1. Central Limit Theorem Example: (NOTE THAT THE ANSWER IS CORRECTED COMPARED TO NOTES5.PPT) –5 chemists independently synthesize a compound 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?

2. Answer: 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)

3. Lab: 1.Let Y = total amount made. Y~N(5*10,5*0.5 2 ) (by CLT) Pr(Y<49.8) = Pr[(Y-50)/1.12 < (49.8-50)/1.12] =Pr(Z < -0.18) = 0.43 2.Let W = average amount made. W~N(10,0.5 2 /5) (by CLT) Pr(W > 10.1) = Pr[Z > (10.1 – 10)/0.22] =Pr(Z > 0.45) = 0.33

4. 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)/0.22] = Pr(Z>4.55) = approximately zero. This suggests that perhaps either the true mean is not 10 or true std dev is not 0.5 (or not normally distributed…)

5. Sample size: 1006 (source: gallup.com)

6. Let X i = 1 if person i thinks the President is hiding something 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 “yesses” = X 1 +…+ X 1006 Y ~ Bin(1006,p) Suppose p = 0.36 (this is the estimate…) What is Pr(Y < 352)? Note that this definition turns three outcomes into two outcomes

7. Normal Approximation to the binomial CDF –Even with computers, as n gets large, computing things like this can become difficult. (1006 is OK, but how about 1,000,000?) –Idea: Use the central limit theorem approximate this probability –Y is approximately N[1006*0.36, (0.36)*(0.64)*1006] = N(362.16,231.8) (by central limit theorem) Pr[ (Y-362.16)/15.2 < (352-362.16)/15.2] = Pr(Z < -0.67) = 0.25 Pr(Y<352) = Pr(Y=0)+…+Pr(Y=351), where Pr(Y=k) = (1006 choose k)0.36 k 0.64 1006-k

8. Normal Approximation to the binomial CDF Black “step function” is plots of bin(1006,0.36) pdf versus Y (integers) Blue line is plot of Normal(362.16,231.8) pdf

9. Normal Approximation to the binomial CDF Area under blue curve to left of 352 is approximately equal to the sum of areas of rectangles (black Stepfunction) to the left of 352

10. Comments about normal approximation of the binomial : Rule of thumb is that it’s OK if np>5 and n(1-p)>5. “Continuity correction” Y is binomial. If we use the normal approximation to the probability that Y<k, we should calculate Pr(Y<k+.5) If we use the normal approximation to the probability that Y>k, we should calculate Pr(Y<k-.5) (see picture on board)

11. Probability meaning of 6 sigma Even if you shift the process mean for the center of the specifications to 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 specification toward which you shifted. (I know it’s convoluted, but that’s the definition…)

12. 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 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

13. Lower specification Upper specification 3.4mm4.6mm 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.) Center of spec = 4mm gap Shifted mean = 3.85mm gap Distribution of gap widths Probability of being out here is Pr( gap is less than 3.4 ) = Pr( (gap – 3.85)/0.1 < (3.4-3.85)/.1) =Pr( Z < -4.5) = 3.4/1,000,000 Arbitrary “magic” number for 6 

14. In general: Assume process mean is 1.5 standard deviations toward the lower 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

15. Control Charts Let X = an average of n measurements. Each measurement has mean  and variance  2. Fact: –By the central limit theorem, almost all observations of X fall in the interval  +/- 3  /sqrt(n) (i.e. mean +/- 3 standard deviations) –  /sqrt(n) is also called  x or standard error

16. Use the “fact” to detect changes in production quality Idea: let x i = average door gap from the n cars made by shift i at the car plant   +3  /sqrt(n) (Upper Control Limit)  -3  /sqrt(n) (Lower Control Limit) shift x1x1 x2x2 x3x3 x4x4 x5x5 x6x6 x7x7 x8x8 Points outside the +/- 3 std error bounds, are called “out of control”. They are evidence that  and or  are not the true mean and std dev any more, and the process needs to be readjusted. Calculate the “false alarm rate”… (= 26/10,000)

17. Assume 100 new people are polled. Assume true pr( a new person says yes) = 0.36. Let P = “P hat” = number say yes/100 What’s an approximation to the distribution of P-hat? Use the approximation to determine a number so that the Pr(p-hat> that number) = 0.95.

18. EXAMPLE OF SAMPLING DISTRIBUTION OF P-HAT X k = 1 if person k says yes and 0 if not. Note that E(X k )=0.36=p and Var(X k )=0.36*0.64=p(1-p) Note that X k is binomial(1,0.36). P-hat = (X 1 +…+X 100 )/100. By CLT, P-hat is approximately N(0.36,0.36*0.64/100). (Rule of thumb is that this approximation is good if np>5 and n(1-p)>5.)

19. Suppose true p is 0.36. If survey is conducted again on 100 people, then P-hat ~ N(.36,(.36)(.64)/100) = N(.36, 0.002304) Want p 0 so that Pr(P-hat<p 0 ) = 0.95 Pr(P-hat<p 0 ) = 0.95 means Pr(Z < (p 0 -.36)/0.048) = 0.95. Since Pr(Z<1.645) = 0.95, (p 0 -.36)/0.048 = 1.645 (p 0 -.36) = 0.07896 p 0 = 0.43896

20. Suppose true p is 0.40. If survey is conducted again on 49 people, what’s the probability of seeing 38% to 44% favorable responses? Pr( 0.38 < P ”hat” < 0.44) = Pr[(0.38-0.40)/sqrt(0.40*0.60/49) < Z < (0.44-0.40)/sqrt(0.40*0.60/49) ] = Pr(-0.29 < Z < 0.57) = Pr(Z<0.57) – Pr(Z<-0.29) = 0.7157-0.3859=0.3298