1. STATISTICS AND PROBABILITY
Raoul LePage
Professor
STATISTICS AND PROBABILITY
click on STT315_F06
Week and some preparation for exam 2.

2. solutions given in text 3-33, 3-41, 3-42 (except b, c, h, m, n),
suggested exercises
solutions given in text
3-33, 3-41,
3-42 (except b, c, h, m, n),
3-43, 3-49, 3-57 (except c, d),
3-59, 3-61, 3-63, 3-65.
textbook exercises are not
comprehensive
Week and some preparation for exam 2.

3. HAVING BROAD APPLICATION
PROBABILITY MODELS
HAVING BROAD APPLICATION
NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION POISSON DISTRIBUTION

4. NORMAL DISTRIBUTION: WHERE ARE THE MEAN AND STANDARD DEVIATION IN
THIS PICTURE?
note the
point of
inflexion
note the
balance point

5. IQ DISTRIBUTION: ~NORMAL, MEAN 100 STANDARD DEVIATION 15 point of
inflexion
SD=15
MEAN = 100

6. DISTRIBUTION OF THE NUMBER OF HEADS IN 100 COIN TOSSES:
APPROXIMATELY NORMAL,
MEAN 50, STD DEVIATION 5 5 50

7. DISTRIBUTION OF THE NUMBER OF ACCIDENTS IN ONE MONTH
IF WE AVERAGE 39.7 PER MONTH:
APPROXIMATELY NORMAL,
MEAN 39.7, STD DEVIATION 6.3
6.3
39.7

8. ~68% NORMAL DISTRIBUTIONS ARE ALIKE IN SD UNITS FROM THE MEAN
~ 68% WITHIN 1 SD OF MEAN
~ 95% WITHIN 2 SD OF MEAN
Illustrated for the
Standard Normal
Mean=0, SD=1
~68%

9. ~95% NORMAL DISTRIBUTIONS ARE ALIKE IN SD UNITS FROM THE MEAN
~ 68% WITHIN 1 SD OF MEAN
~ 95% WITHIN 2 SD OF MEAN
Illustrated for the
Standard normal
Mean=0, SD=1
~95%

10. IQ DISTRIBUTION: ~NORMAL, MEAN 100 STANDARD DEVIATION 15 15 ~68/2 =34%
~95/2=47.5% 85
130
100

11. IQ DISTRIBUTION: ~NORMAL, MEAN 100 STANDARD DEVIATION 15 15 ~68/2 =34%
~95/2=47.5% 85
130
100

12. STANDARD SCORES
CONVERT TO
0 MEAN; SD 1 IQ Z 1 15
Standard Normal
100

13. STANDARD SCORES
CONVERT TO
0 MEAN; SD 1

14. Z - TABLE CUT AND PASTE P(Z > 0) = P(Z < 0 ) = 0.5
= =
P(Z < 1.92) = P(0 < Z < 1.92)
= =

15. BERNOULLI DISTRIBUTION x p(x) p (1 denotes “success”)
q (0 denotes “failure”) __ 1
0 < p < 1
q = 1 - p

16. Notation: BERNOULLI RANDOM VARIABLE X P(success) = P(X = 1) = p
P(failure) = P(X = 0) = q
e.g. X = “sample voter is Democrat”
Population has 48% Dem.
p = 0.48, q = 0.52
P(X = 1) = 0.48

17. INDEPENDENT BERNOULLI-p "S" denotes success "F" denotes failure
P(S1 S2 F3 F4 F5 F6 S7) = p3 q4
just write P(SSFFFFS) = p3 q4
“the answer only depends upon how
many of each, not their order.”
e.g. 48% Dem, 5 sampled, with-repl:
P(Dem Rep Dem Dem Rep) =

18. BINOMIAL DISTRIBUTION FOR THE TOTAL NUMBER OF SUCCESSES IN INDEPENDENT
p-BERNOULLI TRIALS.
e.g. P(exactly 2 Dems out of sample of 4)
= P(DDRR) + P(DRDR) + P(DDRR)
+ P(RDDR) + P(RDRD) + P(RRDD)
= ~
There are 6 ways to arrange 2D 2R.

19. BINOMIAL DISTRIBUTION FOR THE TOTAL NUMBER OF SUCCESSES IN INDEPENDENT
p-BERNOULLI TRIALS.
e.g. P(exactly 3 Dems out of sample of 5)
= P(DDDRR) + P(DDRDR) + P(DDRRD)
+ P(DRDDR) + P(DRDRD) + P(DRRDD)
+ P(RDDDR) +P(RDDRD) + P(RDRDD)
+ P(RRDDD) = ~
There are 10 ways to arrange 3D 2R.
Same as the number of ways to select 3 from 5.

20. COUNTING ARRANGEMENTS 5! ways to arrange 5 things in a line
Do it thus (1:1 with arrangements):
select 3 of the 5 to go first in line,
arrange those 3 at the head of line
then arrange the remaining 2 after.
5! = (ways to select 3 from 5) 3! 2!
So num ways must be 5! /( 3! 2!) = 10.

21. BINOMIAL FORMULA Let random variable X denote the number of
“S” in n independent Bernoulli p-Trials.
By definition, X has a Binomial Distribution
and for each of x = 0, 1, 2, …, n
P(X = x) = (n!/(x! (n-x)!) ) px qn-x
e.g. P(44 Dems in sample of 100 voters) = (100!/(44! 56!)) =

22. Caveats: Binomial Binomial Coefficient
n!/(x! (n-x)!) is the count of how many
arrangements there are of a string of x letters
“S” and n-x letters “F.” .
px qn-x is the shared probability of each string
of x letters “S” and n-x letters “F.”
(define 0! = 1, p0 = q0 = 1 and the formula goes
through for every one of x = 0 through n)
is short for the arrangement count =
Binomial Coefficient

23. Normal Approx of Binomial Poisson and its normal Approx
Aspects of random sampling
Week

24. Normal Approx of Binomial
n = 10, p = 0.4
mean = n p = 4
sd = root(n p q)
~ 1.55
Week

25. Normal Approx of Binomial
n = 30, p = 0.4
mean = n p = 12
sd = root(n p q)
~ 2.683
Week

26. Normal Approx of Binomial
n = 100, p = 0.4
mean = n p = 40
sd = root(n p q) ~
Week

27. p(x) = e-mean meanx / x! for x = 0, 1, 2, ..ad infinitum Poisson
Distribution
Governing
Counts of
Rare Events
p(x) = e-mean meanx / x!
for x = 0, 1, 2, ..ad infinitum
Week

28. e..g. X = number of times ace of spades turns up in 104 tries
Poisson
e..g. X = number of times ace of spades turns up in 104 tries
X~ Poisson with mean 2
p(x) = e-mean meanx / x!
e.g. p(3) = e-2 23 / 3! ~ 0.18
Week

29. Poisson
e.g. X = number of raisins in MY cookie. Batter has 400 raisins and makes 144 cookies.
E X = 400/144 ~ 2.78 per cookie
p(x) = e-mean meanx / x!
e.g. p(2) = e / 2! ~ 0.24
(around 24% of cookies have 2 raisins)
Week

30. note: Poisson sd = root(mean)
THE FIRST BEST THING ABOUT THE POISSON IS THAT THE MEAN ALONE TELLS US THE ENTIRE DISTRIBUTION!
note: Poisson sd = root(mean)
Week

31. E X = 400/144 ~ 2.78 raisins per cookie sd = root(mean) = 1.67
(for Poisson)
Week

32. Poisson
THE SECOND BEST THING ABOUT THE POISSON IS THAT FOR A MEAN AS SMALL AS 3 THE NORMAL APPROXIMATION WORKS WELL.
1.67 = sd = root(mean)
Special to Poisson
Week
mean 2.78

33. WE AVERAGE 127.8
ACCIDENTS PER MO.
E X = accidents
If Poisson then sd = root(127.8) = and the approx dist is:
sd = root(mean) = 11.3
Special to Poisson ~
Week
mean accidents

34. Aspects of
Random Sampling
Week

35. THE GREAT TRICK OF STATISTICS
The overwhelming majority of samples of n from a population of N can stand-in for the population.
ATT Sysco
Pepsico GM Dow
population of N = 5
sample of n = 2

36. THE GREAT TRICK OF STATISTICS
The overwhelming majority of samples of n from a population of N can stand-in for the population.
ATT Sysco
Pepsico GM Dow
ATT
Pepsico
population of N = 5
sample of n = 2

37. GREAT TRICK : SOME CAVEATS
Sample size n must be “large.” For only a few characteristics at a time, such as profit, sales, dividend. SPECTACULAR FAILURES MAY OCCUR!
ATT 12 Sysco 21
Pepsi 42 GM 8 Dow 9
population of N = 5
sample of n = 2

38. With-replacement HOW ARE WE SAMPLING ? ATT 12 Sysco 21
Pepsi 42 GM 8 Dow 9
Pepsi 42
population of N = 5
sample of n = 2

39. With-replacement vs without replacement.
HOW ARE WE SAMPLING ?
With-replacement vs without replacement.
ATT 12 Sysco 21
Pepsi 42 GM 8 Dow 9
population of N = 5
sample of n = 2

40. GREAT TRICK : SOME CAVEATS
This sample is obviously
“not representative.”
ATT 12 Sysco 21
Pepsi 42 GM 8 Dow 9
Sysco 21
Pepsi 42
population of N = 5
sample of n = 2

41. DOES IT MAKE A DIFFERENCE ?
Rule of thumb: With and without replacement are about the same if root [(N-n) /(N-1)] ~ 1.
with vs
without
SAME ?
population of N
sample of n

42. CORRECTION TO PAGE 25 OF TEXT
They would have you believe the population is {8, 9, 12, 42} and the sample is {42}. A SET is a collection of distinct entities.
ATT 12 IBM 42 AAA 9
Pepsi 42 GM 8 Dow 9
WE SAMPLE COMPANIES
NUMBERS COME WITH THEM
Pepsi 42

43. THE ROLE OF RANDOM SAMPLING
IF THE OVERWHELMING MAJORITY OF SAMPLES ARE “GOOD SAMPLES” THEN WE CAN OBTAIN A “GOOD” SAMPLE BY RANDOM SELECTION.

44. SELECTING A LETTER AT RANDOM
HOW TO SAMPLE RANDOMLY ?
SELECTING A LETTER AT RANDOM
Digits are made to correspond to letters.
a = b = …. z = 75-77
Random digits then give random letters.
… (Table 14, pg. 809)
etc… (split into pairs)
f t * w etc… (take chosen letters)
For samples without replacement just pass over any duplicates.

45. The Great Trick is far more powerful than we have seen
The Great Trick is far more powerful than we have seen. A typical sample closely estimates such things as a population mean or the shape of a population density. But it goes beyond this to reveal how much variation there is among sample means and sample densities. A typical sample not only estimates population quantities. It estimates the sample-to-sample variations of its own estimates.

46. EXAMPLE : ESTIMATING A MEAN
The average account balance is $ for a random with-replacement sample of 50 accounts. We estimate from this sample that the average balance is $ for all accounts. From this sample we also estimate and display a “margin of error” $ /- $65.22 = .
s denotes
"sample
standard
deviation"

47. SAMPLE STANDARD DEVIATION
NOTE: Sample standard deviation s
may be calculated in several equivalent ways,
some sensitive to rounding errors, even for n = 2.

48. EXAMPLE : MARGIN OF ERROR CALCULATION
The following margin of error calculation for n = 4 is only an illustration. A sample of four would not be regarded as large enough.
Profits per sale = {12.2, 15.3, 16.2, 12.8}.
Mean = , s = , root(4) = 2.
Margin of error = +/ ( / 2)
Report: /
A precise interpretation of margin of error will be given later in the course, including the role of The interval / is called a “95% confidence interval for the population mean.”
We used: ( )2 + ( )2
+ ( )2 + ( )2 =

49. EXAMPLE : ESTIMATING A PERCENTAGE
A random with-replacement sample of 50 stores participated in a test marketing. In 39 of these 50 stores (i.e. 78%) the new package design outsold the old package design. We estimate from this sample that 78% of all stores will sell more of new vs old. We also estimate a “margin of error +/- 11.5%
Figured: root(pHAT qHAT)/root(n)
=1.96 root( )/root(50)
= in Binomial setup

50. SAMPLING ONLY 600 FROM 500 MILLION ?
A sample of only n = 600 from a population of N = 500 million. (FINE resolution)
sample of n = 600
sample mean = 32.84
POP mean = 32.02
FINE
resolution
densities
very close
population of N = 500,000
with a sample of n = 600