Measures of Central

Tendency

Data can be difficult to perceive in raw form

• 4 5 56 6 4 23 4 5 6 76 7 56 5 54 4 3 5 6 7 8 9 9 00 9 8 78 6 6 5 4 4 3 32 3 4 5 56 6 7 8 8 7 65 6 7 8 7 67 6 5 4 43 3 2 2 2 21 1 2 3 4 5 5 4 3 3 4 5 56 6 7 8 9 90 89 78 7 6 5 4 4 3 3 3 4 5 56 6 7 7 8 8 9 8 7 6 65 54 34 4 5 6 76 7 8 98 99 78 6 5 4 3 4 45 55 6 6 7 7 88 9 9 0 0 0 08 6 5 5 4 5 56 6 5 5 6 7 7 8 89 9 9 8 7 7 6 5 5 45 4 5 65 6

A set of numeric data can be described using a single value.

M of CT are values that describe the center of a body of data.

movie

The Mean

Found by dividing the sum of all the data by the number of pieces of data.

x =x

n

Where n is the number of values in the set.

Imagine 3 blocks on a plank.

Where would you place the fulcrum so the blocks would be balanced?

Weighted Mean

Suppose the blocks from the first example all had different weights.

Where x represents each data value, and w represents it’s weight (usually a percentage), or frequency.

x =xw

w

Find the mean for weighted data

EX. Final marks are weighted accordingly:

Tests: 30%

Quizzes: 20%

Assignments: 10%

The student has 78%, 63%, and 12% respectively.

a) Calculate the final mark:

x = 78(0.3) + 63(0.2) + 12(0.1)

(0.3) + (0.2) + (0.1)

= 62

b) What does the student need on the final exam (40%) to get a mark of 70?

62(0.6) + X(0.4) = 70

0.4X = 70 – 37.2

Let X be the exam percent.

37.2 + 0.4X = 70

0.4X = 32.8

X = 82

The student needs 82% on the final exam to raise a final mark of 62% to 70%

A sample of car owners were asked how old they were when they bought their first car.

The results were reported in the following table

Calculate the mean age of the group.

Find the mean for grouped data.

Age Frequency Midpoint (age)

F X M

16-20 10 18 10 X 18 = 180

21-25 18 23 18 X 23 = 414

26-30 12 28 12 X 28 = 336

31-35 8 33 8 X 33 = 264

36-40 2 38 2 X 38 = 76

Age Frequency Midpoint (age)

F X M

16-20 10 18 10 X 18 = 180

21-25 18 23 18 X 23 = 414

26-30 12 28 12 X 28 = 336

31-35 8 33 8 X 33 = 264

36-40 2 38 2 X 38 = 76

Age Frequency Midpoint (age)

F X M

16-20 10 18 10 X 18 = 180

21-25 18 23 18 X 23 = 414

26-30 12 28 12 X 28 = 336

31-35 8 33 8 X 33 = 264

36-40 2 38 2 X 38 = 76

To calculate the mean

X = 180 + 414 + 336 + 264 + 76

x =(f X m)

f

10 + 18 + 12 + 8 + 2

X = 25.4

The mean age is 25.4

Median: The middle value of an ordered set

Note: If the set has an even number of data points, find the mean of the two middle-most values.

Find the median:

4, 2, 8, 5, 9, 1, 4, 6, 8, 2, 9

Re-order:

1, 2, 2, 4, 4, 5, 6, 8, 8, 9, 9

Identify the median: 5

Mode: The most frequent value

Find the mode:

1 3 5 5 7 8 9 11 14 17

The mode is 5

Outliers

An element of the data set that is very different from the others.

If there is sufficient reason, this element may be ignored.

An example of an outlierA student wanted to determine the most

favorable temperature that people wanted to experience on their holidays.

She took a survey and asked the following question:

What temperature do you prefer when you are looking to travel on the holiday?

Her results were as follows:

All temperatures are in OC

32 31 29 26 4044 33

34 -20 39 41 2829 35

What measure should you use?

If data contains outliers, use the median.

If the data are roughly symmetric, use the mean or the median.

If the data are qualitative (eg election), use the mode

Page 1821a, 2, 5