Download Determine the mean, median and mode for the following

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Assignment 1
BIOSTATISTICS I
30
Question 1
15 Marks
Determine the mean, median and mode for the following frequency distribution.
X
20
30
40
50
60
f
3
11
26
28
12
Solution:
X
20
30
40
50
60
Total
f
fX
cf
3
11
26
28
12
60
330
1040
1400
720
3550
3
14
40
68
80
 f  80
Mean =
 fX
f
Median = l 

3550
 44.375
80

h  f
 C 

f  2

Where,
l  lower class boundry of median class
h  Class interval
f  Frequency of median class
C  Cumulative frequency of the class before median class
f
2

80
 40
2
Class
boundaries
15 – 25
25 – 35
35 – 45
45 – 55
55 – 65
f
 40 in the column of cf and we see that it falls in the class 35 – 45. So, it
2
is the median class
We see
Therefore, we have
l  35
h  10
f  26
C  14
Now,
Median = 35 
10
 40  14 
26
Median = 45
Mode = l 
f m  f1
h
 f m  f1    f m  f 2 
Where,
l  lower class boundry of modal class
h  Class interval
f  Frequency of modal class
f1  Frequency of the class before modal class
f 2  Frequency of the class after modal class
The class which corresponds to maximum frequency is 45 – 55. Therefore it is the model
class.
Therefore, we have
l  45
h  10
f  28
f1  26
f 2  12
Now,
Mode = 45 
28  26
10
 28  26   28 12
Mode = 45 
2
10
2  16
20
18
Mode = 45  1.11
Mode = 45 
Mode = 46.11
Question 2
15 Marks
Calculate mean deviation, variance, standard deviation and coefficient of variation for the
following frequency distribution.
0 – 4 5 – 9 10 – 14 15 – 19 20 – 24 25 – 29
Classes
8
19
23
25
15
10
Frequency
Solution:
Classes f
0–4
5–9
10 – 14
15 – 19
20 – 24
25 – 29
Total
8
19
23
25
15
10
 f  100
X
fX
fX2
2
7
12
17
22
27
16
133
276
425
330
270
32
931
3312
7225
7260
7290
 fX 2  26050
 fX  1450
X X
12.50
7.50
2.50
2.50
7.50
12.50
f X X
100
142.5
57.5
62.5
112.5
125
 f X X =
600
 fX  1450  14.50
 f 100
 f X X
Mean Deviation =
f
 fX    fX 
Variance =
 f   f 
Mean =
2

600
6
100

26050
2
 14.5   260.50  210.25  50.25
100
2
Standard Deviation = Variance = 50.25 = 7.09
SD
7.09
 100 =
 100  48.9%
Coefficient of Variation =
X
14.50