This document provides an overview of key concepts in statistics and probability covered in Chapter 4, including: - Methods for obtaining unbiased samples and representing data in line, bar, and circle graphs. - Calculating the mean, median, and mode of data sets and understanding their relationships. - Applying counting principles and calculating probabilities, including for independent vs. dependent events. - Worked examples are provided to demonstrate computing probabilities and interpreting data distributions and relationships.

1. STATISTICS AND
PROBABILITY
CHAPTER 4

2. STAT. & PROBABILITY
4.1 Sampling, Line, Bar and Circle
Graphs
4.2 The Mean, Median and Mode
4.3 Counting Problems and
Probability

3. Unbiased sample is a
random sample so that each
member has an equal
opportunity of being
selected.
4.1 Unbiased Samples

4. 4.1 Example
1. A college president wants to find out which
courses are popular with students. What
procedure would be most appropriate for
obtaining an unbiased sample of students?
A. Survey a random sample of students from the
English Department.
B. Survey the first hundred students from an
alphabetical listing.
C. Survey random sample of students from list of
entire student body.
D. Survey random sample of students from list of
entire student body.

5. 4.1 Line and Bar Graphs
2. The graph shows
the yearly average
temperature from 1980
to 1985. What is the
difference between the
highest and lowest?
73
77
77 - 73 = 4
74
75
76
A. 73 ºF B. 77 ºF C. 1 ºF D. 4º F
ºF

6. 4.1 Circle Graphs
4. The number of people employed
in different work areas in a
manufacturing plant are
represented by the circle graph.
What percent are represented in
Sales and Administration
combined?
6
3
3
23
Total = 40
8
40 100
40 800 20
  
p
p p
, ,
A. 25%
A
5
S
B. 20% C. 2.5% D. 7.5%

7. 4.1 Relations from Data
0
2
4
6
8
10
1 2 3 4 5 6
Year
Trade-in Value for
A $15000 Car
7. Consider the following
graph showing the value
of a $15,000 car after 1, 2,
3, 4, 5 and 6 years. In
what year did the price of
the car begin to stabilize?
A. 6 B. 5 C. 4 D. 3

8. 4.1 Predictions from Data
Strong Positive Strong Negative None
Never select a choice that says one “caused “ the
other, as the above graphs do not contain sufficient
information to determine cause and effect.
Weak Positive Weak Negative

9. D. There is a positive association between increase
in ads and increase in sales
4.1 Predictions from Data
9. The graph shows number of
TV adds shown & number of
cars sold during a 14 wk.
period. Which best describes
the relationship between the
number of ads and cars sold?
20
14
Ads
Cars
sold
A. No Apparent association
B. Increase in ads caused increase in sales
C. Increase in sales caused increase in ads

10. 4.2 Mean, Median & Mode
Mean - sum of elements in set divided by
number of elements in set.
Median - middle element when arranged in
order or average of two middle elements.
Mode - most frequent element(s). If no
element occurs more than once then there
is no mode.

11. 4.2 Mean, Median & Mode
1. Find mean, median & mode of the data in
this sample: 6, 15, 24, 23, 29, 22, 21, 29, 29
Mode is 29 (most frequent)
Median is 23 (middle)
A. 22, 23,29
B.17.5, 22,29
C. 29, 23,22
D. 23, 22,29
Average too large!
Arrange in order:
6, 15, 21, 22, 23, 24, 29, 29, 29
(6+15+21+22+23+24+29+29+29)/9
198/9=22 the mean

12. 4.2 Relationships & Graphs
NORMAL
Mean = Median = Mode
100
Mean < Med. < Mode
SKEWED Left SKEWED
Right
0 100
Mode < Med.< Mean
0

13. 4.2 Example
0
5
10
15
20
20 65 95
Scores
Test Scores for 40 Students
3. In a literature class, half
scored 95 on a test. Most of
the remaining scored 65,
except for a few who scored
20. Which is true?
Mean < Med. < Mode
A. The mode equals the mean.
B. The median is greater than mode
C. The mode is greater than mean
D. The mean is greater than mode
Half scored
95 means
mode =95

14. 4.2 Applications
Income Level
% of
Families
0 - $4999 6
$5000 - $9999 11
$10,000 - $14,999 10
$15,000 - $24,999 19
$25,000 - $34,999 16
$35,000 - $49,000 17
$50,000 - $74,999 13
$75,000 and over 7
8. The table shows the
percent distribution of
households by income
level in 1990. What
percent of the families
have income of at least
$35,000?
A. 47
17+13+7=37
B. 53 C. 26 D. 37

15. 4.3 The Counting Principle
To count the number of ways a sequence
of events can happen, multiply the amount
of ways each can occur.
1. Students are asked to rank 4 instructors
from best to worst. How many different
ways can the 4 instructors be ranked?
_______ x ________ x __________ x ________
1st 2nd 3rd 4th
4 3 2 1
A. 1 B. 4 C.
64
D. 24

16. 4.3 Computing Probability
It must always be the case: 0≤P(E)≤1
P(not E) = 1- P(E)
P(A or B) = P(A) +P(B) - P(A and B)
A and B are called mutually exclusive when
P(A and B)=0
and then P(A or B) = P(A) +P(B)

17. 4.3 Computing Probability
To calculate P(A and B)
P(A and B)= P(A)·P(B|A)
P(B|A) is the probability of B given A has
occurred.
A and B are called independent events if and only if
P(B|A)=P(B)
and then P(A and B) = P(A)·P(B)
Two events are dependent if and only if the
occurrence of one event affects the probability of
the other.

18. 4.3 Example
A survey at a college indicated that 90% of those
taking the Essay portion of CLAST passed. If only
70% of those taking Math passed, what is the
probability that a randomly selected student will
fail both the Essay and the Math portion?
Since 70% passed math, 30% or 3/10 failed and
Since 90% passed essay, 10% or 1/10 failed
And we will assume the two events are independent

19. 4.3 Example
A survey at a college indicated that 90% of those
taking the Essay portion of CLAST passed. If only
70% of those taking Math passed, what is the
probability that a randomly selected student will
fail both the Essay and the Math portion?
P(failed Math)=3/10 and P(failed Essay)=1/10
P(failed Math and failed Essay)=(3/10)(1/10)
=3/100

20. 4.3 Example
Two common sources of protein for US adults are
beans & meat. If 75% of US adults eat meat, 80%
eat beans and 70% eat both meat & beans, what
is the probability that a randomly selected adult
eats meat or beans?
P(meat or beans)
=P(meat) or P(beans) - P(both)
=75% + 80% - 70% = 85%
20
17
100
85



21. 4.3 Probability Application
8. The table gives the percent of students
at a university by sex and student
classification. Find the probability that a
randomly selected student is a senior.
11% + 9% = 20%,
A. 0.20
20
.
0
100
20
%
20 

Soph. Junior Senior
Fresh.
Male
Female
16% 13% 10% 11%
14% 15% 12% 9%
B. 0.30 C. 0.52 D. 0.49

22. REMEMBER
MATH IS FUN
AND …
YOU CAN DO IT