# MAT 106 LU Algebraic Manipulation with Plug in Some Values Paper

Description

Please read and review pages 38-41. Complete the homework, pg. 42. Please show all work. I attach the textbook below

1 attachmentsSlide 1 of 1attachment_1attachment_1

Unformatted Attachment Preview

Department of Mathematical Sciences
MAT 106
Math for Liberal Arts
Math workbook
(A pilot project on Mat 106 course manual)
Published with funding from a Lincoln University faculty
development grant
Fall 2018
Authors
Dr. Pathak and Dr. Xu
Math for Liberal Arts
MAT 106 course manual
Department of Mathematical Sciences
Lincoln University
Fall 2018
Student’s Name :
Student ID :
Instructor :
• Show work for full credit.
• Points will be given to the correct steps leading to the correct answer.
• All homework assignments should be done in the provided space.
• Calculator use is allowed.
Wish you a happy semester!
For use by instructor only.
CSLO
Points
CSLO Scoring
CSLO #1
CSLO #2
CSLO #3
CSLO #4
Total
2
Assessment
Contents
1 Problem Solving
1.1 Four Step Method .
1.1.1 Tutorial . . .
1.1.2 Examples . .
1.1.3 Homework . .
1.2 Pattern and Puzzles
1.2.1 Tutorial . . .
1.2.2 Examples . .
1.2.3 Homework . .
1.3 Venn Diagrams . . .
1.3.1 Tutorial . . .
1.3.2 Examples . .
1.3.3 Homework . .
1.4 Percents . . . . . . .
1.4.1 Tutorial . . .
1.4.2 Example . . .
1.4.3 Homework . .
1.5 Chapter Test . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
2 Consumer Mathematics
2.1 Simple Interest . . . .
2.1.1 Tutorial . . . .
2.1.2 Examples . . .
2.1.3 Homework . . .
2.2 Compound Interest . .
2.2.1 Tutorial . . . .
2.2.2 Examples . . .
2.2.3 Homework . . .
2.3 Annuity and long term
2.3.1 Tutorial . . . .
2.3.2 Examples . . .
2.3.3 Homework . . .
2.4 Loan Installment . . .
2.4.1 Tutorial . . . .
2.4.2 Examples . . .
2.4.3 Homework . . .
2.5 Chapter test . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
saving
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
3 Descriptive Mathematics
3.1 Describing Data Values
3.1.1 Tutorial . . . . .
3.1.2 Examples . . . .
3.1.3 Homework . . . .
3.2 Central Values . . . . .
3.2.1 Tutorial . . . . .
3.2.2 Examples . . . .
3.2.3 Homework . . . .
3.3 Median And Quartiles .
3.3.1 Tutorial . . . . .
3.3.2 Examples . . . .
3.3.3 Homework . . . .
3.4 Standard Deviation . . .
3.4.1 Tutorial . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
5
5
5
5
7
9
9
9
11
13
13
13
15
17
17
17
18
20
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
plans
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
21
21
21
21
22
24
24
24
26
28
28
28
30
31
31
31
32
33
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
34
34
34
34
36
38
38
38
42
43
43
44
46
48
48
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
3
3.5
3.4.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.4.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
Chapter Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
4 Probability
4.1 Basic concept . . . . . . . . . . .
4.1.1 Tutorial . . . . . . . . . .
4.1.2 Examples . . . . . . . . .
4.1.3 Homework . . . . . . . . .
4.2 Working with Events . . . . . . .
4.2.1 Tutorial . . . . . . . . . .
4.2.2 Examples . . . . . . . . .
4.2.3 Homework . . . . . . . . .
4.3 Counting Methods . . . . . . . .
4.3.1 Tutorial . . . . . . . . . .
4.3.2 Examples . . . . . . . . .
4.3.3 Homework . . . . . . . . .
4.4 Permutations and Combinations
4.4.1 Tutorial . . . . . . . . . .
4.4.2 Examples . . . . . . . . .
4.4.3 Homework . . . . . . . . .
4.5 Chapter Test . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
4
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
56
56
56
57
60
62
62
63
68
72
72
74
76
77
77
77
80
81
1
Problem Solving
1.1
1.1.1
Four Step Method
Tutorial
George Polya (1887-1985) has given four step method for solving problems using Mathematics,
Step 1: Understand the problem
• Restate the problem in a language that is clear and sensible to you.
• Restate the question
• Organize information provided.
Step 2: Device a plan
• Think about a similar problem you may have solved in past and apply that method.
• Gather all the information given in the problem to get an idea to solve the problem.
• Assign variables to unknown quantities and design a step by step plan.
Step 3: Carry out the plan.
• Use algebraic manipulation (such as solving equation).
• Plug in some values and look for a pattern.
• Try a trial and error method to estimate a solution.
• Use a known formula or a concept.
A puzzle: The following puzzle had me confused for a long time, then I realized that there was an error in
calculation!!
On a father’s day, three siblings went to a store to buy a clock for their father. The price tag was \$30, so
each chipped in \$10 from their pocket, paid \$ 30 and walked out. But the clock was \$5 o↵, so the store manager
sent a salesman after them to give back \$5. Salesman thought that \$5 is not divisible by 3, so he gave back \$3.
This way each paid \$9 and the salesman kept \$2. 9 ⇥ 3 = 27, plus 2 equals 29. Where is the 30th dollar?
1.1.2
Examples
1. One side of a rectangular playground is 10 yards longer than the other side. If the perimeter is 120 yards,
what are the dimension of the playground?
! Solution:
Step 1: Perimeter of a rectangle is measurement of the boundary, which is a sum of all four sides.
Therefore sum of all four sides is 120 yards. Suppose one side is x yard long. Then the longer side
is x + 10 yards. In the following figure, we represent this information:
x + 10
x
Perimeter
120 yard
Figure 1: Rectangular playground
5
Step 2: Two of the four sides are x yards long and other two sides are x + 10 yards long. Now sum
of theses sides is 120 yards. Therefore,
(x + 10) + (x) + (x + 10) + (x) = 120
(1)
Step 3: We solve equation given in (1) using our knowledge of linear equations :
(x + 10) + (x) + (x + 10) + (x) = 120
=) 4x + 20 = 120 =) 4x = 100 =) x = 25
So one side is 25 yards and the other one is 35 yards.
Step 4: To check the answer we add four sides and verify that the perimeter is 120 yards.
25 yards + 35 yards + 25 yards + 35 yards = 120 yards
2. Length of a rectangular wall is 3 feet less than twice the length. Perimeter of the wall is 78 feet. Find the
dimension and the area of the wall.
! Solution: Suppose the height is h feet. Then the length is 2h 3 feet. In the following figure, we represent
this information:
2h
h
3
Perimeter
78 feet
Figure 2: Rectangular Wall
Since sum of all the four sides is 78 feet. Therefore,
(2h
3) + (h) + (2h
3) + (h) = 78.Hence, 6h
6 = 78
(2)
Adding 6 on either sides yields 6h = 84 orh = 14.
(3)
We solve equation given in (1) using our knowledge of linear equation from Mat 098:
(x + 10) + (x) + (x + 10) + (x) = 120
=) 4x + 20 = 120 =) 4x = 100 =) x = 25
So one side is 25 yards and the other one is 35 yards. To check the answer we add four sides and verify
that the perimeter is 120 yards.
25 yards + 35 yards + 25 yards + 35 yards = 120 yards.
6
1.1.3
Homework
1. Perimeter of a rectangular room is 270 yards. Longer side of the room is 15 yards longer than the shorter
side. Find the dimensions of the room.
(a) Identify the information provided in the statement and represent the information in a diagram. Use
variable to represent sides.
(b) Construct suitable equation.
(c) Solve the equation.
(d) Check the solution.
2. A rectangular shed is fenced with 120 feet long barbed wire. If length of the shed is twice as long as the
width, what is the length?
7
3. A rectangular football field is 20 feet longer than twice the width. Perimeter of the field is 160 feet. Find
the dimension of the field.
4. A rectangular shed is fenced with 120 feet long barbed wire. If length of the shed is twice as long as the
width, what is the length? Also find the area of the shed.
5. Length of a rectangular dance floor is 22 feet more than the width. Perimeter of the floor is 156 feet. A
designer charges \$12 for a square foot of carpet plus \$99 for carpeting the entire room. Calculate cost of
carpeting the floor.
8
1.2
1.2.1
Pattern and Puzzles
Tutorial
Often finding a pattern is an e↵ective problem solving strategy. Many number games are based on this strategy.
We will study three areas of applications, sequence of numbers, magic squares and Sudoku puzzles.
A sequence is a group of numbers which follow a specific pattern or a rule. For example in
the sequence, 1, 2, 1, 4, 1, 6, . ., 1 is repeats alternatively, and other vales are even numbers in
an ascending order. Once the pattern is recognized, we can guess other terms.
Magic square is an arrangement of integers written in a square formation so that the sum of
each row, each column and each diagonal are equal. Here is an example of a 3 by 3 magic square
which uses numbers 1 through 9:
2
9
4
7
5
3
6
1
8
Sudoku is a logic puzzle. In our course, we will consider 3 by 3 and 4 by 4 puzzles. A square
is divided into 9 or 16 smaller squares (which are called cells). Solver enters 1, 2, 3 or 1, 2, 3,
4 in these cells so that each row and each column contains 1, 2, 3 (or 1, 2, 3, 4). Some cells are
surrounded by thick lines which forms a partition. If you see ‘1 ’ in one cell of any partition,
then the di↵erence of entries in those cells should be 1. But if you see ‘6⇥’ in one cell of any
partition, then the product of entries in those cells should be 6. Here is a 4 by 4 Sudoku puzzle.
7+
1-
4⇥

3-
8+
For the full details on Sudoku puzzles, visit the website of Kenken puzzles at http://www.kenkenpuzzle.com
Research shows that keeping brain active stops or slow downs the progression of Alzheimer.
Cultivate a habit of solving Crossword, Sudoku or some other logic puzzles. If you have any
senior citizens in your family, visit them and help them learn Sudoku puzzles. You can get a
Sudoku or a crossword puzzle book in a dollar store.
1.2.2
Examples
1. Find the next two numbers of the following sequence and explain the pattern:
1, 2, 4, 8,
,
! Solution: We can see that each number is multiplied by 2 to get the next number. We show
our work by making arrows as shown bellow:
1
2
⇥2
4
⇥2
8
⇥2
16
⇥2
32
⇥2
2. Complete the following magic square using numbers 1 through 9:
9
5
2
6
! Solution: Sum of the entries in all the three rows= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45. Since
45
these rows have equal sum, this sum is 3 = 15.
Now suppose the second entry in the last row is x. Then 2 + x + 6 = 15. Therefore, x = 7.
But the column sum is also 15. Therefore, the first entry in the second row is 15 (7 + 5) 3.
Now we use diagonal sum=15 to find the third entry in the first row, which is 15 (2 + 5) = 8.
This way using row sum=column sum=diagonal sum=15, we can fill all the entries as shown
bellow:
4
9
2
3
5
7
8
1
6
3. Solve the Sudoku puzzle given in the tutorial 1.2 .
! Solution: Observe that the second partition in the first column with condition 3 has 2
entries. We must enter two numbers from {1, 2, 3 or 4} whose di↵erence is 3. This can be
done in only one way, 4 1 = 3. Therefore, we can enter 1 and 4 in these two cells. But we
don’t know the order in which these numbers can be entered, so we wait till we get more
information. Now the first column has 1, 2, 3, 4 and 4, 1 are in the last two cells, so first
two cell contains 2 and 3. Now the third partition in the first row with condition 4⇥ has
three entries and product of these entries is 4. This can be done in two ways, 4 ⇥ 1 ⇥ 1 or
1 ⇥ 2 ⇥ 2. In both cases, we have repetition of one entry, which must be written in di↵erent
rows and di↵erent columns. But 2 is in the first or second row, so the third entry in row one
is 1, fourth entry is 4 and the second entry in the last column is 1. Similar logical analysis
and elimination process gives us the following solution:
1-
2
7+
3
4⇥
1

3
4
3-

2
4
4
1
3
1
2
10
4
1
8+
3
2
1.2.3
Homework
1. Find the next two numbers and explain the pattern:
(a) 7, 4, 1, -2,
,
(b) 1, 4 , 9 ,16,
(c)
,
1 1 1 1
2, 4, 6, 8,
,
(d) 0, 1, 8 , 27 ,
,
2. Use the integers from 1 to 9 to complete the following magic squares.
(a)
(b)
3
1
6
6
9
7
3. Use the even integers from 2 to 18 to complete the magic square
4
18
10
11
6
1
5
4. Use 1, 2 and 3 to solve the following Sudoku:

12
6⇥
1
5. Use 1, 2, 3 and 4 to solve the following Sudoku:
7+
1
7+
3+
10+
3
3+
12
1.3
1.3.1
Venn Diagrams
Tutorial
In statistical inquiries, we often see statements connected by ‘AND’, ‘OR’ or ‘NOT’. For example
36 students are enrolled in Mat 111 and Phy 120.
Such information contain di↵erent kind of groups (called sets). We can e↵ectively present these
groups using Venn diagrams. Venn diagram is a diagram representing information pictorially in
circles, one for each group, within an enclosing rectangle. Common elements of the groups are
represented by the areas of overlap among the circles.
Here is an example of a Venn diagram representing three groups, students who drink (C)
co↵ee, (T) students or (J) fruit juice in their breakfast. Area which is common to two circles
represents those who belong to two groups.
J
C
1.3.2
T
Examples
1. In a class of 70 students, 45 are enrolled in Mat 111, 55 are enrolled in Phy 120 and 36
students are enrolled in both Mat 111 and Phy 120. Present this information using a Venn
diagram and answer the following questions:
(a) How many are enrolled in Mat 111 but not in Phy 120?
(b) How many are enrolled in Phy 120 but not in Mat 111?
(c) How many are not enrolled in either Mat 111 or Phy 120?
! Solution: There are two groups of students, one who are enrolled in Mat 111 and two who
are enrolled in Phy 120. So we need two circles in our Venn diagram enclosed by a rectangle
(C) which represents the whole class of 70 students.
C
M
P
Circle M stands for students enrolled in Mat in M , 55 elements in P and 36 in the common
111 and circle P stands for students enrolled area. We enter 36 in the overlap area of the
in Phy 120. Therefor, there are 45 elements two circles as shown here:
13
C
C
6
M
36
P
M
9
36
19
P
Now we answer questions (a), (b) and (c):
Now on the left to the overlap area in M we
write 45 36 = 9 and on the right to the overlap area in P we write 55 36 = 19 remaining
number of students which is 70 (9+36+19) = 6
we write outside of these two circle in C as
show in the following completed Venn diagram.
14
(a) How many are enrolled in Mat 111 but
not in Phy 120? Ans. 9.
(b) How many are enrolled in Phy 120 but
not in Mat 111? Ans. 19.
(c) How many are not enrolled in either Mat
111 or Phy 120? Ans. 6.
1.3.3
Homework
1. Survey of 100 individuals reveal that 58 like Coke, 50 like Pepsi and 18 like both. Complete
the Venn diagram and answer the following questions:
C
P
(a) How many like Coke but not Pepsi?
(b) How many like Pepsi but not Coke?
(c) How many do not like either Coke or Pepsi?
2. A survey asked 170 people what movies they liked. 78 liked action (A); 96 liked horror (H);
67 liked comedy (C); 30 liked comedy & horror; 45 liked action & comedy; 42 liked horror
& action; 29 liked all three. Complete the following Venn diagram and answer how many:
(a) Didn’t like any of these?
(b) Liked action & comedy but not horror?
A
C
H
3. Out of 150 Lincoln students, 90 have a laptop, 70 have a tablet and 30 have both- a laptop
and a tablet. Draw a Venn diagram and answer the following questions:
(a) How many have a laptop but not tablet?
(b) How many have a tablet but not laptop?
(c) How many do not have both a laptop and a tablet?
15
4. A survey asked 150 people what movies they liked. 80 liked action (A); 60 liked horror (H);
75 liked comedy (C); 30 liked comedy & horror; 30 liked action & comedy; 25 liked horror
& action; 10 liked all three. Draw a Venn diagram then answer how many:
(a) Didn’t like any of these?
(b) Liked action & comedy but not horror?
5. A survey asked 155 people where they went on vacation in the last year. 107 the beach; 90
a trip; 76 visiting family; 57 beach & a trip; 54 beach & family; 52 trip & family; 35 beach,
trip, & family. Draw a Venn diagram to answer how many:
(a) went to exactly one of these
(b) went to none of these
(c) went on a trip or to visit family but not to the beach.
16
1.4
1.4.1
Percents
Tutorial
Percent (%) literally means per=divided by (÷) cent=100. Therefore, 5 percent means 5 divided
5
by 100 which is 100 =0.05. As ‘per’ stands for ‘divided by’, ‘of ’ stands for ‘times’. So
5% of \$750 is (5 ÷ 100) ⇥ \$750 = 0.05 ⇥ \$750 = \$37.50.
Absolute Change in a quantity is measured by subtracting original value (called a reference
value) from the current value Thus,
absolute change=current value-reference value.
Relative change is generally written as a percentage of reference value. Thus,
change
relative change= Absolute
reference value ⇥ 100%
Discounts and sales taxes are rate of changes. Discount is a negative relative change and sales
tax is positive relative change.
1.4.2
Example
1. What is 2.8% of 70 feet?
Solution: We replace ‘what’ by x, ‘is’ by =, ‘per’ by ÷, ‘cent’ by 100 and ‘of ’ by ⇥ in the
statement and translate our problem in mathematics as
x = (2.8 ÷ 100) ⇥ 70 ft = 1.96 ft.
2. What percent of \$250 is \$190?
! Solution: As before, we replace ‘what’ by x,‘per’ by ÷, ‘cent’ by 100’, of ’ by ⇥ and ‘is’ by =,
in the statement and translate our problem in mathematics as
(x ÷ 100) ⇥ \$250 = \$190
x ⇥ \$250
= \$190
100
\$2.5x = \$190 or x =
190
= 76%
2.5
3. Find 12% of \$375.
! Solution: 12% of \$375 is

=
4. Find 0.75% of \$500.
! Solution: 0.75% of \$500 is

=
.
Note: Similar question was asked in fall 2016 final and most got it wrong mainly because
students were confused with decimal point! Instead of using 0.75% = 0.75 ÷ 100 = 0.0075, they
used 0.75. This is a common mistake, so be aware of it.
5. What percent of \$300 is \$ 37.50?
! Solution: If x is unknown quantity, then the statement ‘x of \$300 is \$37.50 can be written
as
\$37.50
x ⇥ \$300 = \$37.50 in =) x =
= 0.125
\$300
Solve the above equation. Converting into percentage, we get the answer 12.5%.
6. A table cloth worth \$12.99 is sold for \$ 10. Find the absolute and relative change.
17
Solution: The current value is the final price, which is \$10. Original price is the reference
value. Therefore,
Absolute change = \$10
Relative change =
We can say that discount is 23%.
1.4.3
Homework
1. What is 3.5% of 80 feet?
2. What percent of \$450 is \$75?
3. Find 15% of \$575.
4. Find 0.05% of \$2,500.
5. What percent of \$250 is \$17.50?
6. 12.5% of what amount is \$15?
18
\$12.99 =
\$2.99
⇥ 100% =
\$12.99
\$2.99
23%
7. A laptop worth \$1,250 is sold for \$ 1050. Find the absolute and relative change.
19
1.5
Chapter Test
1. A table cloth is 15% o↵. If the marked price is \$22.99, how much is the final price? (Ignore
taxes).
2. Original cost of a table cloth is \$35, but it was sold for \$29.50. What percentage was the
discount?
3. 25% discount makes the selling price
% of the marked price.
(a) 0.25 (b) 0.75 (c) 75 (d) 25
4. An item worth \$225 will cost \$
after a discount of 12 12 %.
(a) 196.88 (b) 28.13 (c) 212.5 (d) 198.13.
5. I got a TV set for \$754.99 which was 15% o↵ the marked price. Find the marked price.
6. Marked price = \$200, Discount = 15%. Find the selling price.
7. Marked price = \$49.99, Discount = 12 15 %. Find the selling price.
8. Selling price = \$138, Discount = 40%. Find the marked price.
9. Selling price = \$1,260, Discount = 30%. Find the marked price.
10. Jerry bought a game for \$15.60 which was originally sold for \$19.90. Find the discount
percentage.
11. John bought a pair of shoes at a discount of 30%. If the marked price was \$60.00, how much
he need to pay?
12. Agatha paid \$129.48 for a rug which was 30% o↵. Find the original price.
13. Honda electric guitar costs \$250 when it is 25% o↵. If store o↵ered 40% discount, how much
will it cost?
14. Relative change in the original price and final price of a travelers luggage set is 12.5%. If
the final price is \$1,288.24, find the original price
15. Pretax price of a D-link modem is \$59.99. Local tax is 6.7%. Find the original price of the
modem.
16. Jane bought a dress for \$75 which was o↵ 20%. What was the original price of the dress?
20
2
Consumer Mathematics
2.1
2.1.1
Simple Interest
Tutorial
When money is invested, investor earns interest. There are various types of interest, the most
basic model is simple interest. In this model, the interest is calculated on the basis of the invested
amount called principal.
If P0 is the initial deposit or principal R% is the annual percentage to be given as an interest,
then every year the interest amount will be P0 ⇥ r. After t number of years, the interest will be
It = P0 rt and the final balance will be Pt = P0 + It .
Formulas:
Simple Interest It = P0 rt
Final balance Pt = P0 (1 + rt)
2.1.2
Examples
1. Julie invested \$450 in a simple interest plan for 12 years at 6% APR. Find the interest and
the final balance.
Solution: Here, P0 = \$450, APR=6% (hence r = 0.06) and t=12 years. Therefore,
I12 = P0 rt = (\$450)(0.06)(12) = \$324.
Final balance P12 = P0 + I12 = \$450 + \$324 = \$774.
2. Haida borrowed \$800 from Mike with 8% APR of a simple interest. How much will she pay
back after 3 months?
3
Solution: Three months is 12 =0.25 years. So the interest will be calculated with P0 = \$800,
r = 0.06 and t = 0.25.
I0.25 = P0 rt = (\$800)(0.06)(0.25) = \$12 and the final balance = \$812.
She will pay back 812 dollars.
3. Johnny invested \$ 500 in a simple interest plan and received a final balance of \$ 900 after
10 years. Find the rate of interest, APR.
Solution: This time, P0 = \$500, P10 = \$900 and t=10 years. So, I10 = \$900
Therefore,
\$500 = \$400.
I10 = P0 rt =) \$400 = r(\$500)(10) =) \$400 = \$5000r
400
So r =
= 0.08 Or APR = 8%
5000
The annual percentage rate is 8%
ans.
4. How long will it take to get \$500 interest on an investment of \$ 4,000 at 5% APR in a simple
interest account?
21
Solution: We substitute It = \$500, P0 = \$4, 000 and r = 0.05 in our formula.
It = P0 rt =) \$500 = \$4, 000(0.05)t = \$200t
=) t =
\$500
= 2.5
\$200
It will take 2 and half years.
5. Treasury Notes (T-notes) are bonds issued by the federal government to cover its expenses.
Suppose you obtain a \$1,000 T-note with a 4% annual rate, paid semi-annually, with a
maturity in 4 years. How much interest will you earn? First, it is important to know that
interest rates are usually given as an annual percentage rate (APR), the total interest that
will be paid in the year. Since interest is being paid semiannually (twice a year), the 4%
interest will be divided into two 2% payments.
P0 = \$1000 (the principal)
r = 0:02 (2% rate per half-year)
t = 8 (4 years = 8 half-years)
Therefore, I8 = \$1000(0.02)(8) = \$160. You will earn \$160 interest total over the four years.
2.1.3
Homework
1. Jane invested \$500 for 10 years in a simple interest account o↵ering 5% APR. Find the
interest and her final balance.
2. John invested \$3,500 for 8 years in a simple interest account at 2 12 % APR. How much
interest he will earn? Find his final balance.
3. After 12 years, Hardy received an interest of \$1,470 from his investment of \$2,000 in a
simple interest account. Fine his APR.
4. Kate invested \$800 in a simple interest account with 6% APR. After how many years she
will get a final balance of \$3000?
22
5. Find an APR which turns a principal of \$12,000 invested in a simple interest account in 10
years to a final balance of \$21,370.
6. Paris want to earn a simple interest I= \$3,100 on a principal of \$30,000. If the APR is 8%,
how many years she she should invest to reach her goal?
7. On a deposit of \$3,500 for 8 years, John is o↵ered a simple interest of 2%. How much
interest he will earn? Find his Final balance.
8. John borrowed \$760 from a bank with 4% APR. He paid o↵ his loan after 120 days. Calculate
his total due.
23
2.2
2.2.1
Compound Interest
Tutorial
In a simple interest investment, we do not get (or pay) interest on interest earned on our investment. In a compound interest, we add interest to the previous balance. If interest is added N
times in a year, we say that the interest is compounded N times. Thus compounded quarterly
means compounded four times and N = 4. Here is the formula for computing compound interest.
Compound Interest PN = P0 1 +
r Nk
k
Value of k depends on the phrase “compounded —”’ in the statement of a given problem. We
list some values of k:
Phrase
Compounded annually
Compounded half-yearly
Compounded quarterly
Compounded monthly
Compounded daily
2.2.2
k
1
2
4
12
365
Examples
1. Macbeth invested \$1,320 at 6% APR compounded quarterly for 8 years . Find the interest
and the final balance.
! Solution: Observe that P0 = \$1, 320, r = 6% = 0.06, K = 4 (there are 4 quarters in a year) and
N = 8 years. Now,

◆(8)(4)

r ⌘8k
0.06
P8 = P0 1 +
=) P8 = 1, 320 1 +
k
4
=) P8 = 1, 320(1.015)32 = 1, 320(1.61032432) = \$2125.63.
Therefore, interest I=\$2125.63-\$1,320=\$805.63.
2. Joan deposits \$7,500 in a monthly compounding investment with 6% APR. In how many
years the amount will be doubled?
! Solution: Trial and error method In this example, P0 = \$7, 500, PN = \$15, 000 and r=0.056.

◆12N

r ⌘N k
0.06
=) 15000 = 7, 500 1 +
PN = P0 1 +
k
12
Divide both sides by 7,500 to get 2 = (1.005)12N
We try di↵erent values of N (use Excel).
(1.005)120 = 1.8 N=10
(1.005)132 = 1.97 N=11 which is close.
=) N = 11 years.
24
Second solution

◆12N

r ⌘N k
0.06
PN = P0 1 +
=) 15000 = 7, 500 1 +
=) 2 = (1.005)12N
k
12
Now use logarithmic function ln (or LN) on both sides to get,
0.69314718 = (0.004987542)(12N ) or N = 11.58
3. Marc Invested \$125,000 for 12 years at 5% compounded quarterly. Find the accumulated
balance and interest.
! Solution:

r ⌘N k
P12 = P0 1 +
=) P12
k
= \$226, 919.36
Interest I = P12

◆48
0.05
= 125, 000 1 +
= 125, 000(1.0125)48 = 125, 000(1.8153548530)
4
P0 = \$226, 919.36
\$125, 000 = \$101, 919.36
4. How much investment (deposit) guarantees a final balance of \$10,000 in ten years, if the
APR is 4% compounded quarterly?
! Solution: Notice that we need to find P0 . P10 = P0 1 +
r Nk
.
k
Substitution gives,
0.04 40
) = P0 (1.488863734)
4
10, 000
) P0 =
= 6716.531387
1.488863734
10, 000 = P0 (1 +
It requires \$6716.53 to get \$10.000 in 10 years.
25
2.2.3
Homework
1. Jose invested \$1,500 for 5 years in a quarterly compounding account o↵ering 6% APR. How
much interest he will earn? Find his final balance.
2. On a deposit of \$5,000 for 8 years, Joan is o↵ered a monthly compounding of 2 12 %. How
much interest she will earn? Find her final balance.
3. Walter invested \$6,000 in a annual compound interest account for 12 years. The APR
o↵ered to him is 7.5%. Find his gain (interest) and determine how much he will get at the
end.
4. In 10 years, Joseph received a final balance of \$10,353.25 as a result of his investment of
\$2000 in a daily compounding account. Fine his APR.
26
5. Annie invested \$2,500 in a annual compound interest account which gave her 6% APR. In
how many years she can earn an interest of \$1,000?
6. Find an APR which turns a principal of \$12,000 invested in a daily compound interest
account in 10 years to a final balance of \$21,000.
7. Hagar wants to earn \$150,000 from an investment of \$30,000. Government bonds o↵ers 6%
APR with a monthly compounding. In how many years he will reach his goal?
8. Jack and Jill invested \$5,000 each for 10 years. Jack invested in a simple interest accounts
with 8% APR and Jill invested in a a daily compound interest accounts with 4% APR.
Calculate their final balances and determine who earned more.
27
2.3
2.3.1
Annuity and long term saving plans
Tutorial
Investments in which we save a fixed amount every month for number of years are called annuity
or Savings plans. Sometimes annuity refers to the the final balance or a contract from any
financial agency.
There are three formulas in this section:
Accumulated balance
PN =
Monthly deposit
d=
Payo↵ Annuity
P0 =
d[(1+ kr )N k 1]
( kr )
AN ( kr )
[(1+ kr )kN 1]
d[1 (1+ kr )
Nk
]
( )
r
k
Individual retirement accounts (which are also called IRA or 401(K) accounts) are the most
popular savings accounts. We open a savings account with a financial firm of our employees.s
choice. Every month, a certain portion of our income is directly deposited to our account. There
is no income tax charged to the investment, until you make withdrawal. Part of your retirement
income comes from such investment. The retirement age is 65 years. If a 20 years old individual
has an IRA, the first formula computes the total savings after 45 years. If he/she wants to have
certain amount for 20 years after his retirement, the third formula determines the amount he/she
should have at the time of retirement.
2.3.2
Examples
1. Hans saved \$50 every month for 15 years earning 5.7% annual percentage rate (APR). Find
his final balance.
! Solution: Observe that regular deposit d= \$50, APR=5.7% (so r=0.057) and N=15. Substituting these values in the savings plan formula, we get,
PN =
=
d[(1 + kr )N k
1]
r
k
50[(1.00445)180
(0.00475)
=
1]
50[(1 +
=
0.057 (12)(15)
12 )
0.057
12
1]
50(1.346619501)
0.00475
2. Mr. Least Bothered decided to accumulate \$120,000 in 20 years for his daughter’s college
tuition. One investment firm o↵ered him 4.8% APR. Help him find his monthly deposit (d).
! Solution: Observe that P20 = \$120, 000, APR=4.8%, r=0.048, k=12 and N=20. Substituting
these values in the savings plan formula for d, we get,
120000 0.048
12
240
1]
[(1 + 0.048
1]
12 )
120, 000(0.004)
120, 000(0.004)
=
=
240
[(1.004)
1]
1.606700133
480
=
1.606700133
d=
P20 kr
[(1 + kr )kN
=
28
3. Andre wants \$1,000 per month for 30 years after he is retired. His IRA o↵ers him 3.6%
APR. At the time of retirement, how much savings Andre should have in his account?
! Solution: Observe that d = \$1, 000, APR=3.6%, r=0.036, k=12 and N=30. Substituting
these values in the payo↵ annuity formula for P0 , we get,
P0 =
=
d[1
(1 + kr )
Nk
]
r
k
\$1, 000[1 (1.003)
(0.003)
=
\$1, 000[1
360
(1 +
0.036
(12)(30)
]
12 )
0.036
12
]
=
\$1, 000[1 0.3401450]
0.003
4. Jacob wants to save \$200,000 by saving \$200 a month in an investment earning 6% APR.
How long will it take?
! Solution by trial and error method. Observe that PN = \$200, 000, d=\$200 and r=6%=0.06.
Substituting these values (with N as it is) in the savings plan formula (provided), we get,
200, 000 =
200[(1 +
.06 12N
12 )
.06
12
1]
=
200(1.005)(12N )
(0.005)
1]
multiplication by .005 gives 1000 = 200(1.005)(12N )
=) 1001 = 200(1.005)
(12N )
,Or 5.005 = (1.005)
1
12N
For N=25, (1.005)12N = 4.4649698121623049771420869730681
For N=27, (1.005)12N = 5.0327343742406936784902574549
We see that N= 27 years provides a value that is very close to 5.005 compared to N=25.
Hence N=27 years is a correct guess.
29
2.3.3
Homework
1. John saved \$30 a month for 40 years at 4% APR in a savings account. Find his accumulated
balance and calculate the total interest he received.
2. Ann accumulated balance of \$100,000 in 35 years in a savings plan which o↵ered an APR
of 3.6%. Find her monthly deposit d.
3. Nasir wants \$1,200 per month retirement income for 25 years after he is retired. His IRA
o↵ers 4.8% APR. At the time of retirement, how much savings Nasir should have in his
account?
30
2.4
2.4.1
Loan Installment
Tutorial
Loan payment d (see page 212):
d=
[1
P0 ( kr )
(1 + kr )
kN ]
In most cases, we will assume k = 12.
Most common loans are (1) student loans (2) auto finance and (3) mortgage loans. Students
loans are generally for 3 to 4 years. Their APR is lowest among all kind of loans.Auto loans are
also for 4 to 5 years, but the interest rates are very high. Mortgage loans are for 15, 20 or 30
years. Interest rates can be reduces by making a down payment. Good credit history also helps.
2.4.2
Examples
1. Lisa financed a car worth \$30,000 for 4 years with 6.6% APR. Find her monthly payment.
! Solution: Here the loan amount P0 = \$30, 000, APR=6.6%, r=0.066 and N=4. Substituting
these values in the loan formula (provided), we get,
r
P0 12
30, 000 0.066
12
r 12N =
48
[1 (1 + 12 )
]
[1 (1 + 0.066
12 ) ]
30, 000(0.0055)
30, 000(0.0055)
=
=
48
[1 (1.0055) ]
1 0.76852953
165
=
0.231476746
d=
31
2.4.3
Homework
1. Jade mortgage her house for \$200,000 for 30 years with an APR of 4.8%. Find her monthly
installment
2. Walter financed his car worth \$30,000 for 5 years with an APR of 6%. Find his monthly
installment.
32
2.5
Chapter test
1. Henry deposits \$1,300 at 1.5% APR in a simple interest account for 4 years. Find the
interest I and the final balance P4 .
2. With an initial investment of \$2,500, how long will it take to accumulate \$5,000 in an account which is paying simple interest of 8%?
3. After a discount of 15%, I paid \$510 for a TV. Find the original price of the TV.
4. John invested \$12,000 for 8 years at 3.6% compounded quarterly. Find the final balance.
5. What should the initial investment be to accumulate \$8,000 in 10 years in an account with
5.4% APR compounded monthly?
6. Tania saved \$70 per month in a savings plan giving 4% APR for 25 years. Find her accumulated balance and total interest she earned.
7. Rudy is planning for his retirement. He would like to get \$1,500 per month for 25 years
after he retires. What must be his total savings if a 401(K) plan o↵ers him 3% APR?
8. Jerry financed his car worth \$40,000 for four years with 3.6% APR. Find his monthly
installment. Also compute the finance charges.
33
3
Descriptive Mathematics
3.1
3.1.1
Describing Data Values
Tutorial
By describing we mean finding important properties of data that can help us in decision making.
For example, if we know average weight of a Dell desktop, we can decide how many a forklift can
carry. Graphs (such as bar chart, pie chart) of profit of last 10 years can help us with a quick
comparison between two companies. We will study three methods to describe data:
• Graphical representation of data.
– Frequency distribution.
– Bar graphs, pie chart and histograms.
• Measure of central tendency of data.
– Mean, mode, median and quartiles.
• Variation (spread or fluctuation) in data.
– Range and standard deviation.
Frequency Table:
A frequency table is a table with two columns. One column lists the categories, and another
for the frequencies with which the items in the categories occur (how many items fit into each
category).
Bar graph:
A bar graph is a graph that displays a bar for each category with the length of each bar indicating
the frequency of that category.
Pie Chart:
A pie chart is a circle with wedges cut of varying sizes marked out like slices of pie or pizza. The
relative sizes of the wedges correspond to the relative frequencies of the categories.
3.1.2
Examples
1. Construct a frequency table and plot a bar diagram and pie chart for hourly wages of 20
employees.
10, 9, 11, 11, 10, 9, 10, 11, 10, 10, 9, 11, 8, 12, 10, 11, 8, 10, 9, 12.
Also find the mode of the distribution.
! Solution: Frequency of a data value is how often it appears. For example 8 appears 2 times,
so the frequency of 8 is 2 (this was done by observation).
Wages (in \$)
Employees
8
2
34
9
4
10
7
11
5
12
2
Mode of the distribution is a value or values that appears most often. In this example
hourly wage 10 appears 7 times which is more often than any other values. Therefore, 10
is the mode. To plot a bar diagram, we will use x-axis for data values (hourly wages) and
y-axis for the frequencies.
Number of employees
7
6
5
4
4
2
2
2
8
9
10
11
12
Figure 3: Hourly wages (in \$)
We will now draw a pie chart for the same data. First we find relative frequencies of each
data values.
Wages (in \$)
Employees
Rel. freq.
8
2
10%
9
4
20%
10
7
35%
11
5
25%
12
2
10%
Total
20
100%
Relative frequency of value:
Relative F requency =
For example relative frequenvy of 11 =
2.
7
20
Frequency of x
Total Frequency
= 0.35 = 35%. The pie chart is given in the figure
In a separate worksheet we will learn how to make bar and pie chart using MS Excel. We
will also learn to make histograms and line graphs using the software.
35
\$9
20 %
\$10
\$8
10 %
35 %
10 %
25 %
\$12
\$11
Figure 4: Pie chart
3.1.3
Homework
Construct frequency distribution and draw a bar graph and pie chart for each of the following
data set:
(a) Hourly wages: 12, 10, 8, 11, 10, 11, 10, 9, 10, 8, 9, 11, 10, 11, 8, 9, 10, 10, 8, 11.
(b) Annual income (in thousand dollars): 45, 47, 48, 46, 48, 47, 46, 45, 49, 48, 50, 45, 46,
47, 46, 47, 50, 46, 48, 47, 45, 48, 49, 46, 50, 45, 48.
36
(c) Working hours: 7, 5, 10, 8, 9, 8, 6, 7, 9, 9, 6, 9, 8, 10, 5, 7, 7, 5, 8, 9, 10, 6, 9, 8, 7.
37
3.2
3.2.1
Central Values
Tutorial
To describe data by the central theme is vary natural to us. We often say that the average speed
is 60 miles per hour or most prefer ca↵eine free drink. In this worksheet we will learn about
average or mean.
Mean
Mean or average is a measure of central tendency of a distribution. Mean of a population is
denoted by µ and that of a sample is denoted by x̄. When data values are given in row form,
P
sum of x-values
x
µ or x̄ =
=
.
number of x-values
n
P
where n stands for the number of values and
is a sigma symbol which means sum.
In the following example 3, we noticed that we can multiply frequencies and corresponding
data values to find the total of values in a discrete distribution. This will modify our previous
formula for the mean as
P
(x · f )
x̄ =
N
P
where, (x · f ) is a sum of numbers in the x ⇥ f -column.
3.2.2
Examples
1. Find x̄ for the distribution 4, 6, 3, 4, 6, 7, 4, 3, 2, 3.
! Solution: Sum of x values = 4+6+3+4+6+7+4+3+2+3=42.
There are 10 values. Therefore,
x̄ =
42
= 4.2.
10
ans.
2. The average weight of a class of 30 students is 145 lb. Five new students are added to the
class whose weights are 151 lb, 152 lb,154 lb,148 lb and 145 lb. Find the new average weight
of the class.
! Solution: By definition, the average weight of 30 students is:
145 =
total weight
30
So the total weight of 30 students is 145 ⇥ 30 = 4350. Weight added to the class is 151 +
152 + 154 + 148 + 145 = 750 lb. This implies that the total weight of all the 35 students is
4350 + 750 = 5100 lb. Therefore, the new average weight of 35 students is:
5100
= 146.71 lb.
35
38
ans.
3. Find the x̄ for the following distribution:
x
f
0
2
1
2
2
3
3
4
4
1
! Solution: If we write the data values in a row form, we will get the following distribution:
0, 0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4. Therefore,
0+0+1+1+2+2+2+3+3+3+3+4
12
2(0) + 2(1) + 3(2) + 4(3) + 1(4)
=
12
0 + 2 + 6 + 12 + 4
24
=
=
=2
ans.
12
12
x̄ =
4. Mart’s average score for the College basketball season 1 is 32 points and for season 2 is 28
points. In season 1, she played 14 games and in season 2 she played 8 games. What is her
combined average score?
! Solution: Total points for the season 1 are 32 ⇥ 14 = 448 and for the season 2 are 28 ⇥ 8 = 224.
Therefore, she scored 448 + 224 = 672 points in 22 games. This gives us an average of
µ=
672
= 30.5
22
ans.
5. Find the mean of
No. of children
No. of families
0
4
1
7
2
9
3
5
4
3
! Solution We complete the table by adding a column of x ⇥ f .
x
0
1
2
3
4
5
f
4
7
9
5
3
2
N = 30
x⇥f
0
7
18
15
12
P 10
xf = 62
Therefore,
x̄ =
62
= 2.07
30
39
ans.
5
2
Note: Microsoft Excel has a inbuilt function which can find mean. We have solved the above
example using excel:
6. Use excel to find the mean of
No. of children
No. of families
0
4
1
7
2
9
3
5
4
3
5
2
! Solution: Follow the following steps:
(a) Enter the data in a single column of an excel work sheet starting from the second cell
A2.
(b) Last entry will be in A31 cell.
(c) In cell A32, write
=AVERAGE(A2:A31)
(d) Click on AVERAGE. End you will get answer ‘2.066666667’ which is what we got before.
40
41
3.2.3
Homework
1. Find the mean in each cases:
(a) 4, 6, 8, 10, 8.
(b) 3, 5, 6, 8, 7, 9, 6, 3, 5, 6, 8, 7, 9, 6, 3, 5, 6, 8, 7, 9, 6, 3, 5, 6, 8, 7, 9.
(c)
Hourly wages
No. of employees
9
2
10
5
11
6
12
4
13
2
14
1
2. Average weight of 25 students in Mat 106 class was 140 lb. New five students joined the
class. Their weights are 148 lb, 154 lb, 160 lb, 147 lb and 153 lb. Find the new average
weight of the class.
42
3.3
3.3.1
Median And Quartiles
Tutorial
Five number summary: When data values are arranged in ascending order, we can describe data
set by five values located in five key positions. These positions are
• First position, minimum value
• Second positioned at the end of the first quarter called the first quartile Q1 .
• Third positioned in the middle, called the median M .
• Fourth positioned at the end of the third quarter called the third quartile Q3 .
• Fifth position, the maximum value.
Line on either sides are called whiskers. If both whiskers are of equal length, then the data
values form a normal distribution. If left whisker is longer, then the distribution is skewed on
left (or negatively skewed). If right whisker is longer, then the distribution is skewed on right
(or positively skewed). Figure 5 is a box plot of a negatively skewed data.
Median is a middle value of a data distribution. If there are two middle values, then the average
of these two values is the median. To find median, first arrange the values in an ascending order,
then
• Find the median position. If given values are x1 , x2 , . . . xN , then,
the median position = l =
N +1
.
2
• If l is an integer, then xl is the median, otherwise, median is an average of xt and xt+1 where
t is the largest integer smaller than l.
Quartiles are positioned half way on either sides of the median. If we arrange data in a line
(as shown in the following figure) M = median lies in the middle. Q1 in the middle of the first
half and Q3 in the middle of the second half. Therefore Q1 = median of the first half and Q3 =
median of the other half.
whisker

Min
whisker

Q1

M

Q3

Max
Figure 5: Positions of five points
To find Q1 and Q3 , follow the following procedure:
• Find the median M .
• Find the median of all the data values which are smaller than M . This value is Q1 .
• Find the median of all the data values which are larger than M . This value is Q3 .
43
3.3.2
Examples
1. Find M , Q1 and Q3 of: 3, 6, 6, 4, 5, 7, 8.
! Solution: To find the median we need to arrange the data values in the ascending order:
3, 4, 5, 6, 6, 7, 8
N = 7, so the the median position is N2+1 = 82 = 4.
Therefore, median is the 4th data value which is 6.
Now, Q1 is the median of
3, 4, 5 which is 4 and Q3 is the median of 6, 7, 8 which is 7.
Therefore, M = 6, Q1 = 4 and Q3 = 7. The box plot is given as follows:
Q1

3

Q3
M

6
4
7

8
Figure 6: Positions of five numbers
2. Draw the box plot for the data 10, 9.6, 8.2, 12, 11, 10.2.
! Solution: Values in the ascending order are
8.2, 9.6, 10, 10.2, 11, 12.
There are six values. Therefore the median position is N2+1 =
is equal to
x3 + x4
10 + 10.2
=
= 10.1.
2
2
Now, Q1 is the median of
which is 11.
7
2
8.2, 9.6, 10 which is 9.6 and Q3 is the median of 10.2, 11, 12
Therefore, M = 10.1, Q1 = 9.6 and Q3 = 11.
Q1

8.2
= 3.5 Therefore the median
Q3
M

9.6 10.1

11
Figure 7: Positions of five numbers
44

12
3. Find the five number summary of the following distribution and draw the box plot.
x
f
1
4
2
5
3
3
4
2
5
1
! Solution: As in the case of mean, we can write the data values in a row as shown bellow,
and find the median.
1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5.
Since there are 15 values, the median position is
th
Therefore, the median is 8
Now, Q1 is the median of
3, 4, 4, 5 which is 3.
15+1
2
= 8.
value which is 2.
1, 1, 1, 1, 2, 2, 2 which is 1 and Q3 is the median of 2, 3, 3,
Therefore, M in = 1, M ax = 5, M = 2, Q1 = 1 and Q3 = 3.
Q1

1
Q3
M

2

3

5
Figure 8: Positions of five numbers
45
3.3.3
Homework
1. Drow the box plot for the following data. Is the distribution skewed? On which side?.
(a)4, 5, 2, 2, 4, 5, 3, 5, 6, 4.
(b)7.5, 5.5, 7.2, 2.8, 4.5, 3.7, 4.5, 8.6, 9.3, 6.6, 3.4.
(c)4, 3, 2, 2, 5, 3, 5, 6, 9, 7, 5, 1, 2, 4, 3, 5, 6, 10, 8, 5, 7, 5, 4, 5, 9, 8, 6, 4.
46
(d)
x
f
1
2
(e)
x
f
1.5
2
2
5
3
7
2.5
4
4
8
3.5
8
5
4
4.5
10
5.5
6
47
3.4
3.4.1
Standard Deviation
Tutorial
Standard deviation measures how spread the data values are. For example the information “
average score of a class of 32 students is 67 points with a standard deviation of 2 points” can be
interpreted as “at least 24 (75% of 32) students must have scored between 63(average – 2(standard
deviation)) and 71 (average + 2(standard deviation))”.
The simplest measure of variation is a range of values.
Example: Three patients in a hospital has following pulse rates: Patient-1: 70, 85, 100,
Patient-2: 78, 85, 92 and Patient-3: 82, 85, 88. Find range of the pulse rate of these patients.
Notice that all three patients are having same mean x̄ = 85 pulses per minutes and same
median pulse rate (also 85). Do you think that they are in same condition? Why patient 3 is in
better condition than patient 1? We can di↵erentiate three patients using range of their pulse
rate.
! Solution:
1. Patient-1: Range=100-70=30.
2. Patient-2: Range=92-78=14.
3. Patient-1: Range=88-82=6.
Range is easy and a quick way to find the spread of the values. It ignores mean position. For
example, range can not say whether more values are bellow average (x̄) or above the average.
Example: A paint manufacturer claims that the average drying time for his paint is less than
2 hrs. A sample range is 2.5 hrs. This does not give us any indication if most values are below
2 hrs. or above 2 hrs. It is important to know if most values are close to x̄.
Therefore, it is useful to design a unit that measures how far data values are from the mean
value. This parameter must have
• large value if the fluctuations are high.
• small value if the data values are cluster around the mean value.
This can be achieved by considering average distance from the mean value. But the sum of
all such distances is always zero. That is,
X
(x x̄) = 0.
Therefore, the average distance is also zero!
48
Example: Verify the above equality for 2, 4, 4, 6, 7, 7.
!Solution:
x̄ =
2+4+4+6+7+7
30
=
= 5.
6
6
Therefore,
X
(x x̄) = (2
=
5) + (4
3
1
5) + (4
5) + (6
5) + (7
5) + (7
5)
1 + 1 + 2 + 2 = 0.
We define the population standard deviation µ and the sample standard deviation s to be
rP
rP
(x µ)2
(x x̄)2
=
and s =
N
N 1
Standard Deviation: Step by Step
• Calculate the mean, x̄ (or µ).
• Write a table that subtracts the mean from each observed value.
• Square each of the di↵erences.
• Divide by N -1 (or N) where N is the number of items in the sample (or population).
• Take the square root.
3.4.2
Examples
1. Following are four determinations of the specific gravity of aluminum 2.64, 2.70, 2.67 and
2.63. Find
(a) the range.
(b) Population standard deviation.
! Solution:
(a) Range =1.70-2.63=0.07.
(b) Since there are only four values, it must be a sample. So we will calculate sample standard deviation.
First we find mean
x̄ =
In the following table, we find (x
2.64 + 2.70 + 2.67 + 2.63
= 2.66
4
x̄)2 .
49
x
x x̄
(x x̄)2
Therefore,
=
q
0.003
4
= 0.027
2.64
-0.02
0.0004
2.70
0.04
0.0016
2.67
0.01
0.0001
2.63
-0.03
0.0009
Total
0.003
ans.
2. The owner of the Hot & Spice restaurant wants to find out how much people spend at the
restaurant. He examines 10 randomly selected billing receipts and writes down the following
data.
\$44, \$50, \$38, \$96, \$42, \$47, \$40, \$39, \$46, \$50.
! Solution: He calculated the mean by adding and dividing by 10 to get
x̄ = \$49.20
Below is the table to find the standard deviation:
x
44
50
38
96
42
47
40
39
46
50
49.2)2
27.04
0.64
125.44
2190.24
51.84
4.84
84.64
104.04
10.24
0.64
P
(x x̄)2 = 2600.4
x-49.2
-5.2
0.8
11.2
46.8
-7.2
-2.2
-9.2
-10.2
-3.2
0.8
(x
Therefore,
Standard deviation = s =
rP
x̄)2
=
1
(x
N
r
2600.4 p
= 288.7 = \$16.99
9
Statistically, this will be interpreted as follows:
Most customers probably spend between

s = \$49.20
\$16.99 = \$32.21
and
x̄ + s = \$49.20 + \$16.99 = \$66.19
.
50
3.4.3
Homework
Find the population and sample standerd deviation of the following data:
(a) 4, 5, 2, 3, 3, 5, 6, 4.
(b) 1, 7, 4, 4, 6, 5, 8 .
51
3.5
Chapter Test
1. Annual starting salaries (in thousands of dollars) for social workers with a bachelors degree
and no experience are shown below:
28, 29, 30, 32, 34, 29, 29, 30, 32, 34, 29, 29, 31, 33, 34, 29, 30, 31, 33, 35, 29, 30, 31, 33, 35
(a) Construct a frequency distribution and find the mode.
(b) Draw a bar diagram of the frequency distribution.
2. Waiting time of 11 customers at a customer service station are 2, 6, 3, 5, 5, 1, 8, 6, 10, 7,
9 minutes. Find five point summary and draw a box plot of the data. Is the distribution
skewed? On which side?.
52
3. Monthly income (in thousand dollars) of 9 employees are given bellow.
2,2,5,5,6,8,8,9,9.
(a) Find the average income.
(b) Find the value of sample standard deviation (use s =
(c) Find the value of population standard deviation (use
53
q
⌃(x
=
mean )2
).
n 1
q
⌃(x
mean )2
).
n
4. Following pie chart represents the daily wages of 200 employees of a packaging warehouse:
\$7 25%
\$6 35%
\$9
\$10
15%
(a) How many employees have hourly wage of \$9?
(b) Find the mean amount of the distribution.
5. A mean average of 60 on five exams is needed to pass a course. On her first four exams,
(a) What grade must she receive on her last exam to pass the course?
54
(b) An average of 80 is needed to get a B in the course. Is it possible for Tina to get a B?
(c) If her lowest grade of the exams already taken is to be dropped, what grade must she
receive on her last exam to pass the course?
(d) If her lowest grade of the exams already taken is to be dropped, what grade must she
receive on her last exam to get a B in the course?
55
4
Probability
4.1
4.1.1
Basic concept
Tutorial
Statistical inquiry is a study or an experiment with well defined (but random) outcomes. For
example, if we are measuring length of Tsetse flies, we can be sure that the outcome (length)
will be a positive number. When we actually start gathering measurements, it is impossible to
predict the length of a next specimen. This is why we say that the measurements are random
numbers. From our experience we know that it is not likely that this number would ever be
larger than half inch or the probability of finding a fly longer than half inch is zero.
If you roll a die, pick a card from deck of playing cards, or randomly select a person and
observe their hair color, we are executing an experiment or procedure. In probability, we look
at the likelihood of di↵erent outcomes. We begin with some terminology.
Events and Outcomes
• The result of an experiment is called an outcome.
• An event is any particular outcome or group of outcomes.
• A simple event is an event that cannot be broken down further.
• The sample space is the set of all possible simple events.
A die (plural dice) has 6 sides with marks 1 through 6 on it. A single die can roll 1, 2, 3, 4, 5
or 6.
Figure 9: Picture of a die (dice is the plural of die)
Study the following example:
If we roll a standard 6-sided die, describe the sample space and some events.
Solution:
56
• The sample space is the set of all possible simple events: {1,2,3,4,5,6}
• simple events:
We roll a 1: {1}
We roll a 5: {5}
• compound events:
We roll a number bigger than 4: {5,6}
We roll an even number: {2,4,6}
Basic Probability
Given that all outcomes are equally likely, we can compute the probability of an event E using
this formula:
P (E) =
N umber of outcomes corresponding to the event E
T otal number of equally likely outcomes
Certain and Impossible events
• An impossible event has a probability of 0.
• A certain event has a probability of 1.
• The probability of any event must be 0  P (E)  1
4.1.2
Examples
1. If we roll a 6-sided die, calculate
(a) P(rolling a 1)
(b) P(rolling a number bigger than 4)
! Solution: Recall that the sample space is {1,2,3,4,5,6}
(a) There is one outcome corresponding to “rolling a 1”={1}, so the probability is
1
6
(b) There are two outcomes bigger than a 4={5,6}, so the probability is
2
1
=
6
3
2. Let’s say you have a bag with 20 cherries, 14 sweet and 6 sour. If you pick a cherry at
random, what is the probability that it will be sweet?
57
! Solution: There are 20 possible cherries that could be picked, so the number of possible
outcomes is 20.
Of these 20 possible outcomes, 14 are favorable (sweet), so the probability that the cherry
will be sweet is
14
7
=
20
10
3. A coin is tossed twice. Discribe the sample space and find the probabilities of getting (a)
two tails (b) Head first time and then tail (c) only one tail (d) at least one tail.
! Solution: The smple space contains all the possible results. Denote H for head and T for
tail. Then the sample space is {HH, HT, T H, T T }.
(a) Outcome of getting a tail both times is denoted by T T and the probability of getting
two tails is P (T T ) = 14 .
(b) First Head and then Tail can be denoted by HT which is one outcome out of 4. Therefore,
the probability of getting HT is P (HT ) = 14 .
(c) One T can be obtained in two ways, HT and T H. So the probability of getting one tail
is P (one T ) = 24 = 12 .
(d) At least one T can be obtained in three ways, namely HT, T H and T T . Therefore, the
probability of getting at least one tail is P (at least one T ) = 24 = 12 .
58
4. A coin is tossed twice. What is the probability of getting one T ?
! Solution: Here event E = one tail = {HT or T H}. Both P (HT ) and P (T H) are equal to
Therefore,
1 1
2
1
P (one T ) = + = = ! Answer.
4 4
4
2
1
4.
Cards:
A standard deck of 52 playing cards consists of four suits (hearts, spades, diamonds and
clubs). Spades and clubs are black while hearts and diamonds are red. Each suit contains
13 cards, each of a di↵erent rank: an Ace (which in many games functions as both a low
card and a high card), cards numbered 2 through 10, a Jack, a Queen and a King.
5. Compute the probability of randomly drawing one card from a deck and getting an Ace.
! Solution: There are 52 cards in the deck and 4 Aces so
P (Ace) =
4
1
=
t 0.0769.
52
13
We can also think of probabilities as percents: There is a 7.69% chance that a randomly
selected card will be an Ace.
59
4.1.3
Homework
1. A ball is drawn randomly from a jar that contains 6 red balls, 2 white balls, and 5 yellow
balls. Find the probability of the given event.
(a) A red ball is drawn
(b) A white ball is drawn
2. Suppose you write each letter of the alphabet on a di↵erent slip of paper and put the slips
into a hat. What is the probability of drawing one slip of paper from the hat at random
and getting:
(a) A consonant
(b) A vowel
3. A group of people were asked if they had run a red light in the last year. 150 responded
“yes”, and 185 responded “”no”. Find the probability that if a person is chosen at random,
they have run a red light in the last year.
4. In a survey, 205 people indicated they prefer cats, 160 indicated they prefer dots, and 40
indicated they dont enjoy either pet. Find the probability that if a person is chosen at
random, they prefer cats.
5. Compute the probability of tossing a six-sided die (with sides numbered 1 through 6) and
getting a 5.
60
6. Giving a test to a group of students, the grades and gender are summarized below. If one
student was chosen at random, find the probability that the student was female.
Male
Female
Total
A
8
10
18
B
18
4
22
C
13
12
25
Total
39
26
65
7. The table below shows the number of credit cards owned by a group of individuals. If one
person was chosen at random, find the probability that the person had no credit cards.
Male
Female
Total
Zero
9
18
27
One
5
10
15
Two or more
19
20
39
Total
33
48
81
8. Compute the probability of tossing a six-sided die and getting an even number.
9. Compute the probability of tossing a six-sided die and getting a number less than 3.
10. If you pick one card at random from a standard deck of cards, what is the probability it will
be a King?
11. If you pick one card at random from a standard deck of cards, what is the probability it will
be a Diamond?
61
4.2
4.2.1
Working with Events
Tutorial
Now let us examine the probability that an event does not happen. As in the previous section,
consider the situation of rolling a six-sided die and first compute the probability of rolling a six:
the answer is P (six) = 1/6. Now consider the probability that we do not roll a six: there are 5
outcomes that are not a six, so the answer is P (not a six) = 5/6. Notice that
P (six) + P (not a six) =
1 5
6
+ = =1
6 6
6
Complement of an Event
• The complement of an event is the event “E doesn’t happen”.
• The notation E is used for the complement of event E.
• We can compute the probability of the complement using P (E) = 1
• Notice also that P (E) = 1
P (E).
P (E).
Probability of two independent events
Events A and B are independent events if the probability of Event B occurring is the same
whether or not Event A occurs.
• P(A and B) for independent events:
If events A and B are independent, then the probability of both A and B occurring is
P (A and B) = P (A) · P (B)
where P (A and B) is the probability of events A and B both occurring, P (A) is the probability
of event A occurring, and P (B) is the probability of event B occurring.
• P(A or B):
The probability of either A or B occurring (or both) is
P (A or B) = P (A) + P (B)
62
P (A and B)
Conditional Probability
• Conditional Probability
The probability the event B occurs, given that event A has happened, is represented as
P (B | A)
This is read as“the probability of B given A”
• Conditional Probability Formula
If Events A and B are not independent, then
P (A and B) = P (A) · P (B | A)
4.2.2
Examples
1. If you pull a random card from a deck of playing cards, what is the probability it is not a
heart?
! Solution: There are 13 hearts in the deck, so
P (heart) =
13
1
=
52
4
The probability of not drawing a heart is the complement:
P (not heart) = 1
P (heart) = 1
1
3
=
4
4
2. Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting
a head on the coin and a 6 on the die.
! Solution:
We could list all possible outcomes: {H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}.
Notice there are 2 ⇥ 6 = 12 total outcomes. Out of these, only 1 is the desired outcome, so
the probability is
1
12
3. Are these events independent?
(a) A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second
! Solution:
The probability that a head comes up on the second toss is 1/2 regardless of whether
or not a head came up on the first toss, so these events are independent.
(b) The two events (1) ”It will rain tomorrow in Houston” and (2) ”It will rain tomorrow
in Galveston (a city near Houston).
63
! Solution:
These events are not independent because it is more likely that it will rain in Galveston
on days it rains in Houston than on days it does not.
(c) You draw a card from a deck, then draw a second card being red without replacing the
first.
! Solution:
The probability of the second card being red depends on whether the first card is red
or not, so these events are not independent.
64
4. In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which
are white. If you randomly reach in and pull out a pair of socks and a tee shirt, what is the
probability both are white?
! Solution: The probability of choosing a white pair of socks is
6
10
The probability of choosing a white tee shirt is
3
7
The probability of both being white is
6 3
18
9
· =
=
10 7
70
35
5. Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting
a head on the coin or a 6 on the die.
! Solution:
There are still 12 possible outcomes: {H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}.
By simply counting, we can see that 7 of the outcomes have a head on the coin or a 6 on
the die or both we use or inclusively here (these 7 outcomes are H1, H2, H3, H4, H5, H6,
T6), so the probability is
7
12
As we would expect, 12 of these outcomes have a head, and 16 of these outcomes have a 6 on
the die. since it contains both a head and a 6; the probability of both a head and rolling a
1
6 is 12
So
1 1
+
2 6
1
7
=
12
12
6. Suppose we draw one card from a standard deck. What is the probability that we get a red
card or a King?
! Solution:
Half the cards are red, so
P (Red) =
There are four kings, so
P (King) =
26
52
4
52
There are two red kings, so
P (Red and King) =
2
52
We can then calculate
P (Red or King) = P (Red) + P (King)
65
P (Red and King)
P (Red or King) =
26
4
+
52 52
2
28
7
=
=
52
52
13
7. The table below shows the number of survey subjects who have received and not received
a speeding ticket in the last year, and the color of their car. Find the probability that a
randomly chosen person:
Red car
Not red car
Total
Speeding ticket
15
45
60
No speeding ticket
135
470
605
Total
150
515
665
(a) Has a red car and got a speeding ticket
! Solution:
We can see that 15 people of the 665 surveyed had both a red car and got a speeding
ticket, so the probability is
P (had a red car and got a speeding ticket) =
15
t 0.0226
665
(b) Has a red car or got a speeding ticket
! Solution:
We could have found this probability by:
P (had a red car or got a speeding ticket) = P (had a red car)+
P (got a speeding ticket)
P (had a red car and got a speeding ticket)
P (had a red car or got a speeding ticket) =
150
60
+
665 665
15
195
=
t 0.2932
665
665
8. The table below shows the number of survey subjects who have received and not received
a speeding ticket in the last year, and the color of their car. Find the probability that a
randomly chosen person:
Red car
Not red car
Total
Speeding ticket
15
45
60
No speeding ticket
135
470
605
Total
150
515
665
(a) Has a speeding ticket given they have a red car
! Solution:
Since we know the person has a red car, we are only considering the 150 people in the
first row of the table. Of those, 15 have a speeding ticket, so
P (ticket | red car) =
66
15
1
=
= 0.1
150
10
(b) Has a red car given they have a speeding ticket
! Solution:
Since we know the person has a speeding ticket, we are only considering the 60 people
in the first column of the table. Of those, 15 have a red car, so
P (red car | ticket) =
15
1
= = 0.25
60
4
9. If you pull 2 cards out of a deck, what is the probability that both are spades?
! Solution:
The probability that the first card is a spade is
13
52
The probability that the second card is a spade, given the first was a spade, is
12
51
since there is one less spade in the deck, and one less total cards. The probability that both
13 12
156
·
=
t 0.0588
52 51
2652
10. A home pregnancy test was given to women, then pregnancy was verified through blood
tests. The following table shows the home pregnancy test results. Find
Pregnant
Not Pregnant
Total
Positive test
70
5
75
Negative test
4
14
18
Total
74
19
93
(a) P (not pregnant | positive test result)
! Solution:
Since we know the test result was positive, we are limited to the 75 women in the first
column, of which 5 were not pregnant.
P (not pregnant | positive test result) =
5
1
=
t 0.067
75
15
(b) P (positive test result | not pregnant)
! Solution:
Since we know the woman is not pregnant, we are limited to the 19 women in the second
row, of which 5 had a positive test.
P (positive test result | not pregnant) =
67
5
t 0.263
19
4.2.3
Homework
1. Compute the probability of rolling a 12-sided die and getting a number other than 8.
2. If you pick one card at random from a standard deck of cards, what is the probability it is
3. Giving a test to a group of students, the grades and gender are summarized below. If one
student was chosen at random, what is the probability that a student chosen at random did
NOT earn a C?
Male
Female
Total
A
8
10
18
B
18
4
22
C
13
12
25
Total
39
26
65
4. The table below shows the number of credit cards owned by a group of individuals. If one
person was chosen at random, what is the probability that a person chosen at random has
at least one credit card?
Male
Female
Total
Zero
9
18
27
One
5
10
15
Two or more
19
20
39
Total
33
48
81
5. A six-sided die is rolled twice. What is the probability of showing a 6 on both rolls?
6. A fair coin is flipped twice. What is the probability of showing heads on both flips?
68
7. A die is rolled twice. What is the probability of showing a 5 on the first roll and an even
number on the second roll?
8. Giving a test to a group of students, the grades and gender are summarized below. If one
student was chosen at random, find the probability that the student was female and earned
an A.
Male
Female
Total
A
8
10
18
B
18
4
22
C
13
12
25
Total
39
26
65
9. The table below shows the number of credit cards owned by a group of individuals. If one
person was chosen at random, find the probability that the person was male and had two
or more credit cards.
Male
Female
Total
Zero
9
18
27
One
5
10
15
Two or more
19
20
39
Total
33
48
81
10. A jar contains 6 red marbles numbered 1 to 6 and 8 blue marbles numbered 1 to 8. A marble
is drawn at random from the jar. Find the probability the marble is red or odd-numbered.
11. A jar contains 4 red marbles numbered 1 to 4 and 10 blue marbles numbered 1 to 10.
A marble is drawn at random from the jar. Find the probability the marble is blue or
even-numbered.
69
12. Giving a test to a group of students, the grades and gender are summarized below. If one
student was chosen at random, find the probability that a student chosen at random is
female or earned a B.
Male
Female
Total
A
8
10
18
B
18
4
22
C
13
12
25
Total
39
26
65
13. The table below shows the number of credit cards owned by a group of individuals. If one
person was chosen at random, find the probability that a person chosen at random is male
or has no credit cards.
Male
Female
Total
Zero
9
18
27
One
5
10
15
Two or more
19
20
39
Total
33
48
81
14. Compute the probability of drawing the King of hearts or a Queen from a deck of cards.
15. Compute the probability of drawing a King or a heart from a deck of cards.
16. A jar contains 5 red marbles numbered 1 to 5 and 8 blue marbles numbered 1 to 8. A
marble is drawn at random from the jar. Find the probability the marble is
(a)Even-numbered given that the marble is red.
(b)Red given that the marble is even-numbered.
70
17. A jar contains 4 red marbles numbered 1 to 4 and 8 blue marbles numbered 1 to 8. A
marble is drawn at random from the jar. Find the probability the marble is
(a)Odd-numbered given that the marble is blue.
(b)Blue given that the marble is odd-numbered.
18. Compute the probability of flipping a coin and getting heads, given that the previous flip
was tails.
19. Find the probability of rolling a “1 on a fair die, given that the last 3 rolls were all ones.
20. Suppose a math class contains 25 students, 14 females (three of whom speak French) and
11 males (two of whom speak French). Compute the probability that a randomly selected
student speaks French, given that the student is female.
21. Suppose a math class contains 25 students, 14 females (three of whom speak French) and
11 males (two of whom speak French). Compute the probability that a randomly selected
student is male, given that the student speaks French.
71
4.3
4.3.1
Counting Methods
Tutorial
When we get to the probability situations a bit later in this chapter we will need to count some
very large numbers, like the number of possible winning lottery tickets. One way to do this
would be to write down every possible set of numbers that might show up on a lottery ticket,
but believe me: you don’t want to do this.
Basic Counting
Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks) and five choices for a main course (hamburger, sandwich, quiche, fajita or pizza). If you
are allowed to choose exactly one item from each category for your meal, how many di↵erent
meal options do you have?
! Solution 1: One way to solve this problem would be to systematically list each possible
meal:
soup
soup
+
+
+
+
+
hamburger
fajita
sandwich
pizza
quiche
soup
soup
+
+
+
+
+
sandwich
pizza
quiche
hamburger
fajita
soup
+
+
+
+
+
quiche
hamburger
fajita
sandwich
pizza
Assuming that we did this systematically and that we neither missed any possibilities nor
listed any possibility more than once, the answer would be 15. Thus you could go to the
restaurant 15 nights in a row and have a di↵erent meal each night.
! Solution 2: Another way to solve this problem would be to list all the possibilities in a table:
soup
hamburger
soup+burger
bs+burger
sandwich
soup+sw
bs+sw
quiche
soup+qc
bs+qc
fajita
soup+fajita
bs+fajita
pizza
soup+pizza
bs+pizza
if we didn’t really care what the possible meals are, only how many possible meals there
are, we could just count the number of cells and arrive at an answer of 15, which matches
our answer from the first solution.
72
! Solution 3: Let’s draw a tree diagram:
This is called a ”tree” diagram because at each stage we branch out, like the branches on a
tree. In this case, we first drew five branches and then for each of those branches we drew
three more branches. We count the number of branches at the final level and get 15.
Basic Counting Rule
If we are asked to choose one item from each of two separate categories where there are m
items in the first category and n items in the second category, then the total number of available
choices is m · n.
Odds
1. odds in favor of an event E:
P (E)
P (Ē)
or
P (E)
P (not E)
P (Ē)
P (E)
or
P (not E)
P (E)
2. odds against of an event E:
73
3. If the odds in favor of an event E are a to b (or a:b), then
(a) The odds against of the event E are b to a (or b:a).
(b) The probability of event E is
P (E) =
a
a+b
.
(c) The probability of complement of event E is
P (Ē) = P (not E) =
b
a+b
.
4.3.2
Examples
1. There are 21 novels and 18 volumes of poetry on a reading list for a college English course.
How many di↵erent ways can a student select one novel and one volume of poetry to read
during the quarter?
! Solution: There are 21 choices from the first category and 18 for the second, so there are
21 · 18 = 378
possibilities.
2. Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or
breadsticks), five choices for a main course (hamburger, sandwich, quiche, fajita or pasta)
and two choices for dessert (pie or ice cream). If you are allowed to choose exactly one item
from each category for your meal, how many di↵erent meal options do you have?
! Solution: There are 3 choices for an appetizer, 5 for the main course and 2 for dessert, so
there are
3 · 5 · 2 = 30
possibilities.
3. A quiz consists of 3 true-or-false questions. In how many ways can a student answer the
quiz?
! Solution: There are 3 questions. Each question has 2 possible answers (true or false), so the
2·2·2=8
di↵erent ways.
4. If the probability of an event E is
event?
1
4,
what are the odds in favor and odds against of the
74
! Solution:
P (E) =
odds in favor:
1
4
P (not E) = 1
P (E) = 1
1
3
=
4
4
P (E)
1 3
= : =1:3
P (not E)
4 4
odds against:
P (not E)
3 1
= : =3:1
P (E)
4 4
5. If the odds in favor of an event E is 3 to 5, what are the probability of the event E and the
probability of the complement of the event E?
! Solution:
P (E) =
3
3
=
3+5
8
P (Ē) = P (not E) =
75
5
5
=
3+5
8
4.3.3
Homework
1. A boy owns 2 pairs of pants, 3 shirts, 8 ties, and 2 jackets. How many di↵erent outfits can
he wear to school if he must wear one of each item?
2. At a restaurant you can choose from 3 appetizers, 8 entrees, and 2 desserts. How many
di↵erent three-course meals can you have?
3. How many three-letter “words” can be made from 4 letters “FGHI” if
(a) repetition of letters is allowed
(b) repetition of letters is not allowed
4. How many four-letter “words” can be made from 6 letters “AEBWDP” if
(a) repetition of letters is allowed
(b) repetition of letters is not allowed
5. All of the license plates in a particular state feature three letters followed by three digits
(e.g. ABC 123). How many di↵erent license plate numbers are available to the state’s
Department of Motor Vehicles?
76
4.4
4.4.1
Permutations and Combinations
Tutorial
Now, why would we want to use the complicated formula when it’s actually easier to use the
Basic Counting Rule, as we did in the last section? Well, we won’t actually use this formula all
that often, we only developed it so that we could attach a special notation and a special definition
to this situation where we are choosing r items out of n possibilities without replacement and
where the order of selection is important or not important.
Factorial
n! = n · (n
1) · (n
2) · · · 3 · 2 · 1
Permutations
n Pr
= n · (n
1) · (n
2) · · · (n
r + 1)
We say that there are n Pr permutations of size r that may be selected from among n choices
without replacement when order matters.
It turns out that we can express this result more simply using factorials.
n Pr
=
n!
(n
r)!
Combinations
n Cr
=
n Pr
r Pr
We say that there are n Cr combinations of size r that may be selected from among n choices
without replacement when order doesn0 t matter.
We can also write the combinations formula in terms of factorials:
n Cr
4.4.2
=
(n
n!
r)!r!
Examples
1. How many di↵erent ways can the letters of the word MATH be rearranged to form a fourletter code word?
77
! Solution: This problem is a bit di↵erent from previous section counting. Instead of choosing
one item from each of several di↵erent categories, we are repeatedly choosing items from
the same category (the category is: the letters of the word MATH) and each time we choose
an item we do not replace it, so there is one fewer choice at the next stage:
we have 4 choices for the first letter (say we choose A), then 3 choices for the second (M,
T and H; say we choose H), then 2 choices for the next letter (M and T; say we choose M)
and only one choice at the last stage (T). Thus there are 4 · 3 · 2 · 1 = 24 ways to spell a code
word.
2. How many ways can five di↵erent door prizes be distributed among five people?
! Solution: There are 5 choices of prize for the first person, 4 choices for the second, and so
on. The number of ways the prizes can be distributed will be
5! = 5 · 4 · 3 · 2 · 1 = 120
ways.
3. Eight sprinters have made it to the Olympic finals in the 100-meter race. In how many
di↵erent ways can the gold, silver and bronze medals be awarded?
! Solution: Using the Basic Counting Rule, there are 8 choices for the gold medal winner, 7
remaining choices for the silver, and 6 for the bronze, so there are
8 · 7 · 6 = 336
ways the three medals can be awarded to the 8 runners.
4. I have nine paintings and have room to display only four of them at a time on my wall. How
many di↵erent ways could I do this?
! Solution: Since we are choosing 4 paintings out of 9 without replacement where the order of selection is important
there are
9 P4 = 9 · 8 · 7 · 6 = 3, 024
permutations.
5. How many ways can a four-person executive committee (president, vice-president, secretary,
treasurer) be selected from a 16-member board of directors of a non-profit organization?
78
! Solution: We want to choose 4 people out of 16 without replacement and where the order
of selection is important. So the answer is
16 P4
= 16 · 15 · 14 · 13 = 43, 680
6. A group of four students is to be chosen from a 35-member class to represent the class on
the student council. How many ways can this be done?
! Solution: Since we are choosing 4 people out of 35 without replacement where the order of selection is not important
there are
35 · 34 · 33 · 32
= 52, 360
35 C4 =
4·3·2·1
combinations.
7. How many di↵erent combinations of three companies can be selected from the 16 companies?
! Solution: We want to choose 3 companies out of 16 without replacement and where the
order of selection is not important. So the answer is
16 C3
=
16 · 15 · 14
= 560
3·2·1
79
4.4.3
Homework
1. A pianist plans to play 4 pieces at a recital. In how many ways can she arrange these pieces
in the program?
2. In how many ways can first, second, and third prizes be awarded in a contest with 210
contestants?
3. Seven Olympic sprinters are eligible to compete in the 4 x 100 m relay race for the USA
Olympic team. How many four-person relay teams can be selected from among the seven
athletes?
4. In how many ways can 4 pizza toppings be chosen from 12 available toppings?
5. At a baby shower 17 guests are in attendance and 5 of them are randomly selected to receive
a door prize. If all 5 prizes are identical, in how many ways can the prizes be awarded?
6. In the 6/50 lottery game, a player picks six numbers from 1 to 50. How many di↵erent
choices does the player have if order doesnt matter?
7. In a lottery daily game, a player picks three numbers from 0 to 9. How many di↵erent
choices does the player have if order doesnt matter?
80
4.5
Chapter Test
1. If you pick one card at random from a standard deck of cards, what is the probability it will
be a King of hearts?
2. If we toss a coin and roll a 6-sided die at the same time, what is the probability of getting
a tail on the coin and 4 on the die?
3. The table below shows the number of credit cards owned by a group of individuals. If one
Gender
Male
Female
Total
zero
9
18
one
5
10
More than one
16
12
Total
person was chosen at random. Find the following probabilities;
(a) The person was a female.
(b) The person was male and having no credit cards.
(c) The person was male or having one credit card.
(d) Find the probability that the person was a female given that the person had more than
one credit card.
81
4. Two 6-sided dies are rolled.
(a) List all the possible outcomes.
(b) Find the probability by listing all favorable outcomes, that the sum of two rolls is 6.
5. Suppose probability of an event E is
1
4.
What are the odds against of the event E?
6. Suppose the odds in favor of an event E are 12 to 13. What is the probability of the event
E?
7. In a survey, 205 people indicated they prefer cats, 160 indicated they prefer dogs, and 40
indicated they don’t enjoy either pet. A person is chosen at random. Find the following
probabilities;
(a) The person prefer cats.
(b) The person do not prefer dogs.
82
8. A bag contains 5 red balls numbered 1 to 5 and 8 blue balls numbered 1 to 8. A ball is
selected at random from the bag. Find the probability the ball is
(a) odd-numbered given that the ball is blue.
(b) blue given that the ball is odd-numbered.
9. At a restaurant you can choose from 3 appetizers, 5 entrees, and 4 desserts. How many
di↵erent three-course meals can you have?
83

attachment

Tags:
total weight of students

New average weight

rectangular shed

rectangle is measurement

error method to estimate a solution

User generated content is uploaded by users for the purposes of learning and should be used following Studypool’s honor code & terms of service.

## Reviews, comments, and love from our customers and community:

This page is having a slideshow that uses Javascript. Your browser either doesn't support Javascript or you have it turned off. To see this page as it is meant to appear please use a Javascript enabled browser. Peter M.
So far so good! It's safe and legit. My paper was finished on time...very excited! Sean O.N.
Experience was easy, prompt and timely. Awesome first experience with a site like this. Worked out well.Thank you. Angela M.J.
Good easy. I like the bidding because you can choose the writer and read reviews from other students Lee Y.
My writer had to change some ideas that she misunderstood. She was really nice and kind. Kelvin J.
I have used other writing websites and this by far as been way better thus far! =) Antony B.
I received an, "A". Definitely will reach out to her again and I highly recommend her. Thank you very much.  