Random Variables Random variables assigns a number to each outcome of a random circumstance, or equivalently, a random variable assigns a number to each unit in a population.

Random Variables Discrete random variable can only take one of a countable number of distinct values. Examples: Number of siblings, sum of two dice rolled,… Continuous random variable cannot be displayed in a table since there are innumerable values. Examples: Time, height, weight,… There are 2 types of random variables we will use in this statistics course. RV-stands for random variable.

Discrete Random Variable Create a probability distribution table assigning distinct values of the discrete random variable the corresponding relative frequency (probability). This combination of histogram and probability distribution was created in Excel. A probability table includes one row (or column) of possible distinct outcomes of sample space and the assigned probabilities of the outcomes.

Cumulative Probability Distribution Probability Distribution of # of Tails in 2 flips Cumulative Probability Distribution

Notation of Discrete RV Capital letter represents random variable; T stands for number of tails in 2 flips of coin Lower case letter represents a number of the discrete rv could assume; t could equal 0, 1, or 2 P(T = t) is the probability distribution function in which T equals t; P(T = 1) = .5 P(T t) is a cumulative probability function notation of probability that T equals or is less than t; P(T  1) = P(T = 0) + P(T = 1) = .25 + .5 = .75 (pdf) represents probability distribution function (cdf) represents cumulative probability function

Practice Cumulative Probability Probability P(T  t) = ….

How Many Girls Are Likely? BBB BBG BGB GBB BGG GBG GGB GGG Prob 1/8 G=g 1 2 3 List outcomes, assigned probabilities, and cumulative probabilities. Sometimes you may need to determine the possible combination of outcomes, such as how many girls could a family with 3 children have. Since there are 8 different possibilities of 3 children in a family and each are equally likely, each outcome has the likelihood of 1/8. Now, exactly how many girls are in each of the combinations (G=g)? One case has no girls (0); 3 cases have 1 girl; 3 cases have 2 girls; and one case has all girls (3). From there it follows that the probability of a family have no girls is 1/8; the probability that a family has only 1 girl is 3/8; and so on. P(G  g) represents the cumulative probability function (cdf). g 1 2 3 P(G=g) 1/8 3/8 P(Gg) 4/8 7/8

Expected Value of Discrete RV The expected value of a discrete random variable is the MEAN value of the variable X in the sample space of possible outcomes. Formula says to calculate the sum of “outcomes times probability”. E(X) =  = xipi Stress that “mu” and E(X) represent the same thing in the context of the expected value of a discrete random variable.

Example of Expected Value g 1 2 3 P(G = g) 1/8 3/8 gp 0(1/8) 1(3/8) 2(3/8) 3(1/8) 3/8 6/8 Recall probability distribution of number of girls in a family with 3 children. The first table is the probability distribution of this situation. The second table is showing the product of outcomes (0, 1, 2, 3) times the assigned probabilities (1/8, 3/8, 3/8, 1/8). The last line contains the probability and the sum of four products between outcomes and assigned probabilities. Therefore, the expected value of the number of girls in a family with 3 children is 1.75. Which is not possible, you either have 1 or 2. But remember, the expected value of a discrete random variable is the population mean which measures the central tendency of the situation (it doesn’t have to be an actual value of the possible outcomes). E(G) = G = gp = 0 + 3/8 + 6/8 + 3/8 = 12/8 or 1.5

Calculating Variance and Standard Deviation of Discrete RV Variance: V(X) = 2 = (xi - )2pi Standard deviation: square root of Variance First calculate variance, then square root value to find standard deviation. Stress that students need to take their time using the formulas if they want to calculate variance by hand. Or have students read information off computer printout or off calculator screen.

Calculating V(X) and Std dev (X) 1 2 3 g - G 0 – 1.5 1 - 1.5 2 – 1.5 3 – 1.5 (g - G)2 (-1.5)2 (-.5)2 (.5)2 (1.5)2 p 1/8 3/8 (g - G)2p 9/32 3/32 Remember square root of variance is equal to standard deviation. Table demonstrates steps to follow when calculating variance. V(G) = 24/32 = .75 Std dev (X) = .8660

Continuous Random Variable Probabilities of continuous random variables equals the proportion of area shaded under the curve. The total area under the curve is equal to 1.

Uniform Distribution-Wait Time Uniform continuous probability distributions are the easiest to demonstrate area under the curve equates to probability value (concept). Scenario: In Blocker building, suppose the elevator door opens on the first floor every 4 minutes. Now suppose you arrive at the elevators at some random time, how long will you wait for the next bus? This graph displays the wait time between 1 and 2 minutes. The question would be “what is the likelihood you would have to wait between 1 and 2 minutes?”. The proper notation is P(1  X  2). The area shaded is in the shape of a rectangle, so the area of a rectangle is calculated height times width. Now what would be the likelihood you would wait for the elevator between 2 and 4 minutes? (.25)(4-2) = .5 What would be the likelihood you would wait for the elevator between 1.5 and 3.5 minutes? (.25)(3.5-1.5) = .5 P(1  X  2) = (.25)(2-1) = .25 = (height)(width)

P(-1 < Z < 0) To find this area under the curve (probability), you need to remember that Z ~ N(0, 12). Use Z table (back of text). Find the P(Z < 0). Find the P(Z < -1). Subtract as shown below, P(Z < 0) - P(Z < -1). = .5000 - .1587 = .3413 Students must understand table before proceeding with the manipulation of probability statements.

Reading a Z Table P(z < -1.91) = ? .0281 P(z  -1.74) = ? .0409 z .00 .01 .02 .03 .04 -2.0 .0228 .0222 .0217 .0212 .0207 -1.9 .0287 .0281 .0274 .0268 .0262 -1.8 .0359 .0351 .0344 .0336 .0329 -1.7 .0446 .0436 .0427 .0418 .0409 Stress this is a small portion of entire 2 page table AND THE TABLE READS ONLY FOR LESS THAN OR (LESS THAN OR EQUAL TO). The first page of table is for negative z-scores; whereas the second page of table is for positive z-scores.

Reading a Z Table P(z < 1.81) = ? .9649 P(z  1.63) = ? .9484 z .00 .01 .02 .03 .04 1.5 .9332 .9345 .9357 .9370 .9382 1.6 .9452 .9463 .9474 .9484 .9495 1.7 .9554 .9564 .9573 .9582 .9591 1.8 .9641 .9649 .9656 .9664 .9671 Remember the table only reads for less than (less than or equal to).

Other Probability statements for continuous r.v. P(X > b) = 1 – P(X < b) same as P(X  b) = 1 – P(X < b) P(X = b) = 0 P( a  X  b) = P( X  b) – P(X  a) These mathematical manipulation must be used in solving the assigned problems. Really boring slide…

Using Z-Table Find the two tables representing z-scores from –3.49 to +3.49. Notice which direction the area under the curve is shaded (to the left). The assigned probabilities represents the area shaded under the curve from the z-score and far left of the z-score.

.9678 1.85 Find P(z < 1.85) From Table: -3 -2 -1 +1 +2 +3 Have students follow steps as they should do when solving these problems on their own. -3 -2 -1 +1 +2 +3 1.85

.9678 = .0322 1.85 Find P(z > 1.85) From Table: = 1 - .9678 = .0322 .9678 Take the time to carefully go over each step of process. -3 -2 -1 +1 +2 +3 1.85

.2148 -.79 Find P(z < - .79) From Table: -3 -2 -1 +1 +2 +3 Be sure students can locate this probability value off of Z table. -3 -2 -1 +1 +2 +3 -.79

Find P( -.79 < z < 1.85) = P(z < 1.85) - P(z < -.79) = .9678 - .2148 = .7530 -1 +1 +2 -3 -2 +3 -.79 1.85

Calculating Z-scores Z is a standardized score (meaning the distance with standardized deviation from  =0). x is the observed value of the random variable. is the population mean of a normally distributed model. is the population standard deviation of a normally distributed model X ~ N(, ) as Z ~ N(0, 1) Capital letter represents random variable. ~ indicates a distribution of the random variable N represents that the random variable has a normal distribution The numbers in the parentheses indicates the population mean then the variance (standard deviation squared).

Finding z when given P? What does it mean to score at the 80th percentile on the SAT exam? How many standardized standard deviations from the mean was that SAT score? Would the standardized standard deviation be positive or negative? 80th percentile indicates that a student out scored 80% of the other test takers. The students score was higher than 80% of the other scores recorded. Standardized standard deviation implies a z-score needs to be calculated. Since 80% is above the 50%, the z-score will be positive.

P(Z < z*) = .80 What is the value of z*? .03 .04 .05 .06 0.7 .7673 .7704 .7734 .7764 0.8 .7967 .7995 .8023 .8051 0.9 .8238 .8264 .8289 .8315 ?? Could be any number between 1 and 9, we usually use 5 when estimating from the table. P(Z < .84) = .7995 and P(Z < .85) = .8023, therefore P(Z < z*) = .80 has the value of z* approximately .84??

P(Z < z*) = .77 What is the value of z*? .03 .04 .05 .06 0.7 .7673 .7704 .7734 .7764 0.8 .7967 .7995 .8023 .8051 0.9 .8238 .8264 .8289 .8315 Z* = .73something, such as .735 or .738. What is the approximate value of z*?

Using TI-83 to find Z-Score Command invNorm calculates the z-score when you enter the area shaded under the curve to the far left. Ex. invNorm(.5) yields 0, since P(z < 0) = .5 Ex. invNorm(.05) = -1.645, as P(z < -1.645) = .05 Ex. invNorm(.95) = 1.645, as P(z < 1.645) = .95

invNorm( ) .95 .05 .10 Ex. invNorm(.95) = 1.645, since P(z > 1.645) = .95 Ex. invNorm(.05) = -1.645, as P(z > -1.645) = .95 Ex. invNorm(.90) = 1.281, as P(z >1.281) = .10

Using TI-83 To Find Probability of Normal Z Distribution ShadeNorm(lower bd, upper bd, mean, std dev) P(z < -1)ShadeNorm(-4,-1,0,1).1586 P(-2 < z < -1)ShadeNorm(-2, -1, 0, 1).1359 P(z > -1)ShadeNorm(-1, 4, 0, 1).8413

Why is this important?