Exam 1 next Thursday (March 7 th ) in class 15% of your grade Covers chapters 1-6 and the central limit theorem I will put practice problems, old exams, and specific sections that are not included on the web by the end of this week. I’ll also put up solutions to this Thursday’s HW. You will be allowed to bring in one page of notes and a calculator. I’ll provide normal probability tables. Today: Continue with central limit theorem. Announcement

Central Limit Theorem Example One: –Drive through window at a bank –Consider transaction times, X i =transaction time for person i –E(X i ) = 6 minutes and Var(X i ) = 3 2 minutes 2. Transaction time for each person is independent. –Thirty customers show up on Saturday morning. 1.What is the probability that the total of all the transaction times is greater than 200 minutes? 2.What is the probability that the average transaction time is between 5.9 and 6.1 minutes?

Central Limit Theorem Example Two: –5 chemists independently run a synthesis reaction 1 time each. –Each reaction should produce 10ml of a substance. –Historically, the amount produced by each reaction has been normally distributed with std dev 0.5ml. 1.What’s the probability that less than 49.8mls of the substance are made in total? 2.What’s the probability that the average amount produced is more than 10.1ml? 3.Suppose the average amount produced is more than 11.0ml. Is that a rare event? Why or why not? If more than 11.0ml are made, what might that suggest?

Answers: Central limit theorem: If E(X i )=  and Var(X i )=  2 for all i (and independent) then: X 1 +…+X n ~ N(n ,n  2 ) (X 1 +…+X n )/n ~ N( ,  2 /n)

Bank: 1.Pr(total of all the transaction times is greater than 200 minutes) Y = total ~ N(30*6,30*9) (by CLT) Pr(Y > 200) =Pr[(Y-180)/sqrt(54) > ( )/sqrt(270)] =Pr(Z > 1.22) = = Pr(5.9 (6.1-6)/0.55] =Pr(-0.18<Z<0.18) = 2*Pr(0<Z<0.18)=0.14

Follow on Review Question Consider 20 Saturdays. Let X = the number of Saturdays on which 5.9<Average<6.1 What’s probability that 1<= X <= 3? X~Bin(20,0.14) Pr(1<=X<=3) = Pr(X <=3) – Pr(X=0) =Pr(X=1)+Pr(X=2)+Pr(X=3) =0.65 Since X is discrete, be careful about the difference between < and <=

Lab: 1.Let Y = total amount made. Y~N(5*10,5*0.5) (by CLT) Pr(Y<49.8) = Pr[(Y-50)/1.58 < ( )/1.58] =Pr(Z < -0.13) = Let W = average amount made. W~N(10,0.5/5) (by CLT) Pr(W > 10.1) = Pr[Z > (10.1 – 10)/0.32] =Pr(Z > 0.32) = 0.38

Lab (continued) 3.One definition of rare: It’s a rare event if Pr(W > 11.0) is small (i.e. if “Seeing probability of 11.0 or something more extreme is small”) Pr(W>11) = Pr[Z > (11-10)/0.32] = Pr(Z>3.16) = approximately zero. This suggests that perhaps either the true mean is not 10 or true std dev is not 0.1 (or not normally distributed…)

Source: gallup.com Suppose this is based on a poll of 100 people

Let X i = 1 if person i favors NHL players in the Olympics and 0 otherwise. Suppose E(X i ) = p and Var(X i ) = p(1-p) and each person’s opinion is independent. Let Y = total number of “favors” = X 1 +…+ X 100 Y ~ Bin(100,p) Suppose p = 0.72 What is Pr(Y < 70)? Note that this definition turns three outcomes into two outcomes

Normal Approximation to the binomial CDF –Even with computers, as n gets large, computing things like this can become difficult. (100 is OK, but how about 1,000,000?) –Idea: Use the central limit theorem approximate this probability –Y is approximately N[100*(0.72),100*(0.72)*(0.28)] = N(72,20.4) (by central limit theorem) Pr[ (Y-72)/4.5 < (70-72)/4.5] = Pr(Z < -0.44) = 0.33

Rectangles are plots of bin(100,0.72) pdf versus Y (integers) Line is plot of Normal(72,4.5) pdf Normal Approximation to the binomial CDF

Area under curve to left of 70 is approximately equal to the sum of areas of rectangles to the left of 70

What does 6 sigma mean? (example) Suppose a product has a quantitative specification: ex: “Make the gap between the car door and the car body between 3.4 and 4.6mm.” When manufacturing processes are more “in control” they have less variability. (In other words, they produce very close to exact duplicates over and over.) ex: When cars are actually made, the “std dev of car door gap is 0.1mm”. i.e. X 1,…,X n are gap widths. The sqrt(sample variance of X 1,…,X n )= 0.1mm

Let X = car door gap width of a random car of a specific type. Assume process mean is 1.5 standard deviations away from the center of the spec: i.e. E(X)=4-1.5  and assume X has a normal distribution. When the process is in control enough so that the distance between the center of the specs and the lower spec is least 6 , then Pr(X below lower spec) =Pr( X<4- 6  ) =Pr[(X- (4-1.5  -6  (4-1.5  ] =Pr(Z<-4.5) = 3.4/1,000,000 Probability meaning of 6 sigma

Lower specification Upper specification 3.4mm4.6mm (In the car door example, sigma of the manufacturing process is 0.1mm) 4.6 – 3.4 = 1.2 = 12*0.1 = 12*sigma Statistically, six sigma means that Upper Spec – Lower Spec > 12 sigma (i.e. Specs are fixed. Lower the manufactuing process variability.)

Probability meaning of 6 sigma Even if you shift the process mean by 1.5 standard deviations toward one of the specifications, then you will expect no more than 3.4 out of a million defects outside of the spec toward which you shifted. (I know it’s convoluted, but that’s the definition…)