TODAY

1. TODAY • Homework 4 comments • Week in review • Joint Distribution • Binomial Distribution • Central Limit Theorem • Sample problems

2. Sample Problem M&M’s I have 5 M&M’s left in my bag, 2 blue, 2 green and 1 yellow. I plan on eating two more and leaving the others in the bag. Q: How do I interpret the probabilities in each cell? If I sum the rows what do I get? And if I sum the columns?

3. Sample Problem M&M’s Probability distribution table with marginal totals Q: What are the mean and standard deviation of X? EX=ΣxP(X=x) = 0 x 3/15 + 1 x 9/15 + 2 x 3/15 = 1 Var(X) = Σ(X-µx)2P(X=x) = (0-1)2 x 3/15 + (1-1)2 x 9/15 + (2-1)2x3/15 = 6/15 Std Dev(X) = sqrt(6/15) = 0.6324 Similarly – E(Y) = 0.667 ; Var(Y) = 0.355; Std Dev(Y)= 0.596 Q: What is the covariance between X and Y COV(X,Y) = E[(X-µx) (X-µY)] = (0-1)(0-0.667)x 0 + (1-1)(0-0.667) x 3/15 + (2-1)(0-0.667)x 3/15+ ……+(2-1)(2-0.667) x 0 = - 0.2667 Corr(X,y) = -0.2667/(0.6324 x 0.596) = -0.707

4. Sample Problem M&M’s Q: What is the conditional probability distribution of Y given X=1 P(Y=0|X=1) = P(Y=1|X=1) = P(Y=2|X=1) = Q: What is the expected value of Y|X=1 E(Y|X=1)=ΣyP(Y=y|X=1) = 0 x 1/3 + 1 x 2/3 + 2 x 0 =2/3

5. Sample Problem M&M’s Q: What is the conditional probability distribution of Y|X≥1 P(Y=0|X ≥1) = P(Y=1|X ≥1) = P(Y=0|X ≥1) =

6. Sample Problem – Binomial Distribution • An oil exploration firm is formed with enough capital to finance ten explorations. Assume the probability of a particular exploration being successful is 0.1. Assume the explorations are independent. What is the probability at least 2 explorations are successful Let X denote the number of successful explorations X~Bin(10,0.1) Pr(X≥2) = 1-Pr(X=0)-Pr(X=1) = Find the mean and variance of the number of successful explorations E(X) = 10 x 0.1 =1 Var(X) = 10 x 0.1 x 0.9 = 0.9

7. Sample Problem – Binomial Distribution Suppose the firm has a fixed cost of $20,000 prior to doing its first exploration. If each successful exploration costs $30,000 and each unsuccessful exploration costs $15,000, find the expected total cost to the firm of the ten explorations Let Y be the number of unsuccessful operations Y~Bin(10,0.9) Let T = Total Cost = 20,000 +30,000X + 15,000Y E(T) = E[20,000 +30,000X + 15,000Y] = 20,000 + 30,000E[X]+15,000E[Y] =20,000 + 30,000 x 10 x 0.1 + 15,000 x 10 x 0.9 = $185,000 Expected total cost to the firm is $185,000

8. Sample Problem – Binomial Distribution Example: Suppose that a typical radio station plays Madonna songs about 4 times an hour. Q: Assuming that all songs are of length 3 minutes. What percentage of time is spent playing Madonna? Total playing time per hour is 4x3 = 12 mins. Therefore percentage of time spent playing Madonna is 12/60 = 20% Q: What is the probability of turning on the radio at a random time and having to listen to it? 20% Q: What's the probability (Given that you start listening exactly at the end of a song), that you'll have to listen to Madonna if you listen for 9 minutes? Let X be the no. of times I listen to Madonna in 9mins. X~Bin(3,0.2) P(X≥1) = 1-P(X=0) =

9. Sample Problem – Binomial Distribution Q: Assume two radio stations, picking songs independently, both play Madonna at the same rate. What is the probability you can switch stations back and forth for an hour, starting at minute 0, without having to listen to Madonna? Let X be the no. of times I listen to Madonna from Station A in one hour X~Bin(10,0.2) Let Y be the no. of times I listen to Madonna from Station B in one hour Y~Bin(10,0.2) P(X=0,Y=0) = P(X=0)P(Y=0) =

10. Sample Problem – Binomial Distribution – Normal approximation and continuity correction Bob sells life insurance policies for a living. Suppose he contacts 100 potential clients (through referrals, and other leads). With each client there is a 50% chance he/she will purchase (and successfully accepted for) a life insurance contract. To get his quarterly bonus, Bob must have a success rate of at least 60%. What is the probability Bob achieves his bonus. Let be the sample proportion of successes. By normal approximation to sample proportion

11. Sample Problem – Binomial Distribution – Normal approximation and continuity correction Suppose there are 25 other advisers like Bob who all contact 100 people over a 3 month period. The sales team will get a trip to Japan if total sales exceed 1500. What is the probability the sales team will go to Japan?? Let T be total sales by 50 advisors T~Bin(2500,0.5)≈N(1250,25) Pr(T≥1500) ≈ Pr(Z ≥ (1500-0.5-1250)/25)=Pr(Z ≥10) = 0 So the insurance company is offering a prize virtually impossible to achieve!

12. Distribution of the sample mean Suppose I have nindependent and identically distributed random variables Y1 …. Yn from a sample of size n from the population. Eg I have a sample of n males and Yi is the random variable for the height of the ith male is the mean of all the heights in my sample, it is a function of the random variables and therefore is also a random variable Suppose I went out and collected data for Y1 …. Yn . Then I have data y1 …. yn and I can calculate the sample mean Each time I go and collect a different set of data, I have a different set of numbers and can recalculate the mean. So the mean will vary from sample to sample. In statistical terms, we say the mean has a sampling distribution (Note, we say is a statistic, it is a function of the random variables observed in a sample. All statistics have sampling distributions)

13. Sample Problem – Distribution of the sample mean So what is the distribution of ??? First let’s assume Yi~N(µ,σ) Now if Yi~N(µ,σ) We know that sums of normals, are also normally distributed, therefore

14. Sample Problem – Central Limit Theorem But there is a nice theorem that states that even if the variables Yi are not normally distributed, if we have large enough n, then the sampling distribution of is approximately normal The CENTRAL LIMIT THEOREM Given independent and identically distributed random variables Y1 …. Yn from, when n is large, the sampling distribution of the sample mean is approximately Or we can also write (ie the sum of the random variables) is approximately normally distributed with mean nµ and variance nσ2

15. Sample Problem – Central Limit Theorem Risk and insurance An insurance company sells home and contents insurance. The mean claim size is $250 per house and the standard deviation of claim size is $1000. The company has put aside $2.75m in reserve to fund future claims. If the company sells 10,000 (identical) policies, what is the probability that the company will be insolvent at the end of the year? n=10000 Let Yi be the claim size for individual i and T= By the CLT There is a 0.6% chance total claims will exceed $2.75m ie company is insolvent

16. Sample Problem – Central Limit Theorem Test scores Suppose there are 300 students taking a course. Student assessment is based on a midterm (25%) and final exam (75%). Suppose midterm scores have a mean µ=75 (out of hundred) and standard deviation σ=12. Final exam scores have a mean µ=88 and standard deviation σ= 8. Correlation between midterm and final exam scores is 0.5. Teaching fellows get a bonus if the average final grade (of all students) is greater than 80. What is the probability the TFs get a bonus? (Assume scores of different students are independent and identically distributed) Let Xi be the final grade of student i. is the average final grade of all students. We want

17. Sample Problem – Central Limit Theorem By Central Limit Theorem