Measures of Central Location

Week 3

Break Even in Practice (1)

• Contribution per unit = AR – AVC

• c/u = £100 - £40 = £60

• BEP = TFC = £5,000,000

c/u 60

• BEP = 83,333 units

2

Break Even in Practice (2)

• Apple sold 1 million units in 3 days

• Much more than BEP (83,333 units)

• So Apple did break even in first three days of

selling product

• Apple might have set a ‘budgeted output’

(target) of 100,000 units

• TVC £40,000,000

• TFC £5,000,000

• TC = TFC + TVC = £45,000,000

3

Central Location in Practice (1)

Microsoft

• Mean closing price ($) 213.50

8

$26.7

Google

• Mean closing price ($) 4,284.52

8

$535.6 4

Central Location in Practice (2)

• Deviation between opening and closing

price per day

• Return: deviation as a % of opening price

• Microsoft (1st day)

• Return = - 0.28 = - 1.01%

28.03

etc.

5

Central Location in Practice (3) • Average return resulting from same news

events

• Microsoft = - 8% = - 1%

8

• Google = + 0.04% = + 0.01%

8

Using ‘mean’ return suggests that Google is

less risky in share price response to same

news events over this period

6

7

Notation (1)

• Σ = sigma (sum of)

321

3

1

XXXXi

i

432

4

2

XXXXi

i

n

n

i

i XXXX ....21

1

Notation (2)

• 4 throws of dice

4 2 6 1

X1 X2 X3 X4

= 4 + 2 + 6 + 1 = 13

8

Notation (3)

Suppose some numbers occur more than once

Three (F1 ) number 1’s (X1)

Four (F2 ) number 2’s (X2)

Two (F3) number 3’s (X3)

Sum = 3 (1) + 4 ( 2) + 2 (3)

= F1X1 + F2X2 + F3X3

9

10

Notation (4)

Where Fi = Frequency

Xi = Variable value

332211

3

1

XFXFXFXF i

i

i

jji

j

i

i XFXFXFXF ...2211

1

11

Measures of Central Location

• Types of ‘average’

– Mode - Item of data occurring most often

– Median - Item in middle of data when

arranged in order

– Mean - simple average, found by adding all

data and dividing total by number of items

12

Central location: Ungrouped data

• Ungrouped data: data which is available

for each separate item

• Array: items of data arranged in order e.g.

1, 3, 5, 7 ascending array

7, 5, 3, 1 descending array

13

Arithmetic Mean: ungrouped data

• Arithmetic mean is the simple average

where = arithmetic mean

Xi = value of each item

n = number of observations

n

X

X

n

i

i 1

XXX

Example

X1 X2 X3

5 20 8

=

14

3

3

1

i

iX

X

15

Median: ungrouped data

• Median: that value which divides the data into two equal halves; 50% of values lying below and 50% above the median.

• Array: place data in numerical order – whether rising or falling

• Median position is n + 1

2

where n = number of values

• Median value is that value which corresponds to the median position

16

Mode: ungrouped data

• Mode: that value which occurs most often

(i.e. with the highest frequency)

• Modal class interval: that class interval in

which the mode value falls

17

Example: central location for

ungrouped data

• The following data measures the

attention span in minutes of 15

undergraduates in a sociology lecture.

4, 6, 7, 8, 8, 8, 8, 9, 9, 10, 11, 12, 14, 15, 18

a) Find the arithmetic mean

b) Find the median

c) Find the mode

18

Solutions

a) Arithmetic mean = minutes

b) Median position = 15 + 1

2

i.e. 8th in an array

Median value = 9 minutes

c) Mode value = 8 minutes

8.915

147

19

Central location: grouped data

• Grouped data: data which is only

available in grouped form e.g. class

intervals in frequency table

• Class mid-points: we assume that the data

in any class interval all fall on the class

mid-point. Put another way, the data are

equally spread along any given class

interval

20

Mean grouped data

• Where Fi = frequency

of ith class interval

Xi = mid-point of

ith class interval

j = number of

class intervals

j

i

i

j

i

ii

F

XF

X

1

1

Note: simplifying

assumption: all values in

a class interval are

equally spread along that

interval

21

Median: grouped data

LCB + Class Width x No. of observations to median position

Total no. of observations in median class interval

where

LCB = lower class boundary (of median class interval)

22

Example: Find the mean and median

heights of students from the data below X Heights

(cm)

F Frequency

(no of students)

150 and under 155

155 and under 160

160 and under 165

165 and under 170

170 and under 175

175 and under 180

180 and under 185

185 and under 190

1

1

2

3

6

2

4

1

20

23

Xi

(Class mid-points)

Fi FiXi

152.5

157.5

162.5

167.5

172.5

177.5

182.5

187.5

1

1

2

3

6

2

4

1

20

152.5

157.5

325.0

502.5

1,035.0

355.0

730.0

187.5

ΣFiXi = 3,445.0

24

Solution: Mean of grouped data

172.25cm = 1.723 metres

1

i

ii

F

XF

X

20

0.445,3X

X

25

Solution: Median of grouped data (1)

Median position = 10.5

• The class interval in which this median

position lies is 170-175

• Median value = 170cm + (5cm x 3.5)

6

= 172.9cm

2

1

n

2

120

26

Solution: Median of grouped data (2)

Cumulative Frequency

7 10.5 13

3.5

170cm 175cm

LCB UCB

6

27

Mode: in grouped data

• In grouped data we often speak of the

‘model class interval’ i.e. that class interval

which has the highest frequency

28

Normal distribution

• Symmetrical (‘bell shaped’) means one

half is the mirror image of the other half

• Different types of average have the same

value

29

Skewed Distribution

• If the frequency distribution is skewed (i.e.

not symmetrical) then the various types of

average will have a different value

30

Skewed to the right

(positively skewed)

• Here the tail of the distribution is to the

right of the diagram. This means that the

few, extreme observations have a high

value. These ‘outriders’ pull up the simple

average (the mean) above the median

31

Skewed to the left

(negatively skewed)

• Here the tail of the distribution is to the left

of the diagram. The few, extreme

observations have a low value. These

‘outriders’ pull down the simple average

(the mean) below the median