Download BT STAT UNIT-1

BIO-STATISTICS
UNIT – I
Measures of Central tendency and measures of dispersion
Analysis of numerical data, we require certain statistical constants called measures of central tendency.
The various averages in common use are;
Mean
Median
Mode
There are three types of mean:
1. Arithmetic mean
2 .Geometric mean
3. Harmonic mean
I -Arithmetic mean :
Where
=
.
(i) Weighted Arithmetic mean=
[1] Milk is sold at the rate of 8,10, 12 and 15 rupees per liter in four different months. Assuming
that equal amounts are spent on milk by a family in the four months. Find the average in rupees
per month.
Solution:
Average
[2] Calculate the arithmetic mean for the values (L3)
Marks
52
58
60
65
No. of Students
7
5
4
6
Solution:
68
3
70
3
75
2
52
58
60
65
68
70
75
7
5
4
6
3
3
2
30
364
290
240
390
204
210
150
1848
II-MEDIAN
The observation is arranged in ascending order or descending order of magnitude, the middle mot item
is called median.
Median
L=lower limit of the median class
N=Total frequency
m=cumulative frequency of median class
f=frequency of median class
c= class interval
The merits and demerits of median (L1)
Merits: (i) It is easy to understand.
(ii) It is easy to determine.
(iii) It is unaffected by extreme values.
Demerits: (i) It is not well defined.
(ii) It is not based on all the items.
(iii) It is not suitable for algebraic treatment.
(iv) It is affected by sampling fluctuations.
[1] The number of runs scored by 11 players of a cricket team of a school are
5,19,42,11,50,30,21,0,52,36,27. Find the median. (L1)
Solution:
Runs is ascending order : 0,5,11,19,21,27,30,36,42,50,52
Number of observations =11=Odd number
Median
value=27
[2]Find the median of the following term 6,10,4,3,9,11,22,18 (L1)
Solution:
In ascending order: 3, 4,6 ,9,10,11,18,22
Number of observations =8=even number
Median
value=
[3] Calculate median for the following data (L3)
Marks
No. of Students
20
6
09
4
25
16
50
7
40
8
80
2
Solution:
By arranging marks in ascending order
Marks
9
20
25
40
50
80
Here
4
6
16
8
7
2
4
10
26
34
41
43
odd.
Position of median
observation
observation
observation
The above table shows that all items from 11 to 26 have their value 25.So the value of
observation is 25.
Median is 25
[4] Find the mean deviation about the mean for the following data 18, 20, 12, 14, 19, 22,
26, 14, 19, 24. (L1)
Solution:
Mean deviation about the mean
[5] Calculate the median for the following distribution (L1)
Wages (Rs. In
Hundreds
No. of persons
0-10
10-20
20-30
30-40
40-50
22
38
46
35
20
Solution:
Wages
0-10
10-20
20-30
30-40
40-50
22
38
46
35
20
22
60
106
141
161
Here N=161,
Median class is 20-30.
Lower limit of the median class
Frequency of the median class
Cumulative frequency of the class preceeding the median class
Width of the class
Median
Median
Mode
Mode is another measures of central tendency which is commonly used is called the mode.mode is
defined to be that value which occur most often.
Mode
L=lower limit of the modal class
f1=frequency of the modal class
f0=frequency of the pre modal class
f2=frequency of the post modal class
c= class interval
(i) Find the mode of 75, 72, 59, 62, 72, 75, 71, 70, 70, 70 (L1)
Solution:
Here the Mode is 70, since this score was obtained by more students than other.
(ii) Calculate the mode of
(L3)
Size of shoes
5
6
7
8
9
10
11
No. of persons
10
20
25
40
22
15
6
Solution:
Maximum no.of persons=40
Mode is 8.
(iii) Calculate the mode for the following distribution (L3)
Class Interval
0-10
10-20
20-30
30-40
40-50
50-60
Frequency
5
8
7
12
28
20
Solution:
Class Interval
Frequency
Cumulative Frequency
0-10
5
5
10-20
8
13
20-30
7
20
30-40
12
32
40-50
28
60
50-60
20
80
60-70
10
90
70-80
10
100
60-70
10
Mode
(iv) Find the mode from the following data (L1)
Age
0-6
6-12
12-18
Frequency
6
11
25
16-24
24-30
30-36
36-40
35
18
12
6
Solution:
Age
0-6
6-12
12-18
18-24
24-30
30-36
36-42
Frequency
6
11
25
35
18
12
6
Cumulative Frequency
6
17
42
77
95
107
113
Mode
The empirical relations between mean, median and mode
The empirical relations between mean, median and mode is
70-80
10
.
1.In a distribution mean = 65, median = 70, Find mode (L1)
Solution:
We know that
2. Calculate the mean, median and mode for the following data and verify the empirical relation (L3)
Class
0-10
10-20
20-30
3
7
13
Frequenc
y
Solution:
True Class
0.5-10.5
10.5-20.5
20.5-30.5
30.5-40.5
40.5-50.5
50.5-60.5
60.5-70.5
70.5-80.5
80.5-90.5
90.5-100.5
Mid
5.5
15.5
25.5
35.5
45.5
55.5
65.5
75.5
85.5
95.5
3
7
13
17
12
10
8
8
6
6
3040
17
40-50
50-60
12
10
-4
-3
-2
-1
0
1
2
3
4
5
90
A
-12
-21
-26
-17
0
10
16
24
24
30
28
,
Median
Mode
The empirical relations between mean, median and mode is
,which is not true.
The empirical relation is not true.
3
10
23
40
52
62
70
78
84
90
6070
8
70-80
8
8090
6
90100
6
3. Calculate the arithmetic mean, median, mode for the following data (L3)
Profits
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
4
8
18
30
15
10
8
7
No.of
companies
Solution:
Let the assumed mean be A=65
Prifits
Mid value(
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
25
35
45
55
65
75
85
95
Here
No. of
companies(
4
8
18
30
15
10
8
7
100
-40
-30
-20
-10
0
10
20
30
, Median class is 50-60
Lower limit of the median class is
Frequency of the median class is
Cumulative frequency of the class preceeding the median class is
Width of the class is
Median
Modal class is 50-60, since it has the maximum frequency.
Lower limit of the modal class is
Frequency of the modal class is
Frequency of the class preceding the modal class is
Frequency of the class reducing the modal class is
Width of the class is
Mode
-160
-240
-360
-300
0
100
160
210
-590
4
12
30
60
75
85
93
100
4. Compute the arithmetic mean and median for the following data(L3)
Central Wage
15
20
25
30
35
40
45
No. of Wage
earners
Solution:
3
25
19
6
4
5
6
15
20
25
30
35
40
45
Total
3
25
19
6
4
5
6
68
45
500
475
180
140
200
270
1810
3
28
47
53
57
62
68
Mean
Here N
, which is even
Median
5. Compute the arithmetic mean, median and mode for the following data (L3)
Profits
< 10
<20
<30
<40
<50
<60
<70
<80
No. of
Companies
4
16
40
76
96
112
120
125
Solution:
Class
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
Mean
Mid
5
15
25
35
45
55
65
75
4
12
24
36
20
16
8
5
125
20
180
600
1260
900
880
520
375
4735
4
16
40
76
96
112
120
125
Here
, Median class is 30-40
Lower limit of the median class is
Frequency of the median class is
Cumulative frequency of the class preceding the median class is
Width of the class is
Median
Modal class is 30-40, since it has the maximum frequency.
Lower limit of the modal class is
Frequency of the modal class is
Frequency of the class preceding the modal class is
Frequency of the class reducing the modal class is
Width of the class is
Mode
QUARTILES
Quartiles are position values similar to median class. There are three quartiles denoted by Q1,Q2,Q3
where
Li=lower limit of the
fi=frequency of the
class
class
c= class interval
N=Total frequency
mi=cumulative frequency of preceding class
Quartile deviation
Quartile deviation is defined as
(i.e) Q.D
Where
.
are the lower and upper quartiles.
1. Compute the first, second and third quartiles for the following data (L3)
Class Interval
0-50
50-100
100-150
150-200
200-250
250-300
No. of Students
20
60
50
30
24
120
Solution:
Class
interval
0-50
50-100
100-150
150-200
200-250
250-300
No.of students
20
60
50
30
24
16
200
20
80
130
160
184
200
N
2. Calculate the quartile deviation and the quartile coefficient of dispersion for the following data.
(L4)
Class
0-5
5-10
10-15
15-20
20-30
30-40
40-50
50-60
60-70
Frequency
Solution:
3
5
8
12
34
46
28
14
10
Class
Frequency
interval
0-10
10-20
20-30
30-40
40-50
50-60
60-70
8
20
34
46
28
14
10
8
28
62
108
136
150
160
N
Quartile Deviation
Quartile coefficient
1. Calculate the quartile deviation from the following distribution
Value of
the item
Frequency
12
13
14
25
26
27
38
40
2
3
5
8
7
3
2
1
2.
Calculate the quartile deviation and its coefficient for the following
frequency distribution (L4)
Marks
0
10
20
30
40
50
60
70
above
No. of
150
142 130 120
72
30
12
4
Students
Solution:
Class
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70 and above
8
12
10
48
42
18
8
4
8
20
30
78
120
138
146
150
N
Quartile Deviation
Quartile coefficient
MEASURES OF DISPERSION
1. RANGE
2. QUARTILE DEVIATION
3. MEAN DEVIATION
4. STANDARD DEVIATION
RANGE
Definition of Range and its coefficient.
Range is defined to be the difference between the largest and the smallest
of the observations.
(i.e) Range=L-S, where L and S are the largest and the smallest of the
observations.
Coefficient of Range
Calculate the coefficient of range for
13,25,36,22,18,45,21, 26, 30, 22. (L3)
Solution:
Largest Value= L= 45
Smallest Value=S=13
the
set
of observation
Coefficient of range=
STANDARD DEVIATION
Standard deviation is another measures of dispersion and defined aspositive square root of the AM of
the squares of all deviations of observation from their AM . It is denoted by symbol σ
1.The standard deviation (SD) of the set of numbers 3,8,6,10,12,9,11,10,12 and 7. (L1)
Solution:
2. The weekly salaries of a group of employees are given below.
Calculate the mean and median
and standard deviation of the salaries.(L4)
Salary (in Rs.)
75
80
85
90
95
100
No. of persons
3
7
18
12
6
4
Solution:
75
80
85
90
3
7
18
12
-2
-1
0
1
-6
-7
0
12
12
7
0
12
95
100
6
4
50
2
3
12
12
23
24
36
91
3.The first of two samples has 100 items with mean 15 and variance 9. If the whole group has 250
items with mean 15.6, variance 13.44 Calculate the SD of the second sample.(L3)
Solution:
Given
We know that
3360
The SD of the second sample is 4.
4.The following table gives the results of three centers of a public examinations
Centre
No. of Candidates
Mean score
SD
A
200
25
3
B
250
10
4
C
300
15
5
Calculate the mean and standard deviation of the combined score of the three centers. (L3)
Solution:
Given
= 75
=3
=200
= 10
=4
=250
=5
=300
= 15
=
=
=
=
=16
- =25-16=9
- =10-16=-6
- =15-16=-1
=
750
+
+
+
+
+
=(200X9)+(250X16) +(300X25) +(200X81) +(250X36)+(300x1)
=1800+4000+7500+16200+9000+300
=38800
=51.7333
=7.1926
VARIANCE
The mean and variance of a group of 100 items are respectively 15 and 9.
The mean and variance of another group of 150 items are each equal to
16. If the two groups are merged. Calculate the mean and variance of the
combined group. (L3)
Solution:
Mean
Variance Size
15
9
100
16
16
180
Let
be the variance of the combined group.
We know that
900+2400+36+24
Combined mean=15.6 & Combined variance=13.44
3. A factory produces two types of electric bulbs A & B. In an experiment
relating to he life
Length of
500-700
700-900
90011001300Life (in.
1100
1300
1500
hrs)
No. of
A
5
11
26
10
8
bulbs
B
4
30
12
8
6
Compare the variability of the two varieties using the coefficient of variation.
(L4)
Solution:
For A type bulbs:
Mid value of
600
800
1000
1200
1400
5
11
26
10
8
-2
-1
0
1
2
-10
-11
0
10
16
20
11
0
10
32
60
5
73
-8
-30
0
8
12
-18
16
30
0
8
24
78
Mean
For B type bulbs:
Mid value of
600
800
1000
1200
1400
4
30
12
8
6
60
-2
-1
0
1
2
Mean
Coefficient of variation for A =
=
Coefficient of variation for B =
Since c.v for B is more than that of A.
type B bulbs is more variable.
=21.64