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Ex St 801
Statistical Methods
Probability
and
Distributions
a graphic:
Gather data
sample
population
Make inferences
parameters
 ,  ,  ,  ,  , etc .
2
statistics
2
ˆ, etc .

, 
ˆ, 
ˆ ,
ˆ, ˆ
or y , S2 S, p,
b, etc.
some definitions.
Population: Set of all measurements of interest to the
investigator.
Parameter: Characteristic of the population; Greek letter,
usually unknown.
Sample: Subset of measurements selected (usually at
random) from the population.
Statistic: Characteristic of the sample; Latin letter or
Greek letter with a hat ( ^ ).
Random Variable: The characteristic we measure,
usually called ‘y’.
a graphic, again
Probability
Gather data
sample
population
Make inferences
Statistics
More terms:
Probability: Likelihood of occurrence; we know the population,
and we predict the outcome or the sample.
Statistics: We observe the sample and use the statistics
to describe the unknown population.
When we make an inference about the population, it is
desirable that we give a measure of ‘confidence’ in our
being correct (or incorrect). This is done giving a
statement of the ‘probability’ of being correct. Hence,
we need to discuss probability.
So, Probability:
more terms:
event: is an experiment
y: the random variable, is the outcome from
one event
sample space: is the list of all possible outcomes
probability, often written P(Y=y), is the chance that
Y will be a certain value ‘ y ’ and can be computed:
number of successes
p = -------------------------.
number of trials
8 things to say about probability:
1. 0  p  1 ; probabilities are between 0 and 1.
2. pi = 1 ; the sum of all the probabilities of all the
possible outcomes is 1.
Event relations
3. Complement: If P(A=a) = p then A complement is
is P( A not a) = 1-p (= q sometimes).
Note: p + (1-p) = 1 (p+q=1).
A complement is also called
A (not the mean).
8 things continued
4. Mutually exclusive: two events that cannot happen
together. If P(AB)=0, then A
and B are M.E.
5. Conditional: Probability of A given that B has already
happened. P(A|B)
6. Independent: Event A has no influence on the outcome
of event B. If P(A|B) = P(A) or
P(B|A) = P(B) then A and B are
independent.
8 things continued again
two laws about events:
7. Multiplicative Law: (Intersection ; AND)
P(AB) = P(A
 B) = P(A and B) = P(A) * P(B|A)
= P(B) * P(A|B)
if A and B are independent then P(AB) = P(A) * P(B).
8. Additive Law: (Union; OR)
P(A  B) = P(A or B) = P(A) + P(B) - P(AB).
The Venn diagram below can be used to explain
the 8 ‘things’.
A
AB
B
An example: Consider the deck of 52 playing cards:
sample space:
(A,
(A,
(A,
(A,
2, 3,
2, 3,
2, 3,
2, 3,
… , J, Q, K) spades
… , J, Q, K) diamonds
… , J, Q, K) hearts
… , J, Q, K) clubs
Now, consider the following events:
J= draw a J:
P(J)= 4/52=1/13
F = draw a face card (J,Q,K): P(F)= 12/52=3/13
H = draw a heart:
P(H)= 13/52
An example.2: Compute the following:
1. P(F complement) = (52/52 - 12/52) = 40/52
2. Are J and F Mutually Exclusive ?
No: P(JF) = 4/52 is not 0.
3. Are J and F complement M.E. ?
Yes: P(J and F ) = 0
4. Are J and H independent ?
Yes: P(J) = 13/52 = 1/13 = P(J|H)
An example.3: Compute the following:
5. Are J and F independent ?
No: P(J) = 4/52 but P(J|F) = 4/12
6. P(J and H) = P(J) * P(H|J) = 4/52*1/4 = 1/52
7. P(J or H) = P(J) + P(H) - P(JH)
= 4/52 + 13/42 - 1/52 = 16/52.
Two sorts of random variables are of interest:
DISCRETE: the number of outcomes is countable
CONTINUOUS: the number of outcomes in infinite
(not countable).
Random variables are often described with probability
distribution functions. These are graphs, tables or
formula which allow for the computation of
probabilities.
A few common Discrete probability distributions are:
Uniform: P(Y=y) = 1/number of outcomes
(all outcomes are equally likely)
Binomial P(Y=y) = nCy * py * (1-p)(n-y)
where: nCy is the combination of n things
taken y at the time
n is the number of trials
y is the number of successes
p is the probability of succeeding in
one trial
in each trial, the only outcomes are
success and failure (0,1).
y 

e
Poisson P(Y=y) =
;
y!
y=0,1,2,… (for example number of people
waiting in line at a teller)
= the population mean of Y.
These are a few of many and are used when there are
only a few possible outcomes: number of defects on a
circuit board, number of tumors in a mouse, pregnant
or not, dead or alive, number of accidents at an
intersection and so on.
PROBABILITY DEFINITIONS
• An EXPERIMENT is a process by which an
observation is obtained.
• A SAMPLE SPACE (S) is the set of possible
outcomes or results of an experiment.
• A SIMPLE EVENT (Ei) is the smallest possible
element of the sample space.
PROBABILITY DEFINITIONS
• A COMPOUND EVENT is a collection of two
or more simple events.
• Two events are INDEPENDENT if the
occurrence of one event does not affect the
occurrence of the other event.
SAMPLE SPACE FOR THE COINS
IN A JAR EXAMPLE
(P, N1, N2)
(P, N1, D)
(P, N1, Q)
(P, N2, D)
(P, N2, Q)
(P, D, Q) (N1, N2, D) (N1, N2, Q) (N1, D, Q) (N2, D, Q)
INTERPRETATIONS OF
PROBABILITY
Classical:
The number of equally likely wayes E can occur
P(E) =
The total number of equally likely events
Relative Frequency:
The number of times E occurred
P(E) =
The maximum number of times E could have occurred
PROPERTIES OF PROBABILITIES
0 P E i 1
For all E i
P E 1
For all simple events
i
in the sample.
RANDOM VARIABLES
• A RANDOM VARIABLE (r.v.) is a numerical
valued function defined on a sample space
• A DISCRETE RANDOM VARIABLE is a
random variable that can assume a
countable number of values.
RANDOM VARIABLES (CONT.):
• A CONTINUOUS RANDOM VARIABLE is a
random variable that can assume an
infinitely large number of values
corresponding to the points on a line
interval.
A RANDOM VARIABLE FOR THE
COINS IN A JAR EXAMPLE
Let Y be the amount of money taken out of
the jar.
11
16
31
16
31
(P, N1, N2)
(P, N1, D)
(P, N1, Q)
(P, N2, D)
(P, N2, Q)
40
40
36
20
35
(P, D, Q)
(N1, N2, D)
(N1, N2, Q)
(N1, D, Q) (N2, D, Q)
PROBABILITY DISTRIBUTIONS
• A DISCRETE PROBABILITY DISTRIBUTION
is a
formula, table or graph that shows the probability
associated with each value of the discrete random
variable.
• A CONTINUOUS PROBABILITY DISTRIBUTION
is given by an equation f (y) (probability density
function) that shows the density of probability as it
varies with the continuous random variable.
COINS IN A JAR EXAMPLE
PROBABILITY DISTRIBUTION
Y
P(Y)
11
.1
16
.2
20
.1
31
.2
EXPECTED VALUE (MEAN)
E(Y) = 27.6
VARIANCE
VAR (Y) = 105.84
35
.1
36
.1
40
.2
EXPECTATION AND VARIANCE FOR A
DISCRETE RANDOM VARIABLE
  E (Y ) 
 y  p( y)
all y
 2  Var ( y )  E[( y   ) 2 ] 

all y
( y   ) 2  p( y )
EXPECTATION FORMULA
SUPPOSE E(Y) = µY AND E(X)= µX
• E(aY) = a E(Y) = a µY
• E(Y + X) = E(Y) +E(X) = µY + µX
• E(aY + bX) = aE(Y) +bE(X) = aµY + b µX
VARIANCE FORMULA
Suppose Var(Y) = Y2 and Var(X) = X2
• Var(aY) = a2 Var(Y)
• If Y and X are independent, then
– Var(Y + X) = Var(Y) + Var(X)
– Var(aY + bX) = a2 Var(Y) + b2 Var(X)
THE NORMAL DISTRIBUTION
Some properties
1. The area under the entire
curve is always 1.
2. The distribution is
symmetric about the
mean 
3. The mean and the median
are equal.
4. Probabilities may be found
by determining the
appropriate area under
the curve.
SAMPLING DISTRIBUTIONS
• The SAMPLING DISTRIBUTION of a statistic
is the probability distribution for the values
of the statistic that results when random
samples of size n are repeatedly drawn from
the population.
• The STANDARD ERROR is the standard
deviation of the sampling distribution of a
statistic.
DIAGRAM FOR OBTAINING A SAMPLING DISTRIBUTION
POP.
SAMPLE 1
Y1
SAMPLE 2
Y2
SAMPLE 499
Y499
SAMPLE 500
Y500


Y
THE CENTRAL LIMIT THEOREM
• If random samples of n observations are
drawn from a population with a finite mean, ,
and a finite variance 2, then, when n is large
(usually greater than 30), the SAMPLE MEAN,
will be approximately normally distributed
with mean  and variance 2/n.
• The approximation becomes more accurate
as n becomes large.
THE CENTRAL LIMIT THEOREM
if
then
Y ~ ( ,  )
2
2 


 for large n
Y ~ N  ,


n


AN APPLICATION OF THE
SAMPLING DISTRIBUTION
• The Ybar CONTROL CHART can be used to
detect shifts in the mean of a process.
• The chart looks at a sequence of sample
means and the process is assumed to be “IN
CONTROL” as long as the sample means are
within the control limits.
Y CONTROL CHART
UPPER
CONTROL
LIMIT
CENTER
LINE
LOWER
CONTROL
LIMIT
SAMPLE
GLASS BOTTLE EXAMPLE
• A glass-bottle manufacturing company wants
to maintain a mean bursting strength of 260
PSI.
• Past experience has shown that the standard
deviation for the bursting strength is 36 PSI.
• The company periodically pulls 36 bottles off
the production line to determine if the mean
bursting strength has changed.
GLASS BOTTLE EXAMPLE
(CONT)
• Construct a control chart so that 95% of the
sample means will fall within the control
limits when the process is “IN CONTROL.”
• The END