LECTURE : Measure of Dispersion
MEASURE OF DISPERSION
Range
• It is the difference between the highest and lowest values in a set of
data.
• It is an imperfect measure because it is affected by any usual large
or small values
Variance
• It is measure that deals with the spread of values
2
∑(𝑋
−
ẍ)
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 (α2 ) =
𝑁
Where
𝑋 = value of each data set
N = No. of items
ẍ = Mean
Mean Deviation
• It is a simple method of measuring dispersion but not commonly
used because it is unstable
Mean Deviation
MD =
∑/X−ẍ/
N
• But for grouped distribution
∑𝑓/𝑋 − ẍ/
Mean Deviation (MD) =
∑𝑓
Where
𝑋 = mid − value of each group
/𝑋 − ẍ/ = +ve values of 𝑋 − ẍ
N = No. of items
ẍ = Me𝑎𝑛
Standard Deviation
• It determine the extent to which each value is far from the mean
- If the values are close to the mean, then the
standard deviation (SD) is small
- But where many of the values are far from the mean,
then the SD will be large
- While if all values are equal, the SD will be zero.
• The relationship is that SD is the square root of the variance
SD (α) = √α2
SD (α) =
∑(𝑋−ẍ)2
𝑁
Alternative Formula for Standard Deviation
For listed values of X
SD (α) =
∑(𝑋−ẍ)2
𝑁
It can also be written in this format
For grouped data
SD (α) =
∑(𝑋)2
− (ẍ)²
𝑁
When using Assumed Mean
Where X is the mid-value
SD (α) =
∑𝑓𝑑² − ((∑𝑓𝑑)2 /∑𝑓
∑𝑓
Where X is the mid-value
SD (α) =
∑𝑓(𝑋)2
− (ẍ)²
∑𝑓
Work Example 1
Question
The marks of 50 students in an examination is distributed as follows
Mark Interval
5-9
10 - 14
15 - 19
20 - 24
25 - 29
30 - 34
No. of students
4
9
16
12
6
3
Find the (i) mean, (ii) mean deviation, (iii) variance, (iv) standard deviation
Solution
X
F
Xm
FXm
5–9
4
7
28
10 – 14
9
12
108
15 – 19
16
17
272
20 -24
12
22
264
25 – 29
6
27
162
30 - 34
3
32
96
50
930
(i) Mean (ẍ) =
=
∑FXm
∑𝑓
930
50
= 18.6
X
F
Xm
FXm
X-ẍ
(X - ẍ)²
F(X - ẍ)²
/X - ẍ/
F/X - ẍ/
5–9
4
7
28
-11.6
134.56
538.24
11.6
46.4
10 – 14
9
12
108
-6.6
43.56
392.04
6.6
59.4
15 – 19
16
17
272
-1.6
2.56
40.96
1.6
25.6
20 -24
12
22
264
3.4
11.56
138.72
3.4
40.8
25 – 29
6
27
162
8.4
70.56
423.36
8.4
50.4
30 - 34
3
32
96
13.4
179.56
538.68
13.4
40.2
50
930
∑𝑓/𝑋 − ẍ/
ii Mean Deviation (MD) =
∑𝑓
262.8
= 50 = 5.256
2071.56
262.8
∑𝐹(𝑋 − ẍ)2
iii Variance (α ) =
∑𝑓
(iv) Standard Deviation (α) = √α2
= √41.43 = 6.436
2
=
2071.56
= 41.43
50
Standard Deviation: Using Alternative Formula
Assumed mean (XA ) = 17
X
F
Xm
d = Xm - XA
Fd
Fd²
5–9
4
7
-10
-40
400
10 – 14
9
12
-5
-45
225
15 – 19
16
17
0
0
0
20 -24
12
22
5
60
300
25 – 29
6
27
10
60
600
30 - 34
3
32
15
45
675
80
2200
50
Assumed Mean Method
(i) Mean (ẍ) = XA +
∑Fd
∑𝑓
Standard Deviation (α) =
∑𝑓𝑑² − ((∑𝑓𝑑)2 /∑𝑓
∑𝑓
80
= 17 + 50
= 18.6
=
2200 − ((80)2 /50
=
50
6.435
Standard Deviation: Using Alternative Formula
For grouped data
X
F
Xm
FXm
Xm²
FXm²
5–9
4
7
28
49
196
10 – 14
9
12
108
144
1296
15 – 19
16
17
272
289
4624
20 -24
12
22
264
484
5808
25 – 29
6
27
162
729
4374
30 - 34
3
32
96
1024
3072
50
Standard Deviation (α) =
=
930
19370
∑𝑓(Xm )2
− (ẍ)²
∑𝑓
19370
− (18.6)² =
50
6.435
(i) Mean (ẍ)
=
∑FXm
∑𝑓
930
=
50
= 18.6
Assignments
1. The age group of people vaccinated is distributed as follows
Age Group
30 - 39
40 - 49
50- 59
60 - 69
70 - 79
80 - 89
No. of people
8
17
20
11
7
1
Using all methods taught, find the (i) mean, (ii) mean deviation,
(iii) variance, (iv) standard deviation
2. Amount of money deposited by
customers in a day is given as follows.
(i) Find the mode, median and mean
(ii) Given an assumed mean of 275.5
compute the mean and standard
deviation
X
F
X
F
1– 50
5
251-300
40
51– 100
11 301–350
26
101– 150
18 351–400
25
151-200
22
401-450
15
201– 250
•
Submission Date
Scan & submit latest 5th Aug 2021 @ 2pm
29
451-500
7