Download S1 Measures of Dispersion

S1 Measures of Dispersion
The mean, variance and
standard deviation
S1 Measures of Dispersion
Objectives:
To be able to find the variance
and standard deviation for
discrete data
To be able to find the variance
and standard deviation for
continuous data
A factory worker counted the numbers of nuts in a
packet. The results are shown in the table
Number
of nuts
Freq
CF
Calculate P80 and P40
6
7
8
9
10
11
8
12
36
18
15
10
8
20
56
74
89
99
P80 = 99/100x80 = 79.2
P80 = 80th term
P80 = 10 nuts
P40 = 99/100x40 = 39.6
P40 = 40th term
P40 = 8 nuts
Calculate the 40th to 80th interpercentile range
40th to 80th interpercentile range = 10 - 8 = 2 nuts
The lengths of a batch of 2000 rods were measured to the nearest
cm. The measurements are summarised below.
Length
Number of
Cumulative
Q1=74.5 + 500-250 x 5
(nearest cm) rods
frequency
60-64
65-69
70-74
75-79
80-84
85-89
90-94
95-99
11
49
190
488
632
470
137
23
738-250
11
60
250
738
1370
1840
1977
2000
Q1=77.06
Q2=79.5+1000-738 x 5
1370-738
Q2=81.57
Q3=84.5+1500-1370 x 5
1840-1370
Q3=85.88
By altering the formula slightly can you work out how to find the 3rd
decile (D3) and the 67th percentile (P67)?
Answers
D3=74.5 + 600-250 x 5
738-250
D3=78.09
P67=79.5 + 1340-738 x 5
1370-738
P67=84.26
Variance of discrete data
The deviation (difference) of the data (x) from
_
the mean (x) is one way of measuring the
dispersion (spread) of a set of data.
_
Variance = Σ(x – x)²
n
_
Variance = Σx² – Σx ²
n
n
Standard deviation.
As variance is measured in units ² you usually
take the square root. This gives you the
standard deviation.
Standard deviation = √variance
Standard deviation symbol is σ
The marks scored in a test by seven students
are 3,4,6,2,8,8,5. Calculate the variance and
standard deviation.
x
3
4
6
2
8
8
5
x²
9
16
36
4
64
64
25
Σx = 36 Σx² = 218
_
Variance = Σx² – Σx ²
n
n
σ ² = 218 – 36 ² = 4.69
7
7
σ = √4.69 = 2.17
The variance and standard deviation from a
frequency table
Number of
rods (x)
Freq of
rods (f)
35
3
36
17
37
29
38
34
39
26
fx
105
612
1073
1292
1014
fx²
3675
22032
39701
49096
39546
Σf=109 Σfx=4096 Σfx²=154050
Variance = Σfx² – Σfx ²
Σf
Σf
σ ²=1.19805
Variance = 154050 – 4096 ²
109
109
σ = 1.109
Variance and standard deviation from a grouped frequency distribution
Height of
plant (cm)
Freq
(f)
0<h≤5
4
5 < h ≤ 10
15
10 < h ≤ 15
5
15 < h ≤ 20
2
20 < h ≤ 25
0
25 < h ≤ 30
1
Total
σ ² = 6487.5 – 285
27
27
² = 128.85802
σ
σ
= √128.85802
= 11.35