Please Stand By…. Statistics Chapter 4: Probability and Distributions Randomness General Probability Probability Models Random Variables Moments of Random Variables Randomness The language of probability Thinking about randomness The uses of probability Randomness Randomness
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After completing this chapter, you should be able to:
The Sample Space is the collection of all possible outcomes (based on an probabilistic experiment)
e.g., All 6 faces of a die:
e.g., All 52 cards of a bridge deck:
cards
2 Fuel types: Gasoline, Diesel
3 Vehicle types: Truck, Car, SUV
6 possible elementary events:
e1 Gasoline, Truck
e2 Gasoline, Car
e3 Gasoline, SUV
e4 Diesel, Truck
e5 Diesel, Car
e6 Diesel, SUV
Truck
e1
e2
e3
e4
e5
e6
Car
Gasoline
SUV
Truck
Diesel
Car
SUV
E1 = heads on one flip of fair coin
E2 = heads on second flip of same coin
Result of second flip does not depend on the result of the first flip.
E1 = rain forecasted on the news
E2 = take umbrella to work
Probability of the second event is affected by the occurrence of the first event
E2
A card cannot be Black and Red at the same time.
E1
Red
Cards
Black
Cards
Number of ways Ei can occur
Total number of elementary events
P(Ei) =
Number of times Ei occurs
N
Relative Freq. of Ei =
An opinion or judgment by a decision maker about the likelihood of an event
Number of ways Ei can occur
Total number of elementary events
a b c d e …. ___ ___ ___ ___
Rules for
Possible Values
and Sum
Individual Values
Sum of All Values
0 ≤ P(ei) ≤ 1
For any event ei
where:
k = Number of elementary events in the sample space
ei = ith elementary event
Ei = {e1, e2, e3}
then:
P(Ei) = P(e1) + P(e2) + P(e3)
E
E
Or,
P(E1 or E2) = P(E1) + P(E2)  P(E1 and E2)
+
=
E1
E2
E1
E2
P(E1or E2) = P(E1) + P(E2)  P(E1and E2)
Don’t count common elements twice!
P(Red or Ace) = P(Red) +P(Ace)  P(Redand Ace)
= 26/52 + 4/52  2/52 = 28/52
Don’t count the two red aces twice!
Color
Type
Total
Red
Black
2
2
4
Ace
24
24
48
NonAce
26
26
52
Total
P(E1 and E2) = 0
So
E1
E2
= 0
if mutually
exclusive
P(E1 or E2) = P(E1) + P(E2)  P(E1 and E2)
= P(E1) + P(E2)
i.e., we want to find P(CD  AC)
(continued)
20% of the cars have both.
CD
No CD
Total
AC
No AC
1.0
Total
(continued)
CD
No CD
Total
.2
.5
.7
AC
.2
.1
No AC
.3
.4
.6
1.0
Total
Note:If E1 and E2 are independent, then
and the multiplication rule simplifies to
P(E1 and E3) = 0.8 x 0.2 = 0.16
Truck: P(E3E1) = 0.2
Car: P(E4E1) = 0.5
P(E1 and E4) = 0.8 x 0.5 = 0.40
Gasoline P(E1) = 0.8
SUV: P(E5E1) = 0.3
P(E1 and E5) = 0.8 x 0.3 = 0.24
P(E2 and E3) = 0.2 x 0.6 = 0.12
Truck: P(E3E2) = 0.6
Diesel
P(E2) = 0.2
Car: P(E4E2) = 0.1
P(E2 and E4) = 0.2 x 0.1 = 0.02
SUV: P(E5E2) = 0.3
P(E3 and E4) = 0.2 x 0.3 = 0.06
Random
Variables
Discrete
Random Variable
Continuous
Random Variable
xi = Value of Random Variable (Outcome)
P(xi) = Probability Associated with Value
(no overlap)
(nothing left out)
Examples:
Let x be the number of times 4 comes up
(then x could be 0, 1, or 2 times)
Let x be the number of heads
(then x = 0, 1, 2, 3, 4, or 5)
x ValueProbability
0 1/4 = .25
1 2/4 = .50
2 1/4 = .25
Experiment: Toss 2 Coins. Let x = # heads.
4 possible outcomes
Probability Distribution
T
T
T
H
H
T
.50
.25
Probability
H
H
0 1 2 x
(Weighted Average)
E(x) = xi P(xi)
x = # of heads,
compute expected value of x:
E(x) = (0 x .25) + (1 x .50) + (2 x .25)
= 1.0
x P(x)
0 .25
1 .50
2 .25
(continued)
where:
E(x) = Expected value of the random variable
x = Values of the random variable
P(x) = Probability of the random variable having the value of x
(continued)
compute standard deviation (recall E(x) = 1)
Possible number of heads = 0, 1, or 2
E(x + y) = E(x) + E(y)
= x P(x) + y P(y)
(The expected value of the sum of two random variables is the sum of the two expected values)
E(SXi) = SE(Xi) and Var(SXi) = Var(Xi)
σxy = [xi – E(x)][yj – E(y)]P(xiyj)
where:
xi = possible values of the x discrete random variable
yj = possible values of the y discrete random variable
P(xi ,yj) = joint probability of the values of xi and yj occurring
xy > 0 x and y tend to move in the same direction
xy < 0 x and y tend to move in opposite directions
xy = 0 x and y do not move closely together
where:
ρ = correlation coefficient (“rho”)
σxy = covariance between x and y
σx = standard deviation of variable x
σy = standard deviation of variable y
= 0 x and y are not linearly related.
The farther is from zero, the stronger the linear relationship:
= +1 x and y have a perfect positive linear relationship
= 1 x and y have a perfect negative linear relationship
Probability Distributions
Discrete
Probability Distributions
Continuous
Probability Distributions
Binomial
Normal
Poisson
Uniform
Etc.
Etc.
n
!

x
x
n
P(x)
=
p
q
x !
(

)
!
n
x
P(x) = probability of x successes in n trials,
with probability of success pon each trial
x = number of ‘successes’ in sample,
(x = 0, 1, 2, ..., n)
p = probability of “success” per trial
q = probability of “failure” = (1 – p)
n = number of trials (sample size)
Example: Flip a coin four times, let x = # heads:
n = 4
p = 0.5
q = (1  .5) = .5
x = 0, 1, 2, 3, 4
Examples:
n = 10, p = .35, x = 3: P(x = 3n =10, p = .35) = .2522
n = 10, p = .75, x = 2: P(x = 2n =10, p = .75) = .0004
where:
t = size of the segment of interest
x = number of successes in segment of interest
= expected number of successes in a segment of unit size
e = base of the natural logarithm system (2.71828...)
t = 0.50
t = 3.0
where = number of successes in a segment of unit size
t = the size of the segment of interest
(continued)
The Continuous Uniform Distribution:
f(x) =
where
f(x) = value of the density function at any x value
a = lower limit of the interval
b = upper limit of the interval
Example: Uniform Probability Distribution
Over the range 2 ≤ x ≤ 6:
1
f(x) = = .25 for 2 ≤ x ≤ 6
6  2
f(x)
.25
x
2
6
away from the mean is highly unlikely, given
that particular mean and standard deviation
The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below
f(x)
0.5
0.5
μ
x
Also known as the “z” distribution = (xm)/s
Mean is by definition 0
Standard Deviation is by definition 1
f(z)
1
z
0
Values above the mean have positive zvalues, values below the mean have negative zvalues
μ= 100
σ= 50
100
250
x
0
3.0
z
Note that the distribution is the same, only the scale has changed. We can express the problem in original units (x) or in standardized units (z)
(continued)
.0478
.5000
P(x < 8.6)
= P(z < 0.12)
= P(z < 0) + P(0 < z < 0.12)
= .5 + .0478 = .5478
Z
0.00
0.12