Organizing and Displaying Categorical Data Guide Part 5

Chapter 6 1. Definitions: Random Variables// Probability Model Table // Discrete vs. Continuous Random Variables 2. Calculating the Expectation of Random Variables 3. Calculating the Variance of Random Variables 4. Properties of the Variance and Expectation Random Variables & Probability Model Table Definitions Random Variables is the outcome of an experiment ● E.g. We could say X is a random variable assigned to the outcome of the experiment ○ X=1 or 2 or 3 or 4 or 5 or 6 ; X={1,2,3,4,5,6} ● Suppose we want to know how many days it will rain this week we could say Y is the random variable assigned to this experiment ○ Y= 1 or 2 or 3 or 4 or 5 or 6 or 7 ; X={1,2,3,4,5,6,7} Probability Model Table ● A table that list each outcome and the associated probability The Expected Value of a Random Variable = Weighted Average of the outcomes ● E.g. Expected value of X: E(X) = Weighted average of the outcomes {1,2,3,4,5,6} ○ 1x(1/6) +2x(1/6) +3x(1/6) +4x(1/6) +5x(1/6) +6x(1/6) = 3.5 ● E,g, Expected value of Y: E(Y) = Weighted average of the outcomes {1,2,3,4,5,6,7} ○ 0xP(0)+1xP(1) … ● M is equal to the number of meals I eat per day. M is either equal to 0 or 1 or 2 or 3 with probabilities 0.05, 0.2, 0.5, 0.25 respectively. Calculate the E(M) ○ 0(0.05)+1(0.2)+2(0.5)+3(0.25) = 0.2+1+.75 = 1.95 Variance of a Random Variable = Weighted Average of the d^2 from the mean ● E.g. Calculate the variance of X and M, where X is defined as the outcome when you roll a die and M is the number of meals I ate X Variance= d^2(Prob)+d^2(Prob) M Variance Properties of Expectation and Random Variable Weighted Average of the outcome // Weighted average of the d^2 from the mean Method 1: Method 2: Recommended Approach S=Sum of 2 rolls How many random variables do we have? It seems we only have 1, S There are really 2 trials - first roll and second roll S=X1+X2 We previously calculated E(X) and V(X); the average outcome of a roll of a die and the average d^2 from the mean when you roll a die. E(X)= 3.5 ; V(X)= 17.5/6 The Average Outcome when you roll 1 die is 3.5. What do you think the average outcome is when you roll 2 dice? 7 (3.5+3.5) The average d^2 from the mean when you roll 1 die is 17.5/6. What do you think the average d^2 from the mean is when you roll 2 dice? 17.5/6 + 17.5/6 TIPS: Instead of treating S as the random variable we should define S as being X1+X2 Suppose you roll 3 dice. Let S= sum of the 3 dice. Calculate E(S) and V(S). Remember that X= outcome of a single roll and E(X) = 3.5 and V(X) = 17.5/6 A) Calculate E(S)= E( X1+X2+X3 ) a) 3.5+3.5+3.5= 10.5 B) Calculate V(S)= a) 17.5/6 + 17.5/6 +17.5/6 Suppose you roll 4 dice. Let S= sum of the 4 dice A) Calculate E(S)= X1+X2+X3+X4 = 3.5*4 = 14 B) Calculate the S.D. of S. a) Step 1: V(S) = 17.5/6 +17.5/6 + 17.5/6 +17.5/6 = 70/6 b) Step 2: S.D. (S) = SQRT (70/6) HW5 Q3 HW5 Q4 (REFER EXCEL SHEET IN MY DRIVE) Given E(X)=84; SD (X) = 12 ; E(Y)=74 ; SD(Y)= 10 Note: SD= SQRT(V) a) E(X+10) = 84+10 = 94 b) V(X+10) = i) If the average d^2 from the mean of X is 144 (12^2), what do you think the average d^2 from the mean of X+10? 144 is correct. The +10 adds no variation to our data. ii) V(10) = is 0 because it is fixed iii) V(X+10) = V(X)= 144 c) Calculate SD(X+10) i) SD(X)=12 ii) SD(X+10)= 12 iii) +10 adds no variation Another part. (B) d) Calculate E(5Y) = 74*5= 370 e) Calculate V(5Y) = If the variance of Y is 100, what do you think the variance of 5 times Y? i) V(5Y)= 5^2 x V(Y) = 25x100= 2500 ii) Intuitively, the variance measures spread. When you multiply each outcome by 5 your are spreading the data/outcomes even more. The spread increases not by a factor of 5, but by a factor of 25. (Variance measures D^2 - Square distances) f) Calculate the SD(5Y) i) Step 1: Calculate the variance V(5Y) ii) Step 2: Calculate the SD iii) SQRT (2500)= 50 Another part (C) g) Calculate E(X+Y) i) If the average value of X is 84 and the Ave value of Y is 74, What's the average value of X+Y? 1) 84+74= 158 h) V(X+Y) i) If the variance of X is 144 and the variance of Y is 100, ii) 144+100 i) SD(X+Y) i) SD of X is 12 and SD of Y is 10 ii) Step 1: Calculate the V(X+Y) = 244 iii) Step 2: Calculate the SD(X+Y)= SQRT(244) = 15.6 SUMMARY Given E(X)=84, V(X)=144, E(Y)=74, V(Y)=100 E(aX+bY+c) where a,b and c are numbers; Random Variables are always capitalized E(aX+bY+c) = aE(x)+bE(y)+c V(aX+bY+c) = a^2V(X)+b^2V(Y)+ 0 ( c adds no variability) To calculate the standard deviation, X and Y must be independent. HW5 Q12 A company has discovered that a recent batch of batteries had manufacturing flaws, and has issued a recall. In a group of 10 batteries covered by the recall, 5 are dead. Two batteries at random are chosen from the package of 10. a) Has the assumption of independence been met? Explain. No, because the outcome for the first battery affects the probability that the second battery is dead. b) Create a probability model for the number of good batteries chosen. c) What's the expected number of good batteries? 0.99 d) What's the standard deviation? 0.6634 HW5 Q14 A commuter finds that she waits an average of 19.7 seconds at each of 6 stoplights, with a standard deviation of 5.8 seconds. Find the mean and the standard deviation of the total amount of time she waits at all six lights. What, if anything, did you assume? a) Mean time for all 6 lights: 19.7x6= 118.2 b) The standard deviation of the time she waits at all six lights is: (5.8^2)x6 → Squareroot(ans) Qp pq HW5 Q20 A salesman normally makes a sale (closes) on 65% of his presentations. Assuming the presentations are independent, find the probability of each of the following. a) He fails to close for the first time on his 6th attempt./trials b) He closes his first presentation on his 5th attempt. c) The first presentation he closes will be on his second attempt. d) The first presentation he closes will be on one of his first three attempts. THIS IS FOR FAIL // IF SUCCESS SWITCH q and p [q] Close= 0.65 [p] C’=0.35 a) Fail to close for first time on 6th attempt fail i) P(X=6)= 0.65^(6-1) * 0.35 = 0.027216 b) Closes for first time on 5th attempt closes i) P(X=5)=0.35^(5-1)*0.65 c) First presentation he closes will be on his second attempt. i) P(X=2)=0.35^(2-1)*0.65 d) The first presentation he closes will be on one of his first three attempts. i) Complement/Compliment 1-P(X=x) ii) (0.35)^3=0.0429 iii) 1-0.0429= 0.9571 Class 10/18/2020 Midterm: 8-10 November 3 hour Questions drawn from Practice Midterm q^x-1p HW5 Q11 (Probability Tree) A sporting goods manufacturer was asked to sponsor a local boy in two fishing tournaments. They claim the probability that he will win the first tournament is 0.6. If he wins the first tournament, they estimate the probability that he will also win the second is 0.5. They guess that if he loses the first tournament, the probability that he will win the second is 0.2 . Complete parts a through e below. a) Trials independent? i) No, the outcome of the second trial is affected by first trial. b) Calculate the probability you lose both games c) Calculate the probability you lose both games d) Let random variable X be the number of tournaments he wins. Find the probability model for X. e) What are the expected value and standard deviation of X? E(X)= (0*.32)+(1*.38)+(2*.3)=0.98 V(X)= SD(X)= SQRT(V) HW5 Q12(Probability Tree/Model Table/Expectation/Variance) C. E(X)= (0*6/90)+(1*42/90)+(2*42/90)= 1.4 D. V(X)= d^2 from mean * Prob HW5 Q13 (Multiple Repeated Trials) A grocery supplier believes that the mean number of broken eggs per dozen is 0.9 with a standard deviation of 0.5 You buy 8 dozen eggs without checking them. a) How many broken eggs do you expect to get? 0.9x8=7.2 b) What's the standard deviation? Calculate V(x) first. - SD^2= Variance - 0.5^2=0.25 Multiply variance by 8 dozen - 0.25*8= 2 SD for 8 dozen = SQRT(2)= 1.414 c) Is it necessary to assume the cartons of eggs are independent? Why? Did we make an assumption when calculating the expected # of broken eggs? - No Did we make an assumption when calculating the SD of the # of broken eggs in 8 dozen - Yes, it is necessary to assume the cartons are independent to calculate the standard deviation. - Random X * Random X =/= 2*Random X HW5 Q14 (Multiple Repeated Trials) A commuter finds that she waits an average of 18.5 seconds at each of stoplights, with a standard deviation of 5 seconds. Find the mean and the standard deviation of the total amount of time she waits at all 3 lights. What, if anything, did you assume? a. Expected waiting time for the 3 lights? i. 18.5*3=55.5 b. The standard deviation of the time she waits at all 3 lights is i. V(x)= 5^2=25 ii. 25x3=75 iii. SD(x)= SQRT(75)=8.66 c. In parts a and b, did we make any assumptions i. A=No assumptions ii. B=Yes, we assume that trials are independent HW5 Q16 (Multiple Repeated Trials) At a casino, people play the slot machines in hopes of hitting the jackpot, but most of the time, they lose their money. A certain machine pays out an average of 0.95$ (for every dollar played), with a standard deviation of $140. Complete parts a through c. a. Why is the standard deviation so large? i. Most players will lose their money but a few large payouts are expected. b. If a gambler plays a $1 slot machine times, what are the mean and standard deviation of the casino's profit? i. E(Profit)=1-0.95=0.05 ii. SD= 140 c. If a gambler plays 3 games a $1 slot machine times, what are the mean and standard deviation of the casino's profit? i. E(Profit)=1-0.95=0.05x3=0.15 ii. V= 140^2= 19600x3= 58800 iii. SD= SQRT(58800)=242.487113112 d. In part C, assumption for SD/ V i. Independent e. If you play 900 times, calculate E(x) and SD of profits. i. E(x)=900*0.05=$45 ii. V(x)=900*(140^2)=17640000 iii. SD(x)=SQRT(17640000)= $4200 HW5 Q17 (Different random Variables/ Expectation Variance) A bicycle shop plans to offer 2 specially priced children's models at a sidewalk sale. The basic model will return a profit of $100 and the deluxe model $110. Past experience indicates that sales of the basic model will have a mean of bikes with a standard deviation of 5.7, and sales of the deluxe model will have a mean of bikes with a standard deviation of bikes 1.7. The cost of setting up for the sidewalk sale is $200. a. Define the random variables and use them to express the bicycle shop's net profit. P= B*100+D*110-200 b. Expected profit i. E(P)= 5.7(100)+1.7(110)-200=$799 c. Variance profitability i. V(P)= V(100B+110D-200) =100^2(V(B))+110^2(V(D))+0 =10000x5.7^2+12100X1.7^2 =22500+9801=32301 SD= SQRT (32301) HW5 Q18 (Different random Variables/ Expectation Variance) A farmer has 40lbs of apples and 30 lbs of potatoes for sale. The market price for apples (per pound) each day is a random variable with a mean of $0.5 dollars and a standard deviation of $0.2 dollars. Similarly, for a pound of potatoes, the mean price is $0.3 dollars and the standard deviation is $0.1 dollars. It also costs him 4 dollars to bring all the apples and potatoes to the market. The market is busy with shoppers, so assume that he'll be able to sell all of each type of produce at that day's price. Complete parts a) through d). Make sure to consider all necessary assumptions before calculating. A= Price of a lb of Apples P= price of a lb of potatoes Profitability= 40A+30P-4 Expected Profit E(P)= 40(0.5)+30(0.3)-4= $25 Variance of Profitability V(P)= V(40A+30P-4) = V(40A)+V(30P)+V(-4) =40^2V(A)+30^2V(P)+0 1600x0.2^2+900x0.1^2 =73