Transcript Announcement

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.
• 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.
Central Limit Theorem
• Example One:
– Drive through window at a bank
– Consider transaction times, Xi=transaction 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?
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(Xi)=m and Var(Xi)=s2 for all i (and
independent) then:
X1+…+Xn ~ N(nm,ns2)
(X1+…+Xn)/n ~ N(m,s2/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) > (200-180)/sqrt(270)]
=Pr(Z > 1.22) = 1-0.89 = 0.11
2. Pr(5.9<Average<6.1)
W=Average ~N(6,9/30) (by CLT)
=Pr[(5.9-6)/0.55<(W-6)/0.55 > (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 < (49.8-50)/1.58]
=Pr(Z < -0.13) = 0.45
2. 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 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 person’s opinion is independent.
• Let Y
= total number of “favors”
= X1+…+ X100
Note that this definition
• Y ~ Bin(100,p)
turns three outcomes into
two outcomes
• Suppose p = 0.72
• What is Pr(Y < 70)?
Normal Approximation to the
binomial CDF
100 
(0.72) k (0.28)100k
Pr(Y  70)   
k 0  k 
– 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
69
0.08
Normal Approximation to the
binomial CDF
Rectangles are plots of
bin(100,0.72) pdf versus Y (integers)
0.0
0.02
0.04
0.06
Line is plot of Normal(72,4.5) pdf
55
60
65
70
75
80
85
0.08
Normal Approximation to the
binomial CDF
0.02
sum of areas of
rectangles to the
left of 70
0.0
is
approximately
equal to the
0.04
0.06
Area under curve to
left of 70
55
60
65
70
75
Y
80
85
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. X1,…,Xn are gap
widths. The sqrt(sample variance of X1,…,Xn)=
0.1mm
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( X<4- 6s)
=Pr[(X- (4-1.5s))/s  (4-6s(4-1.5s))/s ]
=Pr(Z<-4.5) = 3.4/1,000,000
Statistically, six sigma means that
Upper Spec – Lower Spec > 12 sigma
(i.e. Specs are fixed. Lower the
manufactuing process variability.)
Lower
specification
Upper
specification
4.6 – 3.4 = 1.2 = 12*0.1 = 12*sigma
3.4mm
(In the car door example, sigma of the
manufacturing process is 0.1mm)
4.6mm
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…)