Download section 13.5 and 13.6 class notes

Section 13.5 and 13.6 class notes:
Section 13.5 - Measures of Central Tendency
An “average” is a number that is representative of a group of data. There are at least four different averages; the
MEAN, the MEDIAN, the MODE, and the MIDRANGE.
The Mean, x , is the sum of the data divided by the number of pieces of data. x 
x
n
Example: Five people waiting at a Jack in the Box drive through have ages 27, 18, 48, 34 and 48. What is the Mean
age of this group?
The Median is the value in the middle of a “ranked” set of data.
From above 27, 18, 48, 34 , 48
ranked =
Median = the “middle” value =
(note: the median of a set of data that contains an even number of values is the Mean average of the middle two.
57
 6)
i.e the Median of 2,3,3,5,7,7,9,12 is
2
The Mode is the piece of data that occurs most often. For the ages of the people in the Jack in the Box drive though
…. 27, 18, 48, 34 and 48 the Mode =
If there is more than one value that occurs “most often” then we list both of them as Modes.
i.e the Mode of 2,3,3,5,7,7,9,12
is
Note: 1, 3, 5, 7, 9 has no Mode.
The Midrange = lowest value + highest value
2
So for our Jack in the box drive though
Midrange =
Example: A mean average of 80 or greater for five exams is needed for a final grade of B in a course. Sue’s first
four exam grades were 73,69,85 and 80. What grade does Sue need on the fifth exam to get a B for the course?
Solution:
So Sue must earn a grade of
or better on the fifth exam in order to earn a B for the course.
Let’s try a few problems from the text.
30. Malcom Sander’s mean average on five exams is 76. Find the sum of his scores.
37. (slightly modified with different numbers) A mean average of 70 on 6 exams is needed to pass a course. On
her first 5 exams, Sheryl received grades of 55, 77, 72, 64, and 95.
a)
What grade must she receive on her last exam to pass the course?
b)
An average of 80 is needed to get a B in the course. Is it possible for Sheryl to get a B? If so, what grade
must she receive on the 6th exam?
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 class?
Section 13.6 - Measures of Dispersion
The measures of central tendency by themselves are not always sufficient information to analyze a situation and
make a decision.
Example: Company A makes airplane engines that have an average (mean) life of 1000 hours flying time before
they fail.
Company B makes airplane engines for the same price that have an average (mean) life of 950 hours
flying time before they fail.
If all you look at is the average (mean) life span it may appear that Company A’s is the better deal..BUT..
What if Company A’s engine with an average lifespan of 1000 hours means that half of the engines last 1500 hours
and the other half last 500 hours AND What if Company B’s engines with an average lifespan of 950 hours means
that half of their engines last 975 hours and the other half last 925 hours.
IF this was the case then Company B’s engines are more consistent and consequently more reliable and they may be
the better choice if you don’t want to gamble with reliability.
This example illustrates the importance of knowing something about the SPREAD or VARIABILITY of the data.
Measures of dispersion are used to indicate the spread of the data.
Range: the range of a set of data is the difference between the highest and lowest values in a set of data; it indicates
the total spread of the data.
Example: What is the range of the following set of data? 23,18,30,17,24,21,5,20,23,19
Range =
The second measure of dispersion is called STANDARD DEVIATION, it measures how much data differ from the
mean. The lower case letter “s” or the Greek (sigma)  is used to denote standard deviation. The “s” is used when
the standard deviation of a sample is calculated. The  is used when the standard deviation of the entire population
is calculated.
The larger the spread of data, the larger the standard deviation.
Consider the following two sets of data.
A. 5,8,9,10,12,13
B. 8,9,9,10,10,11
BOTH have a Mean = 9.5 but one set of data is “grouped” much closer to it’s Mean than the other one is.
A.
B.
x
5
6
7
x
x
x
8
9
10
Mean
11
x
x
12
13
5
6
7
x
x
x
x
x
x
8
9
10
11
Mean
12
13
Sometimes only a very small standard deviation is desirable or acceptable.
Example: If a cereal box is supposed to contain 8 oz of cereal then consumers expect 8 oz.
If the amount varies too much from box to box- sometimes underfilling, sometimes overfillingthe manufacturer will soon be in trouble with consumer groups (if underfilling)
or the investors in the company (if overfilling…and losing potential profit).
Other times a larger spread of data might be expected…for example IQ scores are expected to exhibit a considerable
spread about the mean because everyone is different.
How to calculate standard deviation…
s
 X  M 
2

n 1
for a sample
 X  M 
2
n
for a population
(most of the time WE will be analyzing a sample so we will be using the first one)
Let’s look at our two sets of data from before….
A. 5,8,9,10,12,13
B. 8,9,9,10,10,11
Remember BOTH had a Mean, M = 9.5
X
Data
X
Data
(X-M)2
(Data-Mean)2
X-M
Data-Mean
(X-M)2
(Data-Mean)2
X-M
Data-Mean
8
9
9
10
10
11
5
8
9
10
12
13
 X  M 
 X  M 
n 1
Standard deviation = s 
2
 X  M 

 X  M 
2

 X  M 
n 1
n 1
2

Standard deviation = s 
2

2

 X  M 
2
n 1
We can use our calculator to compute standard deviation. I will show you how to do this in class.
Let’s try a few problems from the text.
Section 13.6 problem 16: Determine the range and standard deviation of the set of data, round to the nearest
hundredth when applicable. 10,10,14,16,8,8

Section 13.6 problem 27 with a slight modification.
Six people were asked to determine the amount of money they were carrying, to the nearest dollar. The results
were: $20, $15, $0, $6, $45, $80
a)
Determine the range and standard deviation of the amounts.
b)
Add $5 to each amount. Determine the range and standard deviation of the new amounts.
c)
Compare your answers to part a and b.