Beginning Algebra
Answer/Discussion to Practice Problems
Tutorial 34: Central Tendencies
Answer/Discussion
to 1a
The number of cd's sold by Dave's Discs for the last 6 days are given
in the table:
Day

Day 1

Day 2

Day 3

Day 4

Day 5

Day 6

CD's

15

10

10

10

18

15

The mean is the average of the number of discs
sold.
So we need to sum up all of the CD's sold and then divide by 6, since
there are 6 days: 

*(sum of cd's sold)/(# of days)

The median is the middle value.
We need to list the numbers in numeric order:
10, 10, 10, 15, 15, 18
If we pick 10 (the third one) for our median we have two values below
it and three above it. If we pick 15 (the first one) for our median
then we have three values below it and two above it. So neither of
those values are the median. This does not mean we don't have
a median.
Note how there is an even number of values listed. If that
is the case, we need to draw a line down the middle of the list and take
the mean of the two numbers next to that line.
10, 10, 10  15, 15, 18
The mean of 10 and 15 is 

*Find number exactly in the middle of 10 and
15

12.5 is the median. It is the value that is right smack
dab in the middle of this list of values. 
The mode is the value that occurs the most
often.
It helps to list the numbers in order to find the mode.
10, 10, 10,
15, 15, 18
Note how 10 occurs three times, which is the value that occurs the most.
10 is the mode. 
Answer/Discussion
to 2a
A student received scores of 92, 83, and 71 on three quizzes.
If tests count twice as much as quizzes, what is the lowest score that
the student can get on the next test to achieve a mean of at least 80? Keep in mind that the test score counts twice instead of of time.
So when we set this up we need to make sure that we notate that properly. 

*(sum of scores)/(# of scores) = mean
*Need 2 x's since
tests count twice
*Solve for x (missing
test)
*Inverse of div. by 5 is mult. by 5
*Inverse of add 246 is sub. 246

The student would have to score a 77 on the next test to have a
mean of 80. 
Answer/Discussion
to 3a
Find the range and standard deviation of the list of scores that were
made by a football team during a season:
7, 21, 21, 17, 17, 14, 7, 0 
I don't know about you, but I find it easier to work with a group of
numbers like this when they are in chronological order. Let's put
them in order from lowest to highest: 0, 7, 7, 14, 17, 17, 21, 21.
Let's find the range. What
do you think it is?
Looking at the difference between the largest value, which is 21 and
the smallest value, which is 0, it looks like the range is 21.
Now lets tackle the standard deviation. 
Step 1: Find the mean of
the values of the data set. 
So we need to sum up all of the values and then divide by 8, since
there are 8 numbers: 

*(sum of values)/(# of values)
*Add numerator
*Divide by 8

Step 2: Find the difference between the mean and each
separate value of the data set,
AND
Step 3: Square each difference found in step 2,
AND
Step 4: Add up all of the squared values found in
step 3.
x

x  13


0

13

169

7

6

36

7

6

36

14

1

1

17

4

16

17

4

16

21

8

64

21

8

64


SUM:

402


Step 5: Divide the sum found in step 4 by the
number of data values in the set
AND
Step 6: Find the nonnegative square root of the quotient
found in step 5. 

*Square root of [(sum of diff. squared)/(#
of values)]

The standard deviation is approximately 7.089. 
Answer/Discussion
to 4a
Find the mean of the frequency distribution.
x

f

9

3

10

5

15

4

20

10

Total

22


As requested, I'm going to use the frequency distribution to set up
my mean formula. Instead repeating numbers in my sum, I'm going to
indicate a repetition by taking that value times the number of times it
occurs in the list. For example, 20 occurs 10 times. Instead
of writing it out 10 times in my sum, I will find 20(10) which is the equivalent. 

*(sum of scores)/(# of scores)
*calculate numerator
*divide

Last revised on August 7, 2011 by Kim Seward.
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