Stats Review
Chapters 5-6
Created by Teri Johnson
Math Coordinator, Mary Stangler Center for Academic Success
Examples are taken from Statistics 4 E by Michael Sullivan, III
And the corresponding Test Generator from Pearson
Revised 12/13
Note:
This review is composed of questions the textbook
and the test generator. This review is meant to
highlight basic concepts from the course. It does
not cover all concepts presented by your instructor.
Refer back to your notes, unit objectives, handouts,
etc. to further prepare for your exam. A copy of
this review can be found at www.sctcc.edu/cas.
The final answers are displayed in red and the
chapter/section number is the corner.
Probability Rules
1) P(event) is greater or equal to 0 and less than or
equal to 1.
2) The sum of the probabilities must equal 1.
Is the table a Probability Model?
x
P(x)
x
P(x)
x
P(x)
1
.2
A
.3
Red
.2
2
-.3
B
.2
Green
.2
3
.6
C
.5
Blue
.3
4
.1
D
0
Yellow
.2
No, cannot have
negative
probabilities
Yes
No, the sum of the
probabilities
doesn’t equal 1
5.1
Probability
A survey of 971 investors asked how often they tracked
their portfolio. The table shows the investor responses.
What is the probability that an investor tracks his or her
portfolio daily?
How Frequently
Response
Daily
236
Weekly
261
Monthly
273
Couple times a year
141
Don’t Track
60
Step 1: Find the total of responses = 971
Step 2: Take the amount of daily responses and divide by the total
236/971=0.243
5.1
Addition Rules and Disjoint Events
• What are disjoint events?
– AKA mutually exclusive
– Events that have no outcomes in common
• Addition Rule for Disjoint Events
– If E and F are disjoint events, then
P(E or F)=P(E)+ P(F)
• General Addition Rule
– P(E or F)=P(E)+ P(F)- P(E and F)
5.2
Addition Rules and Disjoint Events
• In the game of craps, two dice are tossed and the up
faces are totaled. Is the event getting a total of 9 and
one of the dice showing a 6 disjoint events? Answer
Yes or No.
– No because you can get a sum of 9 when of the dice is
showing a 6 (the other would be a 3)
• A card is drawn from a standard deck of 52 playing
cards. Find the probability that the card is a queen or a
club. Express the probability as a simplified fraction.
P(Queen or club)=P(Queen)+P(club)-P(Queen and a club)
P(Queen or club)= 4/52 + 13/52 – 1/52
P(Queen or club)= 16/52=4/13
5.2
Compliment Rule
Rooms
Probability
1
.005
2
.011
3
.088
4
.183
5
.23
6
.204
7
.123
8 or More
.156
P(not E)=1-P(E)
or
P(EC)=1-P(E)
• What Is the probability of having less
than 8 rooms?
𝑃 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 8 = 1 − 𝑃 8 𝑜𝑟 𝑚𝑜𝑟𝑒
= 1 − .156 = .844
• What is the probability of having at least
3 rooms?
𝑃 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 3
= 1 − 𝑃 1 𝑟𝑜𝑜𝑚 − 𝑃 2 𝑟𝑜𝑜𝑚𝑠
= 1 − .005 − .011 = .984
5.2
Independence and Multiplication Rule
• When are 2 events independent?
– If the occurrence of event E in a probability
experiment does not affect the probability of
event F
• Multiplication Rule for independent Events
– Events E and F are independent if
P(E and F) =P(E)*P(F)
5.3
Independence and Multiplication Rule
There are 30 chocolates in a box, all identically shaped.
There are 11 filled with nuts, 10 filled with caramel, and 9
are solid chocolate. You randomly select one piece, eat it,
and then select a second piece. Is this an example of
independence? Answer Yes or No.
– No
A single die is rolled twice. Find the probability of getting
a 2 the first time and a 2 the second time. Express the
probability as a simplified fraction.
– These events are independent
P(2 on the first and 2 on second)=P(2 on first roll)(2 on
second roll)=(1/6)*(1/6)=1/16
5.3
Conditional Probability
• P(F|E) is read probability of event F given that
event E has occurred.
• Conditional Probability Rule
–P FE =
P(E and F)
P(E)
=
N(E and F)
N(E)
• General Multiplication Rule
– P(E and F)=P(E)*P(F|E)
5.4
Conditional Probability
Using the given table, given that the car selected was a domestic
car, what is the probability that it was older than 2 years?
Age of car (in years)
Make
0-2
3-5
6-10
Over 10
Total
Foreign
40
30
10
20
100
Domestic
45
27
11
17
100
Total
85
57
21
37
200
N(older than 2 years and Domestic)
P older than 2 𝑦𝑒𝑎𝑟𝑠 𝐷𝑜𝑚𝑒𝑠𝑡𝑖𝑐 =
N(Domestic)
27 + 11 + 17 11
=
=
100
20
5.4
General Multiplication Rule
There are 36 chocolates in a box, all identically shaped.
There 10 are filled with nuts, 12 with caramel, and 14
are solid chocolate. You randomly select one piece, eat
it, and then select a second piece. Find the probability
of selecting solid chocolate then a nut.
P(solid and nut)=P(solid)*P(nut|solid)=
14 10 1
∗
=
36 35 9
5.4
Counting Techniques: Combination
The number of different arrangements of r objects chosen from
in which
1. The n objects are distinct
2. Repetition of objects is not allowed
3. Order is not important
Formula: nCr=
𝑛!
𝑟! 𝑛−𝑟 !
From 9 names on a ballot, a committee of 3 will be elected to
attend a political national convention. How many different
committees are possible?
n=9, r=3
9C3=
9!
3! 9−3 !
= 84
5.5
Counting Techniques: Permutations
• Permutation of Distinct items with replacement
– The selection of r objects from n distinct objects, repetition is
allowed, and order matters
– nr
• Permutation of Distinct items without replacement
– The selection of r objects from n distinct objects, no repetition,
and order matters
– nPr=
𝑛!
𝑛−𝑟 !
• Permutation of nondistinct items without replacement
– The number of ways n objects can be arranged (order matters)
in which there are n1 of one kind, n2 of a second kind and so on
where n=n1+n2+…+nk
–
𝑛!
𝑛1 !𝑛2 !…𝑛𝑘 !
5.5
Counting Techniques
• Outside a home there is a keypad that will open the garage if the
correct 4-digit code is entered. How many codes are possible?
– Permutation of Distinct items with replacement; 104= 10,000
• Suppose 40 cars start at the Indy 500. In how many ways can the
top 3 cars finish?
– Permutation of Distinct items without replacement; 40P3= 59280
• A family has 6 children. If this family has exactly 2 boys, how many
different birth/gender orders are possible?
– Combination; 6C2= 15
• A baseball team consists of 3 outfielders, 4 infielders, a pitcher, and
a catcher. Assuming that the outfielders and infielders are
indistinguishable, how many batting orders are possible?
– Permutation of no distinct items without replacement;
9!
3!4!1!1!
= 2520
5.5
Determine the Appropriate Counting
Technique to Use
Copyright © 2013 Pearson Higher Ed
5.5
Determine the Appropriate Probability
to Rule to Use
5-17
Copyright © 2013 Pearson Higher Ed
5.6
Mean and Standard Deviation of Discrete
Probability Model
In a sandwich shop, the following probability distribution was obtained. The random
variable x represents the number of condiments used for a hamburger. Find the mean
and standard deviation for the random variable x.
x
P(x)
xP(x)
x2P(x)
0
.3
0*.3=0
02*.3=0
1
.4
.4
.4
2
.2
.4
.8
3
.06
.18
.54
4
.04
.16
.64
6.1
Expected Value = Mean
1) Multiply x by P(x)
2) Add the numbers from step 1
0+.4+.4+.18+.16=1.14. This is the mean.
Standard Deviation
1) Find the mean and square it,
=1.14 2=1.2996
2) Take x2 and multiply by P(x)
3) Add the values from step 2
= 2.38
4) Take the number from step 3 and subtract from
value from step 1: 2.38-1.2996=1.0804
5) Take the square root of the number from step 4,
1.04. This is the standard deviation.
Binomial Probability
Criteria for Binomial
1)
2)
3)
4)
Fixed number of trials (times done)
Trials are independent
Outcomes: Success or Failure
The probability of a success is the same
Decide whether the experiment is a binomial experiment. If it is not, explain
why.
1) You draw a marble 350 times from a bag with three colors of marbles. The
random variable represents the color of marble that is drawn.
No, more than two outcomes
2) Selecting five cards, one at a time without replacement, from a standard
deck of cards. The random variable is the number of picture cards obtained.
No, Trials are not independent
3) Survey 50 college students see whether they are enrolled as a new student.
The random variable represents the number of students enrolled as new
students.
Yes
6.2
Binomial Probability
Assume that male and female births are
equally likely and that the births are
independent.
Find the probability that there are exactly 9 girls
out of 10 births
𝑃 𝑋 = 9 = 10𝐶9 .5 9 (1 − .5)10−9 = .0098
Find the probability that there is at least 1 girl
out of 10 births
𝑃 𝑥 ≥ 1 = 1 − 𝑃 0 = .999
6.2
Binomial Mean and Standard Deviation
In 2000, it was reported that 11% of adult
Americans had trust in the federal government’s
handling of domestic issues. Suppose a random
sample of 1100 finds that 84 have a trust in the
government’s handling of domestic issues. Would
these results be considered unusual?
1)
2)
Find the Mean: (np) =1100*.11=121
Find the standard deviation: 𝑛𝑝(1 − 𝑝) =
1100(.11)(1 − .11) =10.38
3) For an event to be unusual, it must be outside the mean
plus/minus two standard deviations
121-2(10.28)=100.44; 121+2(10.28)=141.56
Since 84 is outside this interval, the event is unusual.
6.2
Poisson Distribution
• A stats professor finds that when he schedules
an office hour at the 10:30 a.m. time slot, an
average of three students arrive. Use the
Poisson distribution to find the probability
that in a randomly selected office hour in the
10:30 a.m. time slot exactly seven students
will arrive.
– x=7, λ=3, t=1
–𝑃 𝑥=7 =
(3∙1)7 −3∙1
𝑒
7!
=.0216
6.3