Transcript AP_Statistics_-_Probability_

Mrs. Lerner
Charlotte Catholic High School
With some updates by
Dr. Davidson
Mallard Creek High School
1. Are uncertain in the short run
2. Exhibit a consistent pattern in
the long run
 Note the dual aspect
Note: This is the Condition for the Probability we study


An event is an outcome or a set of
outcomes of random phenomenon (RP).
Outcome(s) of interest are “Success(es)”
[even if they are bad news!!!!]
(note dual aspect)
 Probability of an event =
1. proportion of times a success
occurs in the long run (measure by
experiment or simulation)
2. [number of ways for a success to
occur]/[all possible events
(specified by a model of the RP)

If there are n ways to do a first event
 & m ways to do a second event
 Then the number of

all possible outcomes=nm

This is called the multiplication rule



Consider the sum of two dice that are rolled
Let’s say we are interested in the Probability
the sum is a 4, so SUCCESS = ?
Find # of ways to get a sum = 4 [a success]
is 3:
1+3, 2+2, 3+1 [3 ways]

The probability of a specific outcome is
[# of ways to get a success]/ [all possible
outcomes
Probability of getting a sum =4 is:
3/36 = 1/12 ~ .0833 = 8.33 %
e.g.
1. This is the chance the next outcome is a success
2. This is the proportion of times that successes
occur in a large repetitive # of identical trials


The probability P(A) of any
event A satisfies
0≤P(A)≤1
Thus the probability of
any event is between 0 & 1


The Sample Space is the set of all possible
outcomes
If S is the sample space in a probability
model, then P(S)=1
The probabilities of all
possible outcomes must
add up to 1


The complement of any event A is
1. the event that A does not occur
2. written as Ac
The complement rule states that
P(Ac)=1-P(A)
The probability that an event
does not occur is 1 minus the
probability that it does occur


Two events A and B are disjoint
1. if they have no outcomes in common
2. cannot occur at the same time.
If A and B are disjoint, then
P(A or B) = P(A) + P(B).
If two events have no outcomes in
common, then the probability of either
one occurring is the sum of their
individual probabilities

For any two events A and B:
P(A or B) = P(A) + P(B) – P(A and B)
The probability of either A or B occurring is the
sum of their individual probabilities minus
the probability that they occur at the same
time
Two events A and B are independent if
knowing that one occurs does not change the
probability that the other occurs. If A and B
are independent, then P(A and B) = P(A)P(B).
If two events are independent, then the
probability that they both occur is the
product of their individual probabilities
Note – Disjoint events are always NOT
INDEPENDENT, for if one occurs, then we
know the other can not occur!



And = Joint = Intersection
Or = Both (1, or other, or 2) = Union
P  A and B   P  A  B 
P  A or B   P  A  B 

When P(A)>0, the conditional probability of B
given A is:
The probability of B occurring, given that
A occurs, is the probability of A & B jointly
occurring divided by the probability of the
stated condition (A) occurring

For any two events A and B,
P(A and B) = P(B|A) P(A)
P(A and B) = P(A|B) P(B)
The probability of A and B occurring
jointly is the probability that B occurs given
that A has already occurred multiplied by the
probability that A occurs
Example - 2:
Suppose that for a certain Caribbean island in
any 3-year period:
1. the probability of a major hurricane is .25
2. the probability of water damage is .44
3. and the probability of both a hurricane and
water damage is .22.
What is the probability of water damage
given that there is a hurricane?
Suppose that for a certain Caribbean
island in any 3-year period the
probability of a major hurricane is .25,
the probability of water damage is .44
and the probability of both a hurricane
and water damage is .22. What is the
probability of water damage given that
there is a hurricane?
P  w ater d am ag e|h u rrican e  

P  w ater d am ag e an d h u rrican e 
P  h u rrican e 
.2 2
.2 5
 .8 8
If three people, Joe, Betsy, and Sue,
play a game in which
 Joe has a 25% chance of winning
 Betsy has a 40% chance of
winning
 What is the probability that Sue
will win?
2.
If three people, Joe, Betsy, and Sue, play
a game in which Joe has a 25% chance of
winning and Betsy has a 40% chance of
winning, what is the probability that Sue
will win?
P  S ue   1   .25  .4 
 1  .65
 .35
4.


A summer resort rents rowboats to customers
but does not allow more than four people to a
boat. Each boat is designed to hold no more
than 800 pounds.
Suppose the distribution of the weight of adult
males who rent boats, including their clothes and
gear, is normal with a mean of 190 pounds and
standard deviation of 10 pounds.
If the weights of individual passengers are
independent, what is the probability that a
group of four adult male passengers will
exceed the acceptable weight limit of 800
pounds?
 X  190
 X  10
T  4  X
 4  190 
 760
 X  190
 X  10
T  4  X
T  X X X X
2
2
2
2
2
 4  190 
 10  10  10  10
 760
 400
2
 T  20
2
2
2
 X  190
 X  10
T  4  X
T  X X X X
2
2
2
2
2
 4  190 
 10  10  10  10
 760
 400
2
2
 T  20
n o rm a lcd f  8 0 0,1 E 9 9, 7 6 0, 2 0 
P  exceed 800 lbs   0.023
2
2
The following table shows the frequencies of political affiliations in
the age ranges listed from a random sample of adult citizens in a
particular city.
18-30
31-40
41-50
Over 50
Dem.
25
32
17
14
Repub. Indep.
18
12
21
10
25
17
32
15
_____________________________________________________________
5.
What proportion of the Republicans are over
50?
a.
b.
c.
d.
e.
61/238
32/96
96/238
32/61
Cannot be determined.
The following table shows the frequencies of political affiliations in
the age ranges listed from a random sample of adult citizens in a
particular city.
18-30
31-40
41-50
Over 50
Dem.
25
32
17
14
Repub. Indep.
18
12
21
10
25
17
32
15
_____________________________________________________________
5.
What proportion of the Republicans are over
50?
a.
b. 32/96
c.
d.
e.
The following table shows the frequencies of political affiliations in
the age ranges listed from a random sample of adult citizens in a
particular city.
Dem.
25
32
17
14
18-30
31-40
41-50
Over 50
Repub. Indep.
18
12
21
10
25
17
32
15
_____________________________________________________________
6.
If one adult citizen is chosen at random,
what is the probability that this person is a
Democrat between the ages of 41 and 50?
a.
b.
c.
d.
e.
17/238
17/88
61/238
17/61
88/238
The following table shows the frequencies of political affiliations in
the age ranges listed from a random sample of adult citizens in a
particular city.
18-30
31-40
41-50
Over 50
Dem.
25
32
17
14
Repub. Indep.
18
12
21
10
25
17
32
15
_____________________________________________________________
6.
If one adult citizen is chosen at random,
what is the probability that this person is a
Democrat between the ages of 41 and 50?
a. 17/238
b.
c.
d.
e.
The following table shows the frequencies of political affiliations in
the age ranges listed from a random sample of adult citizens in a
particular city.
Dem.
25
32
17
14
18-30
31-40
41-50
Over 50
Repub. Indep.
18
12
21
10
25
17
32
15
_____________________________________________________________
7.
Given that a person chosen at random is
between 31 and 40, what is the probability
that this person is an Independent?
a.
b.
c.
d.
e.
10/238
10/63
10/54
54/238
63/238
The following table shows the frequencies of political affiliations in
the age ranges listed from a random sample of adult citizens in a
particular city.
18-30
31-40
41-50
Over 50
Dem.
25
32
17
14
Repub. Indep.
18
12
21
10
25
17
32
15
_____________________________________________________________
7.
Given that a person chosen at random is
between 31 and 40, what is the probability
that this person is an Independent?
a.
b. 10/63
c.
d.
e.
The following table shows the frequencies of political affiliations in
the age ranges listed from a random sample of adult citizens in a
particular city.
18-30
31-40
41-50
Over 50
Dem.
25
32
17
14
Repub. Indep.
18
12
21
10
25
17
32
15
_____________________________________________________________
8.
What proportion of the citizens sampled are
over 50 or Independent?
a.
b.
c.
d.
e.
54/238
61/238
100/238
115/238
Cannot be determined
The following table shows the frequencies of political affiliations in
the age ranges listed from a random sample of adult citizens in a
particular city.
18-30
31-40
41-50
Over 50
Dem.
25
32
17
14
Repub. Indep.
18
12
21
10
25
17
32
15
_____________________________________________________________
8.
What proportion of the citizens sampled are
over 50 or Independent?
a.
b.
c. 100/238
d.
e.
Example  Assume an multiple-choice examination
consists of questions, each having five
possible answers.
 Linda estimates that she has probability 0.75
of knowing the answer to any question that
may be asked.
 If she does not know the answer, she will
guess, with conditional probability 1/5 of
being correct. What is the probability that
Linda gives the correct answer to a question?
(Draw a tree diagram to guide the
calculations.)

An examination consists of multiple-choice
questions, each having five possible answers.
Linda estimates that she has probability 0.75
of knowing the answer to any question that
may be asked. If she does not know the
answer, she will guess, with conditional
probability 1/5 of being correct. What is the
probability that Linda gives the correct
answer to a question? (Draw a tree diagram
to guide the calculations.)
P(correct) = .75 +.25*.2
= .8
A
random variable
assumes any of several
different values as a
result of some random
phenomenon

A Discrete RV – has a countable number of possible
values
Value of X x1
x2
x3
…
xk
Probability p1
p2
p3
…
pk
◦ The probabilities must satisfy two requirements
 Every probability is between 0 and 1
 The sum of all probabilities is 1
◦ We can use a probability histogram to look at the
probability distribution.

Mean of a Discrete R. V. – (also called
expected value) –
 X  x1 p1  x 2 p 2  x 3 p 3    x k p k

x
i
pi

Variance of a Discrete R. V. –

2
X
  x1   X

 x
i
2
X
p1  x 2   X
2
pi
2
p 2  x 3   X
2
p 3    x k   X
2
pk

Continuous R. V. – takes all values in an
interval of numbers
◦We look at its distribution
using a density curve
◦The probability of any event
is the area under the density
curve in that interval.
1.
If X is an R. V. and a & b are
fixed numbers, then the
mean
μa+bX = a +bμX
2.
If X and Y are R. V.‘s, then
μX±Y = μX ± μY

If X is an R. V. and a and b are fixed numbers,
then
σ2a+bX = b2σX2
◦ Note that multiplying by a constant changes the
variance but adding a constant does not.

If X and Y are independent R. V.’s, then
σ2X±Y =σX2+σY2
“Pythagorean Theorem of Statistics”
For STANDARD DEVIATION:
square ‘em, add ‘em, take the square root
11. Suppose X and Y are random variables
with μX = 10, σX = 3, μY = 15, and
σY = 4. Given that X and Y are
independent, what are the mean and
standard deviation of the random
variable X+Y?
11. Suppose X and Y are random variables
with μX = 10, σX = 3, μY = 15, and
σY = 4. Given that X and Y are
independent, what are the mean and
standard deviation of the random
variable X+Y?
μX+Y = μX + μY
=10 + 15
= 25
σX+Y = √σ2X + σ2Y
= √9+16
= √25
=5
12.
You roll a die. If it comes up a 6, you win
$100. If not, you get to roll again. If you
get a 6 the second time, you win $50. If
not, you lose.
a. Create a probability model for the amount you
will win at this game.
12.
You roll a die. If it comes up a 6, you win
$100. If not, you get to roll again. If you
get a 6 the second time, you win $50. If
not, you lose.
a. Create a probability model for the amount you
will win at this game.
Winnings
$100
$50
$0
Probability
1/6
(5/6)(1/6)
5/36
(5/6)(5/6)
25/36
12.
You roll a die. If it comes up a 6, you win
$100. If not, you get to roll again. If you
get a 6 the second time, you win $50. If
not, you lose.
b. Find the expected amount you’ll win.
12.
You roll a die. If it comes up a 6, you win
$100. If not, you get to roll again. If you
get a 6 the second time, you win $50. If
not, you lose.
b. Find the expected amount you’ll win.
X  E X

1
 5
 100    50 
6
 36
 $23.61

 25 
 0


 26 

Law of Large Numbers – The long run relative
frequency of repeated independent trials gets
closer and closer to the true relative
frequency as the number of trials increases.

Binomial Distribution – the distribution of the
count X successes in the binomial setting.


B(n,p) where n is the number of observations
and p is the probability of success



Use binompdf(n,p,X) to find the probability
of a single value of X, such as P(X = 3).
Use binomcdf(n,p,X)to find the probability
of at most X successes
for example
P(X ≤ 3).
  np
 
np 1  p 
13. Pepsi is running a sales promotion in
which 12% of all bottles have a “FREE” logo
under the cap. What is the probability
that you find two free cans in a 6-pack?
13. Pepsi is running a sales promotion in
which 12% of all bottles have a “FREE” logo
under the cap. What is the probability
that you find two free cans in a 6-pack?
6
2
4
P  X  2      .12   .88 
2
13. Pepsi is running a sales promotion in
which 12% of all bottles have a “FREE” logo
under the cap. What is the probability
that you find two free cans in a 6-pack?
6
2
4
P  X  2      .12   .88 
2
b in o m p d f  6, .1 2, 2 
P  X  2   .1 3
14. The National Association of Retailers
reports that 62% of all purchases are
now made by credit card; you think this
is true at your store as well. On a typical
day you make 20 sales.
a.
Let X represent the number of customers
who use a credit card on a typical day.
What is the probability model for X?
14. The National Association of Retailers
reports that 62% of all purchases are
now made by credit card; you think this
is true at your store as well. On a typical
day you make 20 sales.
a.
Let X represent the number of customers
who use a credit card on a typical day.
What is the probability model for X?
The model is = B(20, .62) [i.e. B(n,p)]
Please Explain Why
14. The National Association of Retailers
reports that 62% of all purchases are
now made by credit card; you think this
is true at your store as well. On a typical
day you make 20 sales.
b.
Find the mean and standard deviation.
14. The National Association of Retailers
reports that 62% of all purchases are
now made by credit card; you think this
is true at your store as well. On a typical
day you make 20 sales.
b.
Find the mean and standard deviation.
 X  np
 20  .62 
 12.4
X 

np  1  p 
12  .62   .38 
 2.17
14. The National Association of Retailers
reports that 62% of all purchases are
now made by credit card; you think this
is true at your store as well. On a typical
day you make 20 sales.
c.
What is the probability that on a typical
day at least half of your customers use a
credit card?
14. The National Association of Retailers
reports that 62% of all purchases are
now made by credit card; you think this
is true at your store as well. On a typical
day you make 20 sales.
c.
What is the probability that on a typical
day at least half of your customers use a
credit card?
P  X  10   1  P  X  9 
 .9077
1  binom cdf (20, .62, 9)
19. The volumes of soda in quart soda bottles
can be described by a Normal model with
a mean of 32.3 oz and a standard
deviation of 1.2 oz. What is the probability
that a randomly selected bottle has a
volume less than 32 oz?


There are typos in the next slide:
The z score calculation should read:
P(x< 32) = P(z < ([32-32.3]/1.2) = -.25
And
Normalcdf(-E99, 32, 32.3, 1.2)
The volumes of soda in quart soda bottles
can be described by a Normal model
N(32.3, 1.2) a mean of 32.3 oz and a
standard deviation of 1.2 oz.
What is the probability that the volume of a
randomly selected bottle has a less than 32
oz?
32.3  32 

P  x  32   P  z 

19.

1.2

 P  z  .1429 
 .4013
norm alcdf (  1 E 99, 32, 32.1,1.2)
20. A bank's loan officer rates applicants for
credit. The ratings can be described by a
Normal model with a mean of 200 and a
standard deviation of 50. If an applicant is
randomly selected, what is the probability
that the rating is between 200 and 275?
20. A bank's loan officer rates applicants for
credit. The ratings can be described by a
Normal model with a mean of 200 and a
standard deviation of 50. If an applicant is
randomly selected, what is the probability
that the rating is between 200 and 275?
P  2 0 0  x  2 7 5   .4 3 3 2


Sampling distribution – the distribution of
values taken by a statistic in all possible
samples of the same size from the same
population

Provided that the sampled values are
independent and the sample size is large
enough, the sampling distribution of pˆ is
modeled by a Normal model with mean   pˆ  
p .1  p 
and standard deviation
SD  pˆ  
n
p

Assume that 12% of students at a university
wear contact lenses. We randomly pick 200
students.
◦ What is the mean of the proportion of students in
this group who may wear contact lenses?
◦ What is the standard deviation of the proportion of
students in this group who may wear contact
lenses?

Assume that 12% of students at a university
wear contact lenses. We randomly pick 200
students.
◦ What is the mean of the proportion of students in
this group who may wear contact lenses?
◦ What is the standard deviation of the proportion of
students in this group who may wear contact
lenses?
  .1 2
 
 .1 2   .8 8 
200
 .0 2 3


Suppose that x-bar is the mean of an SRS of
size n drawn from a large population with
mean μ and standard deviation σ.
Then the
◦ mean of the sampling distribution of xbar is μ
(hence xbar is an unbiased indicator of μ)
◦ standard deviation of the sampling distribution of
xbar is
σ /√n.

The scores of individual students on the ACT
have a normal distribution with mean 18.6
and standard deviation 5.9. At Northside
High, 76 seniors take the test. If the scores
at this school have the same distribution as
national scores, what are the mean and
standard deviation of the distribution of
sample means for these 76 students?

The scores of individual students on the ACT
have a normal distribution with mean 18.6
and standard deviation 5.9. At Northside
High, 76 seniors take the test. If the scores
at this school have the same distribution as
national scores, what are the mean and
standard deviation of the average (sample
mean) for the 76 students?
  18.6
 
5.9
76
 .6768