Transcript Unit06 PowerPoint for statistics class

Welcome to
Week 06
College Statistics
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Probability
Probability - likelihood of a
favorable outcome
Probability
Probability is defined to be:
P =
# favorable outcomes
total # of outcomes
Usually probabilities are given
in %
Probability
This assumes each outcome is
equally likely to occur –
RANDOM
Probability
IN-CLASS PROBLEMS
What is
P(head on 1 toss of a fair coin)
Probability
IN-CLASS PROBLEMS
P(head on 1 toss of a fair coin)
What are all of the outcomes?
Probability
IN-CLASS PROBLEMS
P(head on 1 toss of a fair coin)
What are all of the outcomes?
H
or
T
What are the favorable
outcomes?
Probability
IN-CLASS PROBLEMS
P(head on 1 toss of a fair coin)
What are the favorable
outcomes?
H
or
T
Probability
IN-CLASS PROBLEMS
So:
P(head on 1 toss of a fair coin)
= 1/2 or 50%
Probability
The Law of Averages
And why it doesn’t work the way
people think it does
Probability
IN-CLASS PROBLEMS
Amy Bob Carlos Dawn Ed
ABC ABD ABE ACD ACE ADE
BCD BCE BDE
CDE
What is the probability that Bob
will be going to the conference?
Probability
IN-CLASS PROBLEMS
ABC ABD ABE ACD ACE ADE
BCD BCE BDE
CDE
How many outcomes total?
How many outcomes favorable?
Probability
IN-CLASS PROBLEMS
ABC ABD ABE ACD ACE ADE
BCD BCE BDE
CDE
How many outcomes total? 10
How many outcomes favorable?
6 have a “B” in them
Probability
IN-CLASS PROBLEMS
ABC ABD ABE ACD ACE ADE
BCD BCE BDE
CDE
So there is a 6/10 or 60%
probability Bob will be going to
the conference
Probability
Because the # favorable
outcomes is always less than
the total # of outcomes
0 ≤ P ≤ 1
or
0% ≤ P ≤ 100%
Probability
And
P(all possible outcomes) = 100%
Probability
The complement of an outcome
is all outcomes that are not
favorable
Called P (pronounced “P prime")
P = 1 – P
or
P = 100% – P
Probability
IN-CLASS PROBLEMS
To find:
P (not 1 on one roll of a fair
die)
This is the complement of:
P (a 1 on one roll of a fair die)
= 1/6
Probability
IN-CLASS PROBLEMS
Since:
P (a 1 on one roll of a fair die)
= 1/6
Then P (not 1 on one roll of a
fair die) would be:
P = 1 – P = 1 – 1/6 = 5/6
Probability
Mutually exclusive outcomes – if
one occurs, then the other
cannot occur
Ex: if you have a “H” then you
can’t have a “T”
Probability
Addition rule - if you have
mutually exclusive outcomes:
P(both) = P(first) + P(second)
Probability
Mutually exclusive events:
Event A: I am 5’ 2” tall today
Event B: I am 5’ 6” tall today
Probability
For tables of data, calculating
probabilities is easy:
Minutes
Internet
Usage
Probability of
Being in Category
1-20
9/52 = 0.17 = 17%
21-40
18/52 = 0.35 = 35%
41-60
15/52 = 0.29 = 29%
61-80
0.15 = 15%
81-100
0.02 = 2%
101-120
0.00 = 0%
121+
0.02 = 2%
The probability
that a subscriber
uses 1-20 min:
P(1-20) = 0.17
Probability
IN-CLASS PROBLEMS
What is the probability of living
more than 12 years after this
diagnosis?
Years After
% of
Diagnosis
Deaths
1-2
3-4
5-6
7-8
9-10
11-12
13-14
15+
15
35
16
9
6
4
2
13
Probability
IN-CLASS PROBLEMS
Since you can’t live BOTH 1314 AND 15+ years after
diagnosis (you fall Years After % of
Diagnosis
Deaths
into one category
1-2
15
or the other) the
3-4
35
5-6
16
events are
7-8
9
mutually exclusive
9-10
6
11-12
13-14
15+
4
2
13
Probability
IN-CLASS PROBLEMS
You can use the addition rule!
P(living >12 years)
= P(living13-14)
+ P(living 15+)
= 2% + 13%
= 15%
Years After
Diagnosis
% of
Deaths
1-2
3-4
5-6
7-8
9-10
11-12
13-14
15+
15
35
16
9
6
4
2
13
Probability
If two categories are “mutually
exclusive”
you add their probabilities to
get the probability of one OR
the other
Probability
Sequential outcomes – one after
the other
first toss: H
second toss: T
third toss: T . . .
Probability
Because one outcome in the
sequence does not affect the
outcome of the next event in
the sequence we call them
“independent outcomes”
Probability
Multiplication rule - if you have
independent outcomes:
P(one then another) =
P(one) * P(another)
Probability
To get the probability of both
event A AND event B occurring,
you multiply their probabilities
to get the probability of both
Probability
IN-CLASS PROBLEMS
What if you had data:
Natural Hair
Color
Blonde
34.0%
Brown
43.0%
Red
7.0%
Black
14.0%
Eye Color
Blue
36.0%
Brown
64.0%
Probability
IN-CLASS PROBLEMS
If hair color and eye color are
independent events, then find:
P(blonde AND blue eyes)
Natural Hair
Color
Blonde
34.0%
Brown
43.0%
Red
7.0%
Black
14.0%
Eye Color
Blue
36.0%
Brown
64.0%
Probability
IN-CLASS PROBLEMS
P(blonde AND blue eyes) =
=.34 x .36 ≈ .12 or 12%
Natural Hair
Color
Blonde
34.0%
Brown
43.0%
Red
7.0%
Black
14.0%
Eye Color
Blue
36.0%
Brown
64.0%
Probability
What would be the probability of
a subscriber being both
(1-20 minutes)
and
Minutes Probability of
Internet Being in Category
Usage
(61-80 minutes)?
1-20
9/52 = 0.17 = 17%
21-40
18/52 = 0.35 = 35%
41-60
15/52 = 0.29 = 29%
61-80
0.15 = 15%
81-100
0.02 = 2%
101-120
0.00 = 0%
121+
0.02 = 2%
Probability
Zero!
They are mutually exclusive
categories
Probability
IN-CLASS PROBLEMS
Check out the gaming apparatus
and answer the questions!
Probability
Empirical vs theoretical probabilities
Empirical – based on observed
results
Theoretical – based on
mathematical theory
Probability
Are the probabilities of coin
tosses empirical or theoretical?
Probability
Are the probabilities of horse
racing empirical or theoretical?
Questions?
Probability
We were studying probability
for things with outcomes like:
H T
1 2 3 4 5 6
Win Lose
Probability
These were all “discrete” outcomes
Probability
Now we will look at probabilities
for things with an infinite
number of outcomes that can
be considered continuous
Probability
You can think of smooth
quantitative data graphs as a
series of skinnier and skinnier
bars
Probability
When the width of the bars
reach “zero” the graph is
perfectly smooth
Probability
SO, a smooth quantitative
(continuous) graph can be
thought of as a bar chart
where the bars have width zero
Probability
The probability for a continuous
graph is the area of its bar:
height x width
Probability
But…
the width of the bars on a
continuous graph are zero, so
P = Bar Area = height x zero
All the probabilities are
P = 0 !
Probability
Yep.
It’s true.
The probability of any specific
value on a continuous graph is:
ZERO
Probability
So…
Instead of a specific value, for
continuous graphs we find the
probability of a range of values
– an
area under
the curve
Probability
Because this would require
yucky calculus to find the
probabilities, commonly-used
continuous graphs are included
in Excel
Yay!
Normal Probability
The most popular continuous
graph in statistics is the
NORMAL DISTRIBUTION
Normal Probability
Two descriptive statistics
completely define the shape of
a normal distribution:
Mean µ
Standard deviation σ
Normal Probability
Suppose we have a normal
distribution, µ = 12 σ = 2
Normal Probability
If µ = 12
12
Normal Probability
If µ = 12 σ = 2
6
8 10 12 14 16 18
Normal Probability
PROJECT QUESTION
Suppose we have a normal
distribution, µ = 10
?
Normal Probability
PROJECT QUESTION
Suppose we have a normal
distribution, µ = 10 σ = 5
?
?
?
?
?
?
?
Normal Probability
Suppose we have a normal
distribution, µ = 10 σ = 5
-5
0
5 10 15 20 25
Questions?
Normal Probability
The standard normal
distribution has a mean µ = 0
and a standard deviation σ = 1
Normal Probability
PROJECT QUESTION
For the standard normal
distribution, µ = 0 σ = 1
?
?
?
?
?
?
?
Normal Probability
PROJECT QUESTION
For the standard normal
distribution, µ = 0 σ = 1
-3 -2 -1 0
1
2
3
Normal Probability
The standard normal is also
called “z”
Normal Probability
We can change any normallydistributed variable into a
standard normal
One with:
mean = 0
standard deviation = 1
Normal Probability
To calculate a “z-score”:
Take your value x
Subtract the mean µ
Divide by the standard
deviation σ
Normal Probability
z = (x - µ)/σ
Normal Probability
IN-CLASS PROBLEMS
Suppose we have a normal
distribution, µ = 10 σ = 2
z = (x - µ)/σ = (x-10)/2
Calculate the z values for
x = 9, 10, 15
Normal Probability
IN-CLASS PROBLEMS
z = (x - µ)/σ = (x-10)/2
x .
9 z = (9-10)/2 = -1/2
Normal Probability
IN-CLASS PROBLEMS
z = (x - µ)/σ = (x-10)/2
x .
9 z = (9-10)/2 = -1/2
10 z = (10-10)/2 = 0
Normal Probability
IN-CLASS PROBLEMS
z = (x - µ)/σ = (x-10)/2
x .
9 z = (9-10)/2 = -1/2
10 z = (10-10)/2 = 0
15 z = (15-10)/2 = 5/2
Normal Probability
-3 -2 -1 0
|
-1/2
1
2
3
|
5/2
Normal Probability
But…
What about the probabilities??
Normal Probability
We use the properties of the
normal distribution to calculate
the probabilities
Normal Probability
IN-CLASS PROBLEMS
What is the probability of
getting a z-score value between
-1 and 1
-2 and 2
-3 and 3
Normal Probability
You also use symmetry to
calculate probabilities
Normal Probability
IN-CLASS PROBLEMS
What is the probability of
getting a z-score value between
-1 & 0
-2 & 0
-3 & 0
Percentage
of the curve
z-score
Normal Probability
IN-CLASS PROBLEMS
What is the probability of
getting a z-score value between
-1 & 2.5 -0.5 & 3 -2.5 & -1.5
Percentage
of the curve
z-score
Normal Probability
Note: these are not exact –
more accurate values can be
found using Excel
Percentage
of the curve
z-score
Questions?
Excel Probability
Normal probability values in Excel
are given as cumulative values
(the probability of getting an x
or z value less than or equal to
the value you want)
Excel Probability
Use the function: NORM.DIST
Excel Probability
Excel Probability
For a z prob, use: NORM.S.DIST
Excel Probability
Excel gives you the cumulative
probability – the probability of
getting a value UP to your x
value:
P (x ≤ your value)
Excel Probability
How do you calculate other
probabilities using the
cumulative probability Excel
gives you?
Normal Probability
IN-CLASS PROBLEMS
If the area under the entire
curve is 100%, how much of the
graph lies above “x”?
Excel Probability
P(x ≥ b) = 1 - P(x ≤ b)
or
= 100% - P(x ≤ b)
or
= 100% - 90% = 10%
Excel Probability
What about probabilities
between two “x” values?
Excel Probability
P(a ≤ x ≤ b)
equals
P(x ≤ b) – P(x ≤ a)
minus
Normal Probability
IN-CLASS PROBLEMS
Calculate the probabilities when
µ = 0 σ = 1:
P(-1 ≤ x ≤ 2)
P(1 ≤ x ≤ 3)
P(-2 ≤ x ≤ 1)
Normal Probability
IN-CLASS PROBLEMS
P(-1 ≤ x ≤ 2) = 34%+34%+13.5%
= 81.5%
Normal Probability
IN-CLASS PROBLEMS
P(-1 ≤ x ≤ 2) = 81.5%
P(1 ≤ x ≤ 3)
= 13.5%+2%
= 15.5%
Normal Probability
IN-CLASS PROBLEMS
P(-1 ≤ x ≤ 2) = 81.5%
P(1 ≤ x ≤ 3)
= 15.5%
P(-3 ≤ x ≤ 1) =
= 2%+13.5%+34%+34%
= 83.5%
Questions?