Give the number of objects described.
1. The number of cards in a standard deck.
2. The number of face cards in a standard deck.
3. The number of vertices of a octogon.
4. The number of faces on a cubical die.
5. The number of possible totals when two dice are rolled.
Slid
e 93
Give the number of objects described.
1. The number of cards in a standard deck. 52
2. The number of face cards in a standard deck. 12
3. The number of vertices of a octogon. 8
4. The number of faces on a cubical die. 6
5. The number of possible totals when two dice are rolled. 11
Slide 9- 4

! Means factorial
0!=1
 1!=1
 n!=(n)(n-1)(n-2)(n-3)….


6!=6*5*4*3*2*1=720
𝑛
 Combination: 𝐶𝑛,𝑟 𝑜𝑟 𝐶
𝑜𝑟 𝑛𝐶𝑟 =
𝑟!
𝑟
𝑛
 Permutation: 𝑃𝑛,𝑟 𝑜𝑟 𝑃
𝑜𝑟 𝑛𝑃𝑟 =
𝑟
𝑛!
𝑛−𝑟 !
𝑛!
𝑛−𝑟 !

Order does not matter! It is not important
› If you have 3,1,2, it is the same as 1,3,2
because they all have 1,2,3

Different Arrangement of things

Combination is choosing
How many 10 person committees can be formed
from a group of 20 people?
Slid
e 99
How many 10 person committees can be formed
from a group of 20 people?
Notice that order is not important. Using combinations,
20
20!
C 
 184, 756.
10! 20  10 !
10
There are 184,756 possible committees.
Slid
e 910

If you have three spots for ASB officers. If
there are 10 people apply to be an
officer, how many different combination
of people can you have?

10C3

Order does matter! It is important

Note: Permutation is position

112 people ran the marathon and only
top 3 runner gets the prize. How many
different ways winners can be
organized?

112P3

16 Actors answer a casting call to try out
for roles as dwarfs in a productions of
Snow White and the Seven Dwarfs. In
how many different ways can the
director cast the seven roles?

16P7

Both of them break down into two different
category:
Combination with repetition (ex: ice cream
scoops)
 Combination without repetition (ex: lottery)

Permutation with repetition (ex: lock in
locker room)
 Permutation without repetition (ex:
marathon race)


You have 8 people, what are the
number of possible ordered seating
arrangement for 5 chairs

Permutation, 𝑃8,5 =
8!
8−5 !
= 6720
Gamestop has 25 new games this month
and you decided to buy 5 of them. How
many different arrangement of game you
can have?

Combination=𝐶25,5 =
25!
5!20!
= 53130

What is the chance of winning the
jackpot for MEGA Million?

You have 5 slots + 1 slot for MEGA
number

The first 5 slots are numbers between 175, mega number is number between 115
75!

∗
70!5!
15 = 258890850

.00000000386

Or .000000386%

1 in 258890850

What is the chance of winning the
Jackpot for Powerball?

You have 5 slots+1 slot for Power number

The first 5 slots are numbers between 156, powerball slot is number between 135

1 in 133,693,560

Even though it’s harder to win the
Jackpot, for overall winning chance, you
have more chance for MEGA million
than Powerball
If a procedure P has a sequence of stages S , S ,..., S and if
1
2
n
S can occur in r ways,
1
1
S can occur in r ways
2
2
S can occur in r ways,
n
n
then the number of ways that the procedure P can occur is the
product rr ...r .
1
2
n
Slid
e 928
If a license plate has four letters followed by three
numerical digits. Find the number of different license
plates that could be formed if there is no restriction on
the letters or digits that can be used.
Slid
e 929
If a license plate has four letters followed by three
numerical digits. Find the number of different license
plates that could be formed if there is no restriction on
the letters or digits that can be used.
You can fill in the first blank 26 ways, the second blank 26 ways, the third blank 26
ways, the fourth blank 26 ways, the fifth blank 10 ways, the sixth blank 10 ways,
and the seventh blank 10 ways. By the Multiplication Principle, there are
26×26×26×26×10×10×10 = 456,976,000 possible license plates.
Slid
e 930

California License Plate has 3 letters and
4 numbers. Find the number of different
license plates that could be formed.

10*26*26*26*10*10*10=175,760,000 ways

You have 7 people with 6 seats. How
many different ways can you make?
(Hint: does order matter?)
There are n! permutations of an n-set.
Slid
e 934
Count the number of different 8-letter “words” that can be
formed using the letters in the word COMPUTER.
Slid
e 935
Count the number of different 8-letter “words” that can be
formed using the letters in the word COMPUTER.
Each permutation of the 8 letters forms a different word. There are 8! =
40,320 such permutations.
Slid
e 936
There are n ! distinguishable permutations of an n-set containing n
distinguishable objects.
If an n-set contains n objects of a first kind, n objects of a second
1
2
kind, and so on, with n  n  ...  n  n, then the number of
1
2
k
distinguishable permutations of the n-set is
n!
.
n !n !n ! n !
1
2
3
k
Slid
e 937
n
There are 2 subsets of a set with n objects (including the
empty set and the entire set).
Slid
e 938

http://www.mathsisfun.com/combinatori
cs/combinations-permutations.html

Read the different examples

P708 #5-42 eoe
Use the distributive property to expand the binomial.
1.  x  y 
2
2. (a  2b)
2
3. (2c  3d )
4. (2 x  y )
5.  x  y 
2
2
3
Slid
e 942
Use the distributive property to expand the binomial.
1.  x  y 
x  2 xy  y
2
2. (a  2b)
2
2
a  4ab  4b
2
2
2
3. (2c  3d ) 4c  12cd  9d
2
2
4. (2 x  y ) 4 x  4 xy  y
2
5.  x  y 
3
2
2
2
x  3x y  3 xy  y
3
2
2
3
Slide 9- 43

What is Pascal’s Triangle?

It is the coefficient for the expansion of
the binomials 𝑎 + 𝑏 𝑛
The binomial coefficients that appear in the expansion of ( a  b)
n
are the values of C for r  0,1, 2,3,..., n.
n
r
A classical notation for C , especially in the context of binomial
n
r
n
coefficients, is   . Both notations are read "n choose r."
r
Slid
e 945
 n   n  1  n  1
 r    r  1    r  or, equivalently, C  C  C
  
 

n
r
n 1
r 1
n 1
Slid
e 946
r
For any positive integer n,
n
n
n
 n
 a  b     a    a b  ...    a b  ...    b ,
0
1
r
 n
n
n!
where    C 
.
r !(n  r )!
r
n
n
n
n 1
nr
r
n
r
Slid
e 947
Expand  a  b  , using a calculator to compute the binomial coefficients.
4
Slid
e 948
Expand  a  b  , using a calculator to compute the binomial coefficients.
4
0,1, 2,3, 4 into the calculator to find the binomial
coefficients for n  4. The calculator returns the list 1,4,6,4,1 .
Enter 4 C
n
r
Using these coefficients, construct the expansion:
a  b
4
 a  4a b  6a b  4ab  b .
4
3
2
2
3
4
Slid
e 949

𝑥+4
5
Expand:
 4𝑥 + 𝑦 3

Expand:
 5𝑥 + 3𝑎

8

Coefficient of 𝑥 10 in the expansion of
𝑥 + 2 15

15𝐶10𝑥 10 25

Find the coefficient of 3𝑥
expansion of 3𝑥 + 4𝑦 7
4
in the

Find the coefficient of 𝑎5 in the
expansion of 𝑎 + 2𝑏 10
For any integer n  1, n !  n  n  1!
For any integer n  0,  n  1!   n  1 n !
Slid
e 957

𝑛
𝑛+1
−
=n
2
2
𝑛
 Remember 𝐶( )
𝑟


𝑛
𝑛+1
−
=
2!
2
2
𝑛+1 !
𝑛!
−
𝑛+1−2 !
2! 𝑛−2 !
𝑛+1 𝑛 𝑛−1 !
𝑛 𝑛−1 𝑛−2 !
−
2 𝑛−1 !
2 𝑛−2 !
=
𝑛2 +𝑛
2
2𝑛
2

=

=

=n
−
𝑛2 −𝑛
2

P715 #1-34 eoe

Probability is a numerical measure
between 0 and 1 that describes the
likelihood that an event will occur.
Probabilities closer to 1 indicate that the
event is more likely to occur.
Probabilities closer to 0 indicated that
the event is less likely to occur.

P(A) = probability of event A; you read it as
“P of A”.

P(A)=1, the event A is certain to occur

P(A)=0, the event A is certain to not occur
› Binary number works like this…1 means it’s true, 0
means false.

“There are only 10 types of people in the
world: those who understand binary, and
those who don't”


1) A probability assignment based on intuition
incorporates past experience, judgment, or
opinion to estimate the likelihood of an event
2) A probability assignment based on relative
frequency uses the formula
› Probability of event=relative frequency=
𝑓
𝑛
› Where f is the frequency of the event occurrence in
a sample of n observations

3) A probability assignment based on equally
likely outcomes uses the formula
› Probability of
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑡𝑜 𝑒𝑣𝑒𝑛𝑡
event=
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

Intuition – NBA announcer claims that Kobe
makes 84% of his free throws. Based on this, he
will have a high chance of making his next free
throw.

Relative frequency – Auto Fix claims that the
probability of Toyota breaking down is .10
based on a sample of 500 Toyota of which 50
broke down.

Equally likely outcome - You figure that if you
guess on a SAT test, the probability of getting it
right is .20

Create a situation for each of the
probability assignments. (intuition,
relative frequency, equally likely
outcome)

Show me

In the long run, as the sample size
increases, the relative frequencies of
outcomes get closer to the theoretical
(or actual) probability value

Example: The more numbers you ask, the
more likelihood that P(getting a girl’s real
number)=1

The more numbers you ask, the more
likelihood that P(getting a (hot) girl’s real
number)=1

Then after collecting all the numbers, the
more girls you ask out on a date, the
more likelihood that P(getting a date)=1
Casino (the more you play, the more you
lose)
 Insurance (the more people you insure,
the less the likelihood the company have
to pay for the insurance benefits)

Statistical experiment or statistical
observation can be thought of as any
random activity that results in a definite
outcome
 An event is a collection of one or more
outcomes of a statistical experiment or
observation
 Simple event is one particular outcome of a
statistical experiment
 The set of all simple events constitutes the
sample space of an experiment

Brown eyes’ genotype is Bb or BB
 Blue eyes’ genotype is bb


If your Dad has Brown eyes (and his dad
has blue eyes) and your Mom has blue
eyes, what’s the probability that you
have blue eyes?
Dad
Mom
B
b
b
Bb
bb
b
Bb
bb
P(blue eyes)=
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
2
1
=4=2

You are running out of time in a true/false quiz. You
only have 4 questions left! How should you guess?

P(all false)=
P(3 false)=

P(all true)=
P(2 false)=

P(1 true)=
P(1 false)=

P(2 true)=

P(3 true)=

Your sample space should have 16 different combinations
TTTT
FTTT
TFTT
TTFT
TTTF
FFTT
FTFT
FTTF
TFFT
TFTF
TTFF
FFFT
TFFF
FTFF
FFTF
FFFF

P(all false)= 1/16
P(3 false)= 4/16

P(all true)= 1/16
P(2 false)= 6/16

P(1 true)=
P(1 false)= 4/16

P(2 true)= 6/16

P(3 true)= 4/16

You will probably choose 2 true and 2 false
4/16

The sum of the probabilities of all simple
events in a sample space must equal 1

The complement of event A is the event
that A does not occur. 𝐴𝑐 designates the
complement of event A. Furthermore,
1) P(A)+P(𝐴𝑐 )= 1
 P(event A does not occur)=P 𝐴𝑐 = 1 − 𝑃(𝐴)


P(getting A in Mr. Liu’s class)+P(not
getting A in Mr. Liu’s class) =1

P(getting A in Mr. Liu’s class)=.15

What’s the P(not getting A in Mr. Liu’s
class)?

P(not getting A in Mr. Liu’s class)= .85

P(having a date on a Friday)=1/7

What’s the P(not having a date on a
Friday)?

6/7
How many outcomes are possible for the following experiments.
1. Two coins are tossed.
2. Two different 6-sided dice are rolled.
3. Two chips are drawn simultaneously without replacement from
a jar with 8 chips.
4. Two different cards are drawn from a standard deck of 52.
5. Evaluate without using a calculator.
4
8
C
C
2
2
Slid
e 981
How many outcomes are possible for the following experiments.
1. Two coins are tossed. 4
2. Two different 6-sided dice are rolled. 36
3. Two chips are drawn simultaneously without replacement from
a jar with 8 chips. 28
4. Two different cards are drawn from a standard deck of 52. 1326
5. Evaluate without using a calculator.
4
8
C
C
2
3/14
2
Slide 9- 82
If E is an event in a finite, nonempty sample space S of equally likely
outcomes, then the probability of the event E is
the number of outcomes in E
P( E ) 
.
the number of outcomes in S
Slid
e 983

P(5 on 1st die and 5 on 2nd die)

P(ace on 1st card and ace on 2nd card)

What is the difference between these
two situation?

In the first situation, the first result does
not effect the outcome of the 2nd result.

In the second situation, the first result
does effect the outcome of the 2nd
result.

Two events are independent if the
occurrence or nonoccurrence of one
does not change the probability that the
other will occur

What does it mean if two events are
dependent?

P(A and B)=𝑃 𝐴 ∗ 𝑃(𝐵)

This means event A AND event B both
have to happen!!! You multiply the
events. You find the probability of two
events happening together.

This is the formula if event A and event B
are independent.

Then we must take into account the
changes in the probability of one event
caused by the occurrence of the other
event.
Sample
Space
A and B
A
B
P(A and B)=𝑃 𝐴 ∗ 𝑃 𝐵 𝐴
 Or
 P(A and B)=𝑃 𝐵 ∗ 𝑃 𝐴 𝐵



It is known as conditional probability
𝑃 𝐴 𝑎𝑛𝑑 𝐵
𝑃 𝐵
𝑃 𝐴𝐵 =
given event B”
𝑃
𝐵𝐴
= “Probability of event A
𝑃(𝐴)

𝑃 𝐴𝐵 =

Quick group work:
𝑃(𝐵)
› What is P 𝐵 𝐴 ?

Your friend has 2 children. You learned
that she has a boy named Rick. What is
the probability that Rick’s sibling is a
boy?

Take a guess 
If you guessed ½ or 50%, that is incorrect.
 First: Think about all the possible outcomes

› S {BB, BG, GB, GG}
What is P(boy and boy)?
 What is P(boy)?


You want to find 𝑃 𝐵 𝐵
𝑃 𝑏𝑜𝑦 𝑎𝑛𝑑 𝑏𝑜𝑦
=
𝑃 𝑏𝑜𝑦
=
1
4
3
4
=
1
3

A machine produce parts that’s either
good (90%), slightly defective (2%) or
obliviously broken (8%). The parts gets
through an automatic inspection
machine that is able to find the oblivious
broken parts and throw them away.
What is the probability of the quality part
that make it through and get shipped?

P(Good given not broken)=
𝑃 𝐺𝑜𝑜𝑑 𝑎𝑛𝑑 𝑛𝑜𝑡 𝑏𝑟𝑜𝑘𝑒𝑛
.90
=
= .978 = 97.8%
𝑃(𝑛𝑜𝑡 𝑏𝑟𝑜𝑘𝑒𝑛)
.92

Conditional Probability can be very
intriguing and complicated. We won’t
go into any more in depth…..or
maybe….
Very important to understand about
probability is that are the events
dependent or independent.

Suppose you are going to throw 2 fair
dice. What is the probability of getting a
3 on each die?
A) Is this situation independent or
dependent?
 B) Create all the sample space (all the
potential outcomes)
 C) What is the probability?

A) Independent because one event
does not affect the second event
 B) You should have 36 total outcomes
 C) 1/36


I took a die away. Now you only have
ONE die! Again you toss the die twice.
What is the probability of getting a 1 on
the first and 4 on the second try?
It is still an independent event!
 1/36


The last two examples are considered
multiplication rule, independent events.

Mr. Liu has a 85% probability of teaching
statistics next year. Mr. Riley has a 15%
probability of teaching statistics next
year. What is the probability that both
Mr. Liu and Mr. Riley teach statistics next
year?

.85*.15=.1275 or 12.75% probability

Suppose you have 100 Iphones. The
defective rate of iphone is 10%. What is
the probability that you choose two
iphones and both are defective?
P(1st defective iphone)=10/100
 P(2nd defective iphone)=9/99


P(1st defective iphone and 2nd defective
1
1
1
iphone)= ∗ =
= .9%
10
11
110

What is the probability of getting tail and
getting a 3 on a die and getting an ace
in a deck of cards?
P(tail)=1/2
 P(3)=1/6
 P(ace)=4/52=1/13

1

2
1
6
∗ ∗
1
13
=
1
156

You use addition when you want to
consider the possibility of one event OR
another occurring

1) Satisfying the humanities requirement
by taking a course in the history of Japan
or by taking a course in classical
literature
 2) Buying new tires and aligning the tires
 3) Getting an A in math but also in
biology
 4) Having at least one of these pets: cat,
dog, bird, rabbit

1) or
 2) and
 3) and
 4) or


Two events are mutually exclusive or
disjoint if they cannot occur together. In
particular, events A and B are mutually
exclusive if P(A and B)=0

P(A or B)=P(A)+P(B)
General Rule for any events A and B
 P(A or B)=P(A)+P(B)-P(A and B)


Remember in mutually exclusive events
P(A and B)=0
Employee
type
Democrat
(D)
Republican
(R)
Independent
(I)
Row Total
Executive (E)
5
34
9
48
Production
Worker (PW)
63
21
8
92
Column Total
68
55
17
140 grand
total
a)
b)
c)
d)
e)
Compute P(D) and P(E)
Compute 𝑃 𝐷 𝐸
Are events D and E independent?
Compute P(D and E)
Compute P(D or E)


A) P(D)=68/140
B) 𝑃 𝐷 𝐸 =
P(E)=48/140
5
48
C) Determine if P(D)=𝑃 𝐷 𝐸 , they are not
the same! So they are not independent
 D) P(D and E)=5/140


E) P(D or
68
48
5
E)= +
−
140
140
140
=
111
140
A tree diagram shows all the possible
outcomes of an event.
 All possible outcomes of an event are
shown by a tree diagram.


If a coin and a dice are tossed
simultaneously, what is the probability of
getting tail and even number?

1/4

You are on a sports team. What is the
probability that out of three games, you
win two of them?

You have 7 balls, 4 are blue and 3 are
green. What is the probability that when
you pick the balls, you get green on 1st
and blue one 2nd?

You make free throws 85% of the time.
What is the probability of making at least
one out of the three?

P(make 1 out of 3)=99.66% of the time
Outcome
2
3
4
5
6
7
8
9
10
11
12
Probability
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
Slide
9- 125
Find the probability of rolling a sum divisible by 4 on a
single
roll of two fair dice.
Slide
9- 126
Find the probability of rolling a sum divisible by 4 on a
single
roll of two fair dice.
The event E consists of the outcomes 4,8,12 . To get the probability
of E we add up the probabilities of the outcomes in E:
3
5
1
9 1
P( E ) 
 

 .
36 36 36 36 4
Slide
9- 127
Dylan opens a box of a dozen chocolate cremes and offers three of them
to
Russell. Russell likes vanilla cremes the best, but all the chocolates look alike
on the outside. If five of the twelve cremes are vanilla, what is the
probability
that all of Russell’s picks are vanilla?
Slide
9- 128
Dylan opens a box of a dozen chocolate cremes and offers three of them
to
Russell. Russell likes vanilla cremes the best, but all the chocolates look alike
on the outside. If five of the twelve cremes are vanilla, what is the
probability
that all of Russell’s picks are vanilla?
The experiment in question is the selection of three chocolates,
without regard to order, from a box of 12. There are C  220
12
3
outcomes of this experiment. The event E consists of all possible
combinations of 3 that can be chosen, without regard to order, from
the 5 vanilla cremes available. There are C  10 ways.
5
3
Therefore, P( E )  10 / 220  1/ 22.
Slide
9- 129
Suppose an event A has probability p1
and an event B has probability p2 under
the assumption that A occurs. Then the
probability that both A and B occur is
p1p2.
Slide
9- 130
Dylan opens a box of a dozen chocolate cremes and offers three of them
to
Russell. Russell likes vanilla cremes the best, but all the chocolates look alike
on the outside. If five of the twelve cremes are vanilla, what is the
probability
that all of Russell’s picks are vanilla?
Slide
9- 131
Dylan opens a box of a dozen chocolate cremes and offers three of them
to
Russell. Russell likes vanilla cremes the best, but all the chocolates look alike
on the outside. If five of the twelve cremes are vanilla, what is the
probability
that all of Russell’s picks are vanilla?
The probability of picking a vanilla creme on the first draw is 5/12.
Under the assumption that a vanilla creme was selected in the first
draw, the probability of picking a vanilla creme on the second draw is
4/11. Under the assumption that a vanilla creme was selected in the first
and second draw, the probability of picking a vanilla creme on the third
draw is 3/10. By the Multiplication Principle, the probability of picking
5 4 3
60
1
a vanilla creme on all three picks is
  
 .
12 11 10 1320 22
Slide
9- 132
If the event B depends on the event A, then P( B | A) 
P( A and B)
.
P( A)
Slide
9- 133
Suppose an experiment consists of n-independent repetitions of an
experiment with two outcomes, called "success" and "failure." Let
P (success)  p and P (failure)  q. (Note that q  1  p.)
Then the terms in the binomial expansion of ( p  q ) give the respective
n
probabilities of exactly n, n  1,..., 2, 1, 0 successes.
Number of successes out of
Probability
n independent repetitions
n
n
p
n 1
 n 
 n  1 p q


1
0
n 1
n
 r  pq
 
q
n 1
n
Slide
9- 134
Suppose Tommy makes 92% of his free throws. If he shoots 15 free
throws, and if his chance of making each one is independent of
the other shots, what is the probability that he makes all 15?
Slide
9- 135
Suppose Tommy makes 92% of his free throws. If he shoots 15 free
throws, and if his chance of making each one is independent of
the other shots, what is the probability that he makes all 15?
P(15 successes)   0.92   0.286
15
Slid
e 9136
Suppose Tommy makes 92% of his free throws. If he shoots 15 free
throws, and if his chance of making each one is independent of
the other shots, what is the probability that he makes exactly 10?
Slid
e 9137
Suppose Tommy makes 92% of his free throws. If he shoots 15 free
throws, and if his chance of making each one is independent of
the other shots, what is the probability that he makes exactly 10?
 15 
P(10 successes)=    0.92   0.08   0.00427
10 
10
5
Slid
e 9138

Pg 728 #1-40 eoe
Evaluate each expression when a  3, r  2, n  4 and d  2.
1. a  (n  1)d
2. a  r
n 1
Find a .
10
k 1
k
4. a  2  3
3. a 
k
k 1
k
5. a  a  3 and a  10
k
k 1
9
Slide 9- 141
Evaluate each expression when a  3, r  2, n  4 and d  2.
1. a  (n  1)d 9
2. a  r
n 1
24
Find a .
10
k 1
k
4. a  2  3
3. a 
k
k 1
k
11
10
39,366
5. a  a  3 and a  10
k
k 1
9
13
Slide 9- 142
Let a  be a sequence of real numbers, and consider lim a .
n
n 
n
If the limit is a finite number L, the sequence converges and L
is the limit of the sequence. If the limit is infinite or nonexistent,
the sequence diverges.
Slid
e 9143
Determine whether the sequence converges or diverges. If it converges,
give the limit.
2 1 2
2
2,1, , , ,..., ,...
3 2 5
n
Slid
e 9144
Determine whether the sequence converges or diverges. If it converges,
give the limit.
2 1 2
2
2,1, , , ,..., ,...
3 2 5
n
lim
n
2
 0, so the sequence converges to a limit of 0.
n
Slid
e 9145

Converges or diverges
3𝑛

𝑛+1
5𝑛2
 3
𝑛 +1
𝑛3 +2
 2
𝑛 +𝑛

Converges to 3

Converges to 0

Diverges

Converges or diverges
1 1 1 1
1
 , , , ,…
1 2 3 4
𝑛
2 3 4 5
 , , , ,…
1 2 3 4
2,4,6,8,10, …
 -1, 1, -1, 1, -1, 1…


Converges to 0

Converges to 1

Diverges

Diverges
A sequence a  is an arithmetic sequence if it can be written in the
n
form a, a  d , a  2d ,..., a  (n  1)d ,... for some constant d .
The number d is called the common difference.
Each term in an arithmetic sequence can be obtained recursively from
its preceding term by adding d : a  a  d (for all n  2).
n
n 1
Slid
e 9150
Recursive formula:
 𝑎𝑛 = 𝑎𝑛−1 + 𝑑

Explicit formula:
 𝑎𝑛 = 𝑎1 + 𝑛 − 1 𝑑

Find (a) the common difference, (b) the tenth term, (c) a
recursive rule for the
nth term, and (d) an explicit rule for the nth term.
-2, 1, 4, 7, …
Slid
e 9152
Find (a) the common difference, (b) the tenth term, (c) a
recursive rule for the
nth term, and (d) an explicit rule for the nth term.
-2, 1, 4, 7, …
(a) The common difference is 3.
(b) a  2  (10  1)3  25
10
(c) a  2 a  a  3 for all n  2
1
n
n 1
(d) a  2  3( n  1)  3n  5
n
Slid
e 9153

6, 10, 14, 18, …

A) Find the common difference

B) the tenth term

C) a recursive rule for the nth term

D) Explicit rule for the nth term

23, 27, 31, …

A) Find the common difference

B) the tenth term

C) a recursive rule for the nth term

D) Explicit rule for the nth term
A sequence a  is a geometric sequence if it can be written in the
n
form a, a  r , a  r ,..., a  r ,... for some nonzero constant r.
2
n 1
The number r is called the common ratio.
Each term in a geometric sequence can be obtained recursively from
its preceding term by multiplying by r : a  a  r (for all n  2).
n
n 1
Slide 9- 156
Recursive:
 𝑎𝑛 = 𝑎𝑛−1 ∗ 𝑟

Explicit:
 𝑎𝑛 = 𝑎1 ∗ 𝑟 𝑛−1

Find (a) the common ratio, (b) the tenth term, (c) a recursive rule
for the
nth term, and (d) an explicit rule for the nth term.
2, 6, 18,…
Slid
e 9158
Find (a) the common ratio, (b) the tenth term, (c) a recursive rule
for the
nth term, and (d) an explicit rule for the nth term.
2, 6, 18,…
(a) The ratio is 3.
(b) a  2  3
10 1
10
 39,366
(c) a  2 and a  3a
1
n
n 1
for n  2.
(d) a  2  3 .
n 1
n
Slid
e 9159

3, 6, 12, 24, 48…

A) Find the common ratio

B) the tenth term

C) a recursive rule for the nth term

D) an Explicit rule for the nth term

1, 3, 9, 27…

A) Find the common ratio

B) the tenth term

C) a recursive rule for the nth term

D) an Explicit rule for the nth term

The second and fifth terms of a
sequence are 3 and 24.

Find the explicit and recursive formulas
for the sequence if

A) arithmetic

B) geometric

The fourth and seventh term of an
arithmetic sequence are -8 and 4.

Find the explicit rule and recursive rule
and the first term.

2nd and 8th term of a geometric
sequence are 3 and 192

Find the explicit rule, common ratio, first
term
The Fibonacci sequences can be defined recursively by
a 1
1
a 1
2
a  a a
n
n2
n 1
for all positive integers n  3.
Slid
e 9165

Pg 739 #1-32 eoo
a  is an arithmetic sequence. Use the given information to find a
n
10
.
1. a  5; d  4
1
2. a  5; d  2
3
a  is a geometric sequence. Use the given information to find a
n
10
.
3. a  5; r  4
1
4. a  5; r  4
3
5. Find the sum of the first 3 terms of the sequence n .
2
Slid
e 9168
a  is an arithmetic sequence. Use the given information to find a
n
10
1. a  5; d  4
41
2. a  5; d  2
19
1
3
a  is a geometric sequence. Use the given information to find a
n
10
.
.
3. a  5; r  4 1,310,720
1
4. a  5; r  4
3
81,920
5. Find the sum of the first 3 terms of the sequence n  . 14
2
Slide 9- 169
In summation notation, the sum of the terms of the sequence a , a ,..., a 
1
2
n
is denoted  a which is read "the sum of a from k  1 to n."
n
k 1
k
k
The variable k is called the index of summation.
Slid
e 9170
Let a , a ,..., a  be a finite arithmetic sequence with common difference d .
1
2
n
Then the sum of the terms of the sequence is
 a  a  a  ...  a
n
k 1
k
1
2
n
a a 
 n

 2 
n
  2a  (n  1)d 
2
1
n
1
Slid
e 9171
A corner section of a stadium has 6 seats along the front row. Each
successive row has 3 more seats than the row preceding it. If the top
row has 24 seats, how many seats are in the entire section?
Slid
e 9172
A corner section of a stadium has 6 seats along the front row. Each
successive row has 3 more seats than the row preceding it. If the top
row has 24 seats, how many seats are in the entire section?
The number of seats in the rows form an arithmetic sequence with
a  6, a  24, and d  3. Solving
1
n
a  a  (n  1)d
n
1
24  6  3(n  1)
n7
Apply the Sum of a Finite Sequence Theorem:
 6  24 
Sum of chairs  7 
  105. There are 105 seats in the section.
 2 
Slid
e 9173

Write each sum using summation
notation and find the sum

-7-1+5+11+……….53

Write each sum using summation
notation and find the sum

111, 108, 105, … 27

Write each sum using summation
notation and find the sum

5, 13, 21, …..45
Let a , a ,..., a  be a finite geometric sequence with common ratio r.
1
2
n
Then the sum of the terms of the sequence is
 a  a  a  ...  a
n
k 1
k
1

2
a 1  r
1
n
n

1 r
Slide 9- 177

4, -4/3, 4/9, -4/27, …4
1 10
−
3

1 11
4(1− −
)
3
1
1−(− )
3
≈ 3.000016935

Find the sum of the geometric
sequences.

3, 6, 12, ….12,288

Find the sum of the geometric
sequences.

5, 15, 45, …98,415

An infinite series is an expression of the form  a  a  a  ...  a  ...
k 1
k
1
2
n
Slid
e 9182

The geometric series  a  r
k 1
k 1
converges if and only if | r | 1.
a
If it does converge, the sum is
.
1 r
Slid
e 9183
Determine whether the series converges. If it converges, give the sum.
 2  0.25 

k 1
k 1
Slid
e 9184
Determine whether the series converges. If it converges, give the sum.
 2  0.25 

k 1
k 1
Since |r | 0.25  1, the series converges.
The sum is
a
2
8

 .
1  r 1  0.25 3
Slid
e 9185

Find the sum

5, 1, 1/5, ….

Find the sum

7, 7/5, 7/25, …

Find the sum of the first n terms of the
sequence. The sequence is either
arithmetic or geometric

6, -3, 3/2, -3/4, ….n=11

Find the sum of the first n terms of the
sequence. The sequence is either
arithmetic or geometric

2, 5, 8, …n=10

Find the sum of the first n terms of the
sequence. The sequence is either
arithmetic or geometric

-1, 11, -121, … n=9

Pg 749 #1-40 eoe
1. Expand the product k (k  2)(k  4).
Factor the polynomial.
2. n  7 n  10
2
3. n  3n  3n  1
3
2
x
.
x 1
5. Find f (t  1) given f ( x)  x  1.
4. Find f (t ) given f ( x) 
2
Slid
e 9193
1. Expand the product k (k  2)(k  4). k  6k  8k
3
2
Factor the polynomial.
 n  2  n  5
 3n  1  n  1
2. n  7 n  10
2
3. n  3n
3
2
3
x
t
.
x 1 t 1
5. Find f (t  1) given f ( x)  x  1. t  2t  2
4. Find f (t ) given f ( x) 
2
2
Slid
e 9194
The Tower of Hanoi Problem
 Principle of Mathematical Induction
 Induction and Deduction

… and why
The principle of mathematical induction is a
valuable technique for proving combinatorial
formulas.
Slid
e 9195
The minimum number of moves required
to move a stack of n washers in a Tower
of Hanoi game is 2n – 1.
Slid
e 9196
Let Pn be a statement about the integer n.
Then
Pn is true for all positive integers n provided
the
following conditions are satisfied:
1. (the anchor) P1 is true;
2. (inductive step) if Pk is true, then Pk+1 is
true.
Slid
e 9197
Bars can be vertical or horizontal.
 Bars are of uniform width and uniformly
spaced.
 The lengths of the bars represent values of
the variable being displayed, the frequency
of occurrence, or the percentage of
occurrence. The same measurement scale
is used for the length of each bar.
 The graph is well annotated with title, lables
of each bar, and vertical scale or actual
value for the length of each bar.

Age
Life Expectancy
82
80
78
76
74
72
70
68
66
64
Men
Women
1980
1990
2000
Year of birth
2010

Look at the number where y-axis started.

You might see the graph with squiggle
on the changed axis. Sometimes, if a
single bar is unusually long, the bar
length is compressed with a squiggle in
the bar itself. (look at pg 51 example 211b with the graph)
Life Expectancy
Year of birth
2010
2000
Women
1990
Men
1980
60
65
70
75
Age
80
85
# of families
Survey of the type of car families own
9
8
7
6
5
4
3
2
1
0
Amount
Toyota
Nissan
Haundai
Type of car
Ford

Use the info below to create a bar graph.

Average annual income (in thousands) of a
household headed by a person with the
stated education level is as follows: 16.1 for
highschool, 34.1 for highschool graduates,
48.6 for associated degrees, 62.1 for
bachelor’s degrees, 71.0 for master’s
degrees and 84.1 for doctoral degrees

What can you conclude?

Pareto chart is a bar graph in which the
bar height represents frequency of an
event. In addition, the bars are
arranged from left to right according to
decreasing height.
Consider this situation:
Causes for lack of sleep(two month study
61 days)

Cause
Frequency
Playing x-box or ps3
14
Texting
9
Watching movie/TV
5
Talking on the phone
10
Doing homework/project
20
Other
3
Frequency
Frequency
20
14
10
9
5
3

Use the info below to create a pareto
chart.

Here are a list of the most common
stolen items per 100000 cases: 10.1
electronics; 15.6 jewelries; 7.3 cars; 20.4
cash; 26.7 identity

What can you conclude?

Circle graph or pie chart, wedges of a
circle visually display proportional parts
of the total population that share a
common characteristic.
Consider the situation:
Monthly Financial Budget (based on $4000
monthly)

Categories
Amount spent Fraction
Percentag Degree of the pie
e
Food
800
800/4000
0.2
.2*360°=72°
Investment
500
500/4000
0.125
.125*360°=45°
Bills/debt
1750
1750/4000
.4375
.4375*360°=157.5°
Rent
950
950/4000
.2375
.2375*360°=85.5°
Monthly Financial Budget ($4000)
Rent
24%
Food
20%
Investment
12%
Bills/debts
44%

Is the chart consistent with our data?

Create a circle graph with the following
info:

Gamestop took a survey on the first 500
customers to see what genre of games
they bought. 70 Fighting, 123 shooter,
150 action-adventure,53 role-playing, 12
strategy, 92 others.

What can you conclude?
Cause of Death
 Heart Disease 700,142
 Cancer 553,768
 Stroke 163,538
 Other 1,018,977


Create a circle graph, and bar graph

Stem-and-leaf is a method of exploratory
data analysis that is used to rank-order
and arrange data into groups.

Similarity: Both display frequency
distributions

Difference: In histogram, we lose most of
the specific data values (because of
intervals). Stem-and-leaf display is a
device that organizes and groups data
but allow us to recover the original data
if desired.

Write out all the numbers

Put this chart into a stem-and-leaf
display
30
27
12
42
35
47
38
36
27
35
22
17
29
3
21
0
38
32
41
33
26
45
18
43
18
32
31
32
19
21
33
31
28
29
51
12
32
18
21
26

A frequency table partitions data into
intervals and shows how many data
values are in each interval. The intervals
are constructed so that each data value
falls into exactly one interval.

Note: intervals are known as classes. The
book uses the word “classes”, but I use
“intervals” because it makes more sense.

Consider this situation: You are collecting
how many minutes each student study
for a particular class. You interviewed 50
students and here is the chart.
15
47
10
5
30
20
18
40
3
1
12
45
1
8
4
15
10
19
25
13
34
16
17
7
16
17
13
10
17
35
7
6
25
27
18
31
4
12
48
14
6
2
14
13
7
15
46
12
9
18

1) Determine how many intervals you want.
› between 5-15 is usually preferred
› Anything less than 5, you risk losing information
› Anything more than 15, data might not be
sufficiently analyzed
Let’s use 6 intervals for this case. (remember
you can any number between 5 and 15)
 With this, you can find the width of each
interval.

𝐿𝑎𝑟𝑔𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒−𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒
 Interval width=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠
48−1
47
 So in our case:
= ≈ 7.8 = 8
6
6
› Note: You always round to the next whole
number, even if the number is 2.3. 2.3 would
become 3

So in each interval, it will include 8
numbers and this tells you the limit of
each interval

The lower interval limit is the lowest data
value that can fit in an interval.

The upper interval limit is the highest data
value that can fit in an interval.

The interval width is the difference between
the lower class limit of one interval and the
lower class limit of the next interval.
› In our case, our lowest number is 1, so 1+8=9,
therefore, 9 would be the start of the next
interval (remember we will have 6 intervals total)

Find the starting number of each interval
Start of 1st interval=1
 Start of 2nd interval=9
 Start of 3rd interval=17
 Start of 4th interval=25
 Start of 5th interval=33
 Start of 6th interval=41
 Start of 7th interval=49

Lower interval
limit
Upper interval
limit
1
8
9
16
17
24
25
32
33
40
41
48

Tally is the mark that is used to count the
amount of numbers that lies in each
interval.

Frequency (represented by 𝑓) is the
number of tally marks corresponding to
that interval

Now tally up all the numbers that fall in
each interval. Find the frequency also
Lower interval
limit
Upper interval
limit
Tally
Frequency 𝒇
1
8
13
9
16
17
17
24
8
25
32
5
33
40
3
41
48
4
Midpoint=
𝐿𝑜𝑤𝑒𝑟 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑙𝑖𝑚𝑖𝑡 + 𝑢𝑝𝑝𝑒𝑟 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑙𝑖𝑚𝑖𝑡
2


Find the midpoint of each interval
Lower interval limit
Upper interval limit
Interval midpoint
1
8
4.5
9
16
12.5
17
24
20.5
25
32
28.5
33
40
36.5
41
48
44.5

Upper interval boundaries, add 0.5 to the
upper interval limit.

Lower interval boundaries, subtract 0.5
from the lower interval limits.

Find the interval boundaries for all
interval.
Lower interval
limit
Upper interval
limit
Lower interval
boundary
Upper interval
boundary
1
8
0.5
8.5
9
16
8.5
16.5
17
24
16.5
24.5
25
32
24.5
32.5
33
40
32.5
40.5
41
48
40.5
48.5


Relative Frequency shows the probability
of data values that falls in each interval
Relative
𝑓
frequency=
𝑛
=
𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑇𝑜𝑡𝑎𝑙

Find the relative frequency of each
interval
Frequency 𝒇
13
Relative frequency
𝒇
𝒏
13/50=0.26
18
18/50=0.36
7
7/50=0.14
5
5/50=0.1
3
3/50=0.06
4
4/50=0.08
This is a data represent glucose blood level
after 12 hour fast for a random sample of
70 women. Use 6 intervals (classes)
45
66
83
71
76
64
59
59
76
82
80
81
85
77
82
90
87
72
79
69
83
71
87
69
81
76
96
83
67
94
101
94
89
94
73
99
93
85
83
80
78
80
85
83
84
74
81
70
65
89
70
80
84
77
65
46
80
70
75
45
101
71
109
73
73
80
72
81
63
74
Lower
interval
limit
Upper
interval
limit
Lower
interval
boundar
y
Upper
Tally
interval
boundar
y
Frequenc
y𝒇
45
55
44.5
55.5
3
Interval Relative
midpoi frequenc
nt
y
𝒇
𝒏
50
0.04
56
66
55.5
66.5
7
61
0.1
67
77
66.5
77.5
22
72
.31
78
88
77.5
88.5
26
83
.37
89
99
88.5
99.5
9
94
.13
100
110
99.5
110.5
3
105
.04

You use the frequency table to graph a
histogram (use the example we did
together in class about study minutes
with 50 students)
You use lower/upper interval boundaries
for the x axis because you don’t want
any gaps.
 Let’s graph both frequency histogram
and relative-frequency histogram

Frequency
Frequency f
20
15
10
5
0
Minutes

Time-series graph, data are plotted in
order of occurrence at regular intervals
over a period of time

Consider this situation:
Points Scored in a game (49er 2012)
Wee
k
1
2
3
4
5
6
7
8
9
Point
s
30
27
13
34
45
3
13
24
0
Wee
k
10
11
12
13
14
15
16
17
Point
s
24
32
31
13
27
41
13
27
Points for
Points Scored
50
40
30
20
Points for
10
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Weeks

Is there a pattern? Is there anything you
can conclude?
Create a time-series graph from the
following data
 What can you conclude?

Week
1
2
3
4
5
6
7
8
9
Distanc
e
1.5
1.4
1.7
1.6
1.9
2.0
1.8
2.0
1.9
Week
10
11
12
13
14
15
16
17
18
Distanc
e
2.0
2.1
2.1
2.3
2.3
2.2
2.4
2.5
2.6
Bar graphs are useful for quantitative or
qualitative data.
 Pareto Charts identify the frequency of
events or categories in decreasing order of
frequency of occurrence.
 Circle graph display how a total is dispersed
into several categories.
 Time-series graph display how data change
over time.
Note: Make sure you provide title, label axes
and identify units of measure in all type of
graphs!!


Think back to middle school and high
school (before taking Mr. Liu’s awesome
statistics class). What do you think of
when you heard of statistics?

The 3 M’s. Mean, median and mode

Mode is the value that occurs most
frequently

25, 37, 12, 27, 69, 64, 44, 30, 50, 52, 65, 25,
10, 12, 60, 27, 12, 65, 25

12 and 25

What is the mode out of this data set?

537244248343486245345987
1

4

Median can be represented as 𝑥 “x
tilde”. Median is the central value of an
ordered distribution (the exact middle)
1) Order the data from smallest to largest
 2) For an odd number of values in the
distribution, Median= Middle data value
 3) For an even number of data values in
the distribution, 𝑀𝑒𝑑𝑖𝑎𝑛 =

𝑆𝑢𝑚 𝑜𝑓 𝑚𝑖𝑑𝑑𝑙𝑒 𝑡𝑤𝑜 𝑣𝑎𝑙𝑢𝑒𝑠
2
Find the Median
 19 20 25 36 48 35 21 18


18 19 20 21 25 35 36 48

Because it is even, I add (21+25)/2 = 23

So median = 23
Find the median
 36, 4, 69, 47, 81, 84, 6, 38, 17, 93, 46


Median uses the position rather than the
specific value of each data entry. If the
extreme values of a data set change,
the median usually does not change.

It is often used as the average for house
prices.

For an ordered data set of size n,

𝑛+1
2
𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑖𝑑𝑑𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 =

If n = 77, then the middle value is
(77+1)/2= 39th number.

If n=90, then the middle value is
(90+1)/2=45.5th value. Which is in
between 45th and 46th number.
𝑠𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑒𝑛𝑡𝑟𝑖𝑒𝑠
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑛𝑡𝑟𝑖𝑒𝑠

𝑀𝑒𝑎𝑛 =

Mean means average.

It is aka arithmetic mean

Find the mean: 58, 92, 50, 68, 80, 64

Mr. Liu is a Paladin Tank that’s low in HP.
He only has 2 hits left before the boss
unleash his ultra super attack and pwns
Mr. Liu. The boss has a total of 750 HP
and Mr. Liu has damaged the boss with
the following damage points: 98, 50, 65,
71, 120, 84, 93, 27. What is the mean
damage of the next 2 hits so that Mr. Liu
survives?

Many ways to solve this… depend on
how you think about it.

The mean is 71 damage points for the
next 2 hits.


(𝑠𝑖𝑔𝑚𝑎) 𝑚𝑒𝑎𝑛𝑠 𝑠𝑢𝑚
𝑥 𝑚𝑒𝑎𝑛𝑠 𝑠𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑥, 𝑜𝑟 𝑎𝑙𝑙 𝑑𝑎𝑡𝑎 𝑣𝑎𝑙𝑢𝑒𝑠
Sample statisic 𝒙 (x bar)
𝑥
𝑥=
𝑛
n=number of data values in
the sample
𝑥 = 𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛
Population parameter 𝝁 (Mu)
𝑥
𝜇=
𝑁
N=number of data values in
the population
𝜇 = 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑚𝑒𝑎𝑛

Resistant measure is one that is not
influenced by extremely high or low data
values.

Which of the following data has more of
a resistant measure, mean or median?

Median, although the disadvantage of
the median is that it is not sensitive to the
specific size of a data value.

Mean is too sensitive to the data value.

It is called 10% trimmed mean

Procedures:
› 1) Order the data from smallest to largest
› 2) Delete the bottom 5% or the data and the top
5% of the data. Note: If the calculation of 5% of
the number of data values does not produce a
whole number, round down to the nearest
integer
 Ex: if you get .75, you round to 0
 Ex: if you get 1.95, you round to 1
› 3) Compute the mean of the remaining 90% of
the data







Here are some sample class sizes for introductory
lecture courses
14, 30, 45, 11, 20, 80, 65, 34, 18, 60, 100, 42, 40, 40, 78,
31, 52, 35, 120, 19
A) compute the mean
B) compute the 10% trimmed mean
C) Find the median
D) Find the median of the 10% trimmed data set.
Does the median change when you trim the data?
E) Is the trimmed mean or the original mean closer to
the median?
A) 46.7
 B) 44.61
 C) 40
 D)40
 E) Trimmed mean is closer to the median


How old are basketball players? Here are some of
the ages: 40, 38, 21, 18, 24, 23, 28, 25, 21, 27, 26, 25,
31, 32, 19, 29, 33, 24, 23, 35, 24, 23, 25, 32, 39, 26, 31,
28, 30, 31, 22, 23, 24, 25, 26, 27, 38, 27, 26, 22

A) compute the mean
B) compute the 10% trimmed mean
C) Find the median
D) Find the median of the 10% trimmed data set.
Does the median change when you trim the data?
E) Is the trimmed mean or the original mean closer to
the median?




A) 27.275
 B) 27.083
 C) 26
 D) 26
 E) Closer


This deal with my class especially!!!

𝑊𝑒𝑖𝑔ℎ𝑡𝑒𝑑 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =

w= weight
𝑥𝑤
𝑤

If your midterm is worth 40% of your total
grade and your final is worth 60% of your
total grade, you got 70 on your midterm
and 90 on your final. What is your final
grade?
𝑊𝑒𝑖𝑔ℎ𝑡𝑒𝑑 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
82
 you would have a B

.40 70 +.60(90)
.40+.60
=
28+54
1
=






Test/Quizzes
Participation/bell work
Notebook
Projects
Homework
Finals
30 %
10 %
15 %
20 %
15 %
10 %






You scored 85% on test/quizzes
You scored 60% on Homeworks
You scored 100% on participation/bell work
You scored 80% on projects
You scored 98% on notebook
You scored 50% on finals

What is your grade for the class?

.85(.30)+.60(.15)+1(.10)+.80(.20)+.98(.15)+
.50(.10)=.255+.09+.10+.16+.147+.05=.802
or 80.2% which is a B-
These are a measure of the distribution or spread of data
around an expected mean (𝑒𝑖𝑡ℎ𝑒𝑟 𝜇 𝑜𝑟 𝑥)
Variance is a measure of how far a set of numbers is
spread out.
 A variance of zero indicates that all the values are
identical.
 A non-zero variance is always positive:
› a small variance indicates that the data points tend to be
very close to the mean (expected value) and hence to
each other.
› a high variance indicates that the data points are very
spread out from the mean and from each other.
Standard Deviation shows how much variation
or dispersion from the average (mean, also called

Are you prepared?

Note: I will be covering both sample
statistics and population parameters


Sample mean=𝑥 =
𝑥
𝑛
Sample variance=𝑠 2 =
𝑥−𝑥 2
𝑛−1
𝑥−𝑥 2
𝑛−1

Sample standard deviation=𝑠 =

Where x is a member of the data set, 𝑥 is
the mean, and n is the number of data
Random Sample
 5 7 8 12 15 15


Find the variance and standard
deviation
𝑥
𝑥−𝒙
𝒙−𝒙
𝟐
4
4-10=-6
−6
7
7-10=-3
−3
2
=9
7
7-10=-3
−3
2
=9
11
11-10=1
1
15
15-10=5
5
2
= 25
16
16-10=6
6
2
= 36
𝑥 = 60
𝑥 𝟔𝟎
𝒙=
=
= 𝟏𝟎
𝐧
𝟔
2
2
=1
𝑥−𝑥
𝑠2
=
𝑥−𝑥 2 𝟏𝟏𝟔
= 𝟓
𝑛−1
𝑠=
= 36
2
= 116
= 𝟐𝟑. 𝟐
𝑥−𝑥
𝑛−1
2
= 23.2 ≈ 4.82
Random Sample
5 5 5 6 7 8


Find the variance and standard
deviation
𝑠 2 = 1.6
 𝑠 ≈ 1.26


Population mean=𝜇 =
𝑥
𝑁
𝑥−𝜇 2
𝑁

Population Variance=𝜎 2 =

Population standard deviation=𝜎 =
𝑥−𝜇 2
𝑁

Where N is the number of data values in

Samples are usually in “english
alphabets”

Populations are usually in “greek
alphabets”

Samples are in lower case

Population are in upper case

Have you guys taken your SAT yet?
Remember how you look at the score
and it says your “percentile”?

If you are in the 70th percentile, it means
that you scored the same or better than
70% of the people who took the SAT.
30% Scored the same or above you.

For whole numbers P (where 1 ≤ 𝑃 ≤
99), the Pth percentile of a distribution is
a value such that P% of the data fall at
or below it and (100-P)% of the data fall
at or above it.

It gives indication of “relative” position.
Basically, it tells you how you scored
against others that took the same test.

A) If your score is at the 27th percentile,
what percentage of scores are at or
below yours?

B) If the score ranged from 1 to 100 and
your score is 90%, does that necessarily
mean that your score is at the 90th
percentile?

A) 27%

B) No because percentile gives an
indication of relative position of the
scores.

Median is an average computed by
using relative position of the data.
Median is an example of a percentile; it
is the 50th percentile.

What is a Quartiles? (Look at the word
itself, what does it sound like?)

Quartile sounds like “quarters” so it has 4
intervals
1) Order the data from smallest to largest
 2) Find the median (Q2). This is the second
quartile (Q2)
 3) The first quartile Q1 is then the median of
the lower half of the data; that is, it is the
median of the data falling below the Q2
position (and not including Q2)
 4) The third quartile Q3 is the median of the
upper half of the data; that is, it is the
median of the data falling above the Q2
position (and not including Q2)

16
17
18
20
25
33
33
35
37
38
40
47
50
50
50
63
84
97
97
99
100
103
107
107
107
108
123

1) Where is the median (Q2)? Since there are 27
data, It is at the14th number, which is 50

2) Where is Q1? Since Q2 is at 14th spot, there are 13
numbers below. The median between Q1 and Q2 is
at the 7th spot, which is 33

3) Where is Q3? Since Q2 is at the 14th spot, there are
13 numbers above it. The median between Q2 and
Q3 is at the 21th spot, which is 100

4) Interquartile is Q3-Q1, which is 100-33=67
7
15
23
6
12
35
30
14
6
21
25
39
5
16
9
25
21
7
12
10
31
Q1=8
 Q2=15
 Q3=25
 IQR=17

540
570
520
550
490
430
490
490
570
380
510
640
310
400
380
390
410
340
390
350
330
350
Q1=380
 Q2=420
 Q3=520
 IQR=140


Quartiles together with the low and high
data values give us a very useful fivenumber (no…not $5 foot long) summary
of the data and their spread.
Max
 Q3
 Q2
 Q1
 Min


The longer the whisker is, the more
skewed it is toward that value.

If the median is closer to the lower part
of the box, it means that data are more
concentrated toward the lower end.

If the median is closer to the upper part
of the box, it means that data are more
concentrated toward the upper end.
1) Draw a vertical/horizontal scale to
include the lowest and highest data values.
 2) To the right/top of the scale, draw a box
from Q1 to Q3
 3) Include a solid line through the box at
the median level
 4) Draw vertical/horizontal lines, called
whiskers, from Q1 to the lowest value and
from Q3 to the highest value


(Look at previous slide for examples)
Lower outlier = 𝑄1 − 1.5 𝑥 𝐼𝑄𝑅
 Upper outlier = 𝑄3 + 1.5 𝑥 (𝐼𝑄𝑅)


SAT example:
Q1=380
 Q2=420
 Q3=520
 IQR=140

Lower Outlier: 380 − 1.5 𝑥 140 = 170
 Upper Outlier: 520 + 1.5 𝑥 140 = 730


Points Mr. Liu scored in basketball games:
Q1=8
 Q2=15
 Q3=25
 IQR=17

Q1=8
 Q2=15
 Q3=25
 IQR=17

Lower Outlier: 8 − 1.5 17 = −17.5
 Upper Outlier: 25 + 1.5 17 = 50.5


Mr. Liu’s points never goes below -17.5 or

65 72 68 4 64 60 55 73 71 52 63 61
75 62 89 64
Pg 769 #1-24 eoe
 Pg 782 #1-39 eoo
