Exam 1 next Thursday March 7th in class

1
Announcement

• Exam 1 next Thursday (March 7th) 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.
• Ill also put up solutions to this Thursdays HW.
• You will be allowed to bring in one page of notes
  and a calculator. Ill provide normal probability
  tables.
• Today Continue with central limit theorem.

2
Central Limit Theorem

• Example One
• Drive through window at a bank
• Consider transaction times, Xitransaction time
  for person i
• E(Xi) 6 minutes and Var(Xi) 32 minutes2.
  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?

3
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.
• Whats the probability that less than 49.8mls of
  the substance are made in total?
• Whats 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?

4
Answers

• Central limit theorem
• If E(Xi)m and Var(Xi)s2 for all i (and
  independent) then
• X1Xn N(nm,ns2)
• (X1Xn)/n N(m,s2/n)

5
Bank

• Pr(total of all the transaction times is greater
  than 200 minutes)Y total N(306,309) (by
  CLT)Pr(Y gt 200)Pr(Y-180)/sqrt(54) gt
  (200-180)/sqrt(270)Pr(Z gt 1.22) 1-0.89
  0.11
• Pr(5.9ltAveragelt6.1)WAverage N(6,9/30) (by
  CLT)Pr(5.9-6)/0.55lt(W-6)/0.55 gt
  (6.1-6)/0.55Pr(-0.18ltZlt0.18)
  2Pr(0ltZlt0.18)0.14

6
Follow on Review Question

• Consider 20 Saturdays. Let X the number of
  Saturdays on which 5.9ltAveragelt6.1
• Whats probability that 1lt X lt 3?
• XBin(20,0.14)
• Pr(1ltXlt3) Pr(X lt3) Pr(X0)Pr(X1)Pr(X2
  )Pr(X3)0.65

Since X is discrete, becareful about the
difference between lt and lt
7
Lab

• Let Y total amount made. YN(510,50.5) (by
  CLT)Pr(Ylt49.8) Pr(Y-50)/1.58 lt
  (49.8-50)/1.58Pr(Z lt -0.13) 0.45
• Let W average amount made.WN(10,0.5/5) (by
  CLT)Pr(W gt 10.1) PrZ gt (10.1
  10)/0.32Pr(Z gt 0.32) 0.38

8
Lab (continued)

• One definition of rareIts a rare event if Pr(W
  gt 11.0) is small(i.e. if Seeing probability of
  11.0 or something more extreme is
  small)Pr(Wgt11) PrZ gt (11-10)/0.32
  Pr(Zgt3.16) 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)

9
Source gallup.com Suppose this is based on a
poll of 100 people
10

• Let Xi 1 if person i favors NHL players in the
  Olympics and 0 otherwise.
• Suppose E(Xi) p and Var(Xi) p(1-p) and each
  persons opinion is independent.
• Let Y total number of favors X1 X100
• Y Bin(100,p)
• Suppose p 0.72
• What is Pr(Y lt 70)?

Note that this definition turns three outcomes
intotwo outcomes
11
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 N100(0.72),100(0.72)(0.28)
  N(72,20.4) (by central limit theorem)
• Pr (Y-72)/4.5 lt (70-72)/4.5 Pr(Z lt -0.44)
  0.33

12
Normal Approximation to the binomial CDF
Rectangles are plots of bin(100,0.72) pdf versus
Y (integers)
Line is plot of Normal(72,4.5) pdf
13
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
14
What does 6 sigma mean?(example)

• Suppose a product has a quantitative
  specificationex 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. X1,,Xn are gap widths. The
  sqrt(sample variance of X1,,Xn) 0.1mm

15
Probability meaning of 6 sigma

• 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)4-1.5s 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 6s, then
• Pr(X below lower spec) Pr( Xlt4- 6s)Pr(X-
  (4-1.5s))/s lt (4-6s-(4-1.5s))/s Pr(Zlt-4.5)
  3.4/1,000,000

16
Statistically, six sigma means that Upper Spec
Lower Spec gt 12 sigma (i.e. Specs are fixed.
Lower the manufactuing process variability.)
Lower specification
Upper specification
4.6 3.4 1.2 120.1 12sigma
(In the car door example, sigma of the
manufacturing process is 0.1mm)
3.4mm
4.6mm
17
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 its convoluted, but thats the
  definition)