Biostatistic Tool

1. Dr. M. MARIAPPAN
M.Sc., M.Phil., M.Ed., Ph.D.
Assistant Professor,
Department of Zoology,
Government Arts College,
Melur – 625 106

2.  The average of data is called a measure of central
tendancy.
 Average lies in the centre of the lowest and highest
values of data. So, it is called a measure of central
tendancy.
 The value of average describe the entire series of
data.
MEASURES OF CENTRAL TENDANCY

3. Characteristics of Central Tendancy:
 It is not depend on the personal prejudice and bias of
the investigator.
 It should be rigidly defined.
 It should depend on each and every item of the series.
 It should be easily understood.
 It should lead for further algebraic treatment.

4. Mode
Median
TYPES OF CENTRAL TENDANCY
Mean (or) Arithmetic Mean

5.  It is simply referred as Average.
 Its value is obtained by adding all the values and by
dividing the total by the number of items.
 Mean is represented by the symbol X (X bar).
 Mean can be obtained for ungrouped and grouped
data.

6. Ungrouped Data
Discrete Data
Continuous Data
DATA

7. S.
No. 1 2 3 4 5 6 7 8 9 10
Eggs 33 35 44 34 41 45 39 46 38 47
No. of Potatoes
per plant (x)
4 6 3 8 9 5
No. of Plants 17 9 5 20 15 12
Class
Interval
10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
Frequency 3 5 10 15 5 12
UNGROUPED
DATA
DISCRETE
DATA
CONTINUOUSD
ATA

8.  In ungrouped data, frequencies are not given, the
calculation of mean is very simple.
 Add all the values or the variable and divide the total
by the number of items.
Symbolically:
1. Arithmetic Mean for Ungrouped Data:
X
where,
X = Sum of all the values of variables
N = Number of observations

9. Problem – 1
The following table gives the number of eggs laid by 10
lizards in a season. Find out the mean for the data.
S.
No.
1 2 3 4 5 6 7 8 9 10
Eggs 33 35 44 34 41 45 39 46 38 47
Steps:
1. Construct two columns table.
2. In first column write the serial number (S.No.). This is
the number of observation (N).
3. In second column write the values of the variables X
(Eggs) and find out total (X).
4. Divide X by their number of observation (N).

10. S. No. Eggs
1 33
2 35
3 44
4 34
5 41
6 45
7 39
8 46
9 38
10 47
N = 10 X = 402
X
X = 402
N = 10
X
= 40.2
Mean ( X ) = 40.2

11.  In this, the values of the variable are multiplied by
their respective frequencies (fx) and totaled (fx).
 Total (fx) is divided by the total number of
frequencies (f).
Symbolically:
2. Arithmetic Mean for Discrete Series:
Where,
fx = the sum of products
f = total of frequency
X

12. Problem – 2
Find the mean number of potatoes per plant given the
following frequencies of occurance:
Steps:
 Construct three columns table.
 In first column enter variable values (x) and in second
column enter frequency (f).
 In third column, multiply each variable by its frequency
(fx).
 Add all the fx (fx).
 Divide fx by the total of frequency f.
No. of Potatoes
per plant (x)
4 6 3 8 9 5
No. of Plants 17 9 5 20 15 12

13. fx = 492
f = 78
= 6.3
Mean ( X ) = 6.3
X f fx
4 17 68
6 9 54
3 5 15
8 20 160
9 15 135
5 12 60
f = 78 fx = 492
X
X

14. 1. Find out the Mean for the following Data
165, 155, 175, 165, 154, 185, 163
Length of the
Fish (cm) 5 8 9 12 13 14 20
No. of Fishes 3 6 8 12 5 4 2
2. Find out the Mean for the following Data

15.  In continuous series, the value of each individual
frequency distribution is unknown.
 Symbolically:
3. Arithmetic Mean for Continuous Series:
Where,
m = midpoint of classes
f = frequency
f = total of frequency
X

16. Problem – 3
Calculate the arithmetic mean for the following data:
Steps:
 Construct 4 columns table.
 In first column write the Class interval.
 Find out the mid value (m) and write in the second column
 In third column write the frequency (f) and find the total (f).
 Multiply the mid-points (m) by respective frequency (f) of
each class and sum them up to fm.
 Divide fm by sum of frequencies (f).
Class
Interval
10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
Frequency 3 5 10 15 5 12

17. Class Interval
Mid-point
(m)
Frequency
(f)
fm
10 – 20 10+20/2 = 15 3 45
20 – 30 20+30/2 = 25 5 125
30 – 40 30+40/2 = 35 10 350
40 – 50 40+50/2 = 45 15 675
50 – 60 50+60/2 = 55 5 275
60 – 70 60+70/2 = 65 12 780
f = 50 fm = 2250
X
Mean ( X ) = 45

18. Wages
(Rs) 10-20 20-30 30-40 40-50 50-60 60-70
No. of
Persons 3 6 8 12 5 4
3. Compute the arithmetic mean for the following
data:

19. Merits of Mean:
1. It is easily understandable and easy to compute.
2. Its calculation is based on all the observations.
3. It is the best measure to compare two or more series.
4. It is easy for further calculations.
5. Sampling fluctuations do not affect the value of mean.
Demerits of Mean:
1. It cannot be calculated if all the values are not known.
2. Mean value is very much influenced by very big or very
small numbers.
3. It cannot be determined for the qualitative data such as
love, honesty, beauty, etc.
4. Some times it may give a false conclusion.

20. Median is the middle value of a data when the values
are arranged in the ascending (or) descending orders.
Median divides a distribution into two equal halves,
one half consisting of all values less and the other all
values greater than it.
Median is represented by the symbol md.

21.  Arrange the data in ascending (or) descending order.
 If the number of observations were odd, the value of
the middle most items is the median.
 If the number be even, the average of the two middle
most items is taken as median.
 Symbolically:
1. Median for Ungrouped Data:
where,
n = Number of observations
Median (md)

22. Problem – 4
Find the median of the following data:
21, 12, 49, 37, 88, 46, 55, 74, 63
Steps:
 It is an ungrouped data with odd numbers (xw;iwg;gil
vz;).
 Count the number of observation (N).
 The data are arranged in an ascending order.
 Apply the formula.

23. Median (md)
12, 21, 37, 46, 49, 55, 63, 74, 88
= 5th Value
Now,
the 5th Value in the data is 49
Median (md) = 49

24. Problem – 5
Find the median of the following data:
88, 72, 33, 29, 70, 86, 54, 91, 61, 57
Steps:
 It is an ungrouped data with even numbers (,ul;ilg;gil
vz;).
 The data are arranged in ascending order.
 The average of the two middle observations is
median.
29, 33, 54, 57, 61, 70, 72, 86, 88, 91
Middle two values = 61 & 70

25. Median (md)
Median (md) = 65.2

26. • Find the cumulative frequencies (cf).
• Apply the formula
2. Median for Discrete Series:
Where,
n = frequency

27. Problem – 6
Calculate the median for the following data:
Steps:
 Draw a table with three columns.
 Enter the marks in the first column (X).
 Enter the no. of students in the second column (f)
and find the total (N).
 Find out cumulative frequency (cf) and enter in the
third column.
Marks 9 20 25 40 50 80
No. of
Students
4 6 16 8 7 2

28. Marks
(X)
No. of
Students
(f)
Cumulative
frequency
(cf)
9 4 4
20 6 10
25 16 26
40 8 34
50 7 41
80 2 43
N = 43
Here,
N = 43
In the data, 22nd value lies in between 10 and 26 of cumulative frequency.
So, we take the highest cumulative frequency 26 and its corresponding X value 25.
Median (md) = 25
= 22nd Value

29. 1. Find out the Median weight for the following
students weights are in Kg.
67, 69, 66, 68, 63, 76, 72, 74, 70, 65
Marks 30 40 50 60 70 80 90
No. of
Students 15 20 10 15 20 15 5
2. Find out the Median from the frequency table.

30.  In this, data is given in the form of a frequency with
class interval.
 Cumulative frequencies are found out for each value.
 Median class is then calculated, where cumulative
frequency N/2 lies is called median class.
 Symbolically:
3. Median for Continuous Series:
Where,
L = Lower limit of the median class
N = Total number of items
c.f. = Cumulative frequency prior to the median class
f = Actual frequency of the median class
C.I. = Class Interval of the median class

31. Problem – 7
Find the median of the data given below:
Steps:
 Draw a table with four vertical columns.
 Enter the class interval in the first column (C.I.).
 Find out mid value and write in the second column
(m).
 Enter frequency in the third column (f) and totaled
(N).
 Find out cumulative frequency and enter in the
fourth column (cf).
Class
Interval
15 – 25 25 – 35 35 – 45 45 – 55 55 – 65 65 – 75
Frequency 4 11 19 14 0 2

32. Class Interval
(C.I.)
Mid Value
(m)
Frequency
(f)
Cumulative
Frequency
(cf)
15 – 25 15 + 25/2 = 20 4 4
25 – 35 25 + 35/2= 30 11 15
35 – 45 35 + 45/2= 40 19 34
45 – 55 45 + 55/2= 50 14 48
55 – 65 55 + 65/2= 60 0 48
65 – 75 65 + 75/2= 70 2 50
N = 50

33.  Find the median class using the formula N / 2 = 50 / 2 =
25.
 The value 25, is falls in between 15 and 34 of
cumulative frequency.
 So, the class interval 35 – 45 have the highest frequency
34 is selected as the median class.
 35 is the lower limit of the median class (L).
 19 is the actual frequency of the median class (f).
 15 is the cumulative frequency prior to the
median class (cf).
 10 is the class interval of the median class (C.I).

34. Apply the formula,
Median (md) = 40.26

35. Find out the Median for the following data
Marks 0 – 10 10 -20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
No. of
Students
5 7 10 15 13 10 6

36. Merits of Median:
1. It is easily understood.
2. It is not affected by the extreme values.
3. The value of the median can be graphically
presented.
4. It is the best measure for qualitative data such as
beauty, intelligence, etc.
5. It can be determined by inspection in many cases.
Demerits of Median:
1. Data must be arranged for calculation.
2. It is not subject to algebraic treatment.
3. Median is more influenced by sampling fluctuations.

37.  The observation which occurs more often than others
is called Mode.
 It is repeated the highest number of items in the
series.
 When a distribution has one repeated of frequency, it
is called “UNIMODAL”, when it has 2 repeated of
frequency it is termed as “BIOMODAL”.
 Similarly, if 3 repeated of frequencies it is termed as
“TRIMODAL”.
 It is represented by the Mo.

38.  Mode for ungrouped data can be determined by locating that
value which occurs the maximum number of times.
 It can be determined by inspection only.
1. Mode for Ungrouped Data:
Problem – 8
Determine mode from the following data
34, 36, 45, 40, 27, 65, 45, 48, 62, 25
Steps:
 First arrange the data in the ascending order.
 Find out the repeated data.
25, 27, 34, 36, 40, 45, 45, 48, 62, 65
In this data 45 is repeated 2 times. So, Mode is 45.

39.  Mode is the value which has occurred the maximum number
of times.
 Straight X value is Mo.
2. Mode for Discrete Data:
Problem – 9
Find out the mode of the distribution:
Solution:
From the above data we can clearly say that modal length is
31 cm because the value 31 has occurred the maximum
number of times i.e. 65.
So, Mode is 31 cm.
Length
of Fish
(cm)
28 29 30 31 32 33
No. of
Fishes
10 20 40 65 50 15

40.  Modal class in a grouped frequency distribution is determined
by inspection.
 When mode is ill – defined (or) bi-modal , its value is
ascertained by the following formula:
Mode = 3 Median – 2 Mean
 All classes are of equal width, determine the mode by
applying the following formula,
3. Mode for Continuous Data:

41. Where,
L = Lower limit of modal class (i.e. the class
containing the largest frequency)
d1 = Difference between the frequency of modal
class and frequency of Pre-modal class
(f1 – f0).
d2 = Difference between the frequency of modal
class and frequency of Post-modal class
(f1 – f2).
CI = Class interval of the modal class
f1 = Frequency of the modal class
f0 = Frequency of the preceding modal class
f2 = Frequency of the succeeding modal class

42. Problem – 10
From the following data calculate mode.
Steps:
 Construct two column table.
 In first column write the class intervals (C.I.).
 In second column write the frequencies (f).
 Find out the highest frequency by observation and
fix it as f1.
 The preceeding frequency of f1 is f0.
 The succeeding frequency of f1 is f2.
 Apply the formula and find out the mode.
Weight
of Fishes
(mg)
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
No. of
Fishes
5 15 25 15 3 4 3

43. Maximum frequency in the
distribution is 25 & it lies in the class
interval 20 – 30. Thus modal class is
20 – 30.
Class Interval
(CI)
Frequency
(f)
0 – 10 5
10 – 20 15 f0
20 – 30 25 f1
30 – 40 15 f2
40 – 50 3
50 – 60 4
60 – 70 3
Where,
L = 20
d1 = f1 – f0 = 25 – 15 = 10
d2 = f1 – f2 = 25 – 15 = 10
C.I. = 10

45. 1. Determine the mode from the following data.
50 62 48 50 63 65 50 45 43 61 50 49
2. Determine the mode from the following data.
X 20 25 30 35 40 45 50
f 1 2 1 5 1 2 1
3. Find out the mode from the following data.
Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
No. of
Students
4 9 13 15 12 8 3

46. Merits of Mode:
1. Mode is not affected by extreme values.
2. It can be obtained by inspection.
3. It can be easily understood.
4. It can be used to describe qualitative phenomenon.
5. The value of mode can also be found graphically.
Demerits of Mode:
1. The value of mode cannot always be determined. In
some cases, we may have bimodal or multimodal
series.
2. It cannot be treated algebraically.
3. It is not based on all observations.

47.  The percentile values divide the distribution into 100
parts.
 There are 99 percentiles P1, P2, ……. P99, called the
first percentile, second percentile, etc.
I. Percentiles for Ungrouped Data:
 Arrange the data in ascending order.
 Apply the formula,
Where,
X = Percentile Number
n = Total number of observation

48. Problem – 11
Calculate P15 for the data given below:
5, 24, 36, 12, 20, 8
Steps:
 Arrange the values in the ascending order.
5, 8, 12, 20, 24, 36
 Apply the formula
X = 15
n = 6

49. = 1st item + 0.05 (2nd item – 1st item)
= 5 + 0.05 (8 – 5)
= 5 + 0.05 (3)
= 5 + 0.15
P15 = 5.15

50. II. Percentiles for Continuous Data:
 Construct three columns table.
 Write the class interval in the first column.
 In second column write the frequency.
 Find cumulative frequency and write in the third column.
 Apply the formula and findout the percentile class
interval
Where,
X = Percentile Number
n = Total of frequency

51.  Apply the following formula and findout the
percentile,
Where,
m = cumulative frequency of preceeding
class interval
L = lowest value of percentile class
f = frequency of percentile class
C.I = class interval value of percentile class

52. Problem – 12
Find P53 for the following frequency distribution
Class
Interval
0 – 5 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 - 40
Freqency 5 8 12 16 20 10 4 3
C.I. f cf
0 – 5 5 5
5 – 10 8 13
10 – 15 12 25
15 – 20 16 41
20 – 25 20 61
25 – 30 10 71
30 – 35 4 75
35 – 40 3 78
n = 78
Percentile
Class
Solution:

53. X = 53
n = 78
Value 41.34 lies in between the cumulative frequency 41 and
61. So, take the highest cumulative frequency 61 as
percentile class interval

54. X = 53 L = 20 f = 20
n = 78 C.I. = 5 m = 41

55. 1. Determine the 35th Percentile of the scores
7 3 12 15 14 4 20
2. The income of 70 persons are given as follows. Find out P65
value.
Income 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
Person 10 15 25 12 8