ???????.???? 1,7,3,12,???? x n r N varies # successes selected 3 selection size 7 # successes in pool 12 pool size x P(x) 0 0.0455 0.0455 1 0.3182 0.3636 2 0.4773 0.8409 3 0.1591 sum P(X<=x) 1 1 18