Measures of Central Location Week 3
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
$26.7
• Mean closing price ($) 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