BIOSTATISTICS MEAN MEDIAN MODE SEMESTER 8 HOW TO LEARN BIOSTATISTICS AND RESEARCH METHODOLOGY

1. BIOSTATISTICS
By Payaam Vohra
Gold Medalist in MU
NIPER AIR 11
GPAT AIR 43
ICT MTECH SCORE RANK 01
CUET-PG AIR 01
MET AIR 08
IIT BHU AIR 07
GATE AND BITS HD QUALIFIED

2. UNIT – I
Introduction: Statistics, Biostatistics, Frequency distribution.
Measures of central tendency: Mean, Median, Mode- Pharmaceutical examples.
Measures of dispersion: Dispersion, Range, standard deviation, Pharmaceutical problems.
Correlation: Definition, Karl Pearson’s coefficient of correlation, Multiple correlation -Pharmaceuticals
examples.
UNIT – II
Regression: Curve fitting by the method of least squares, fitting the lines
y = a + bx and x = a + by, Multiple regression, standard error of regression– Pharmaceutical Examples.
Sample, Population, large sample, small sample, Null hypothesis, alternative hypothesis, sampling, essence of
sampling, types of sampling, Error-I type, Error-II type, Standard error of mean (SEM) - Pharmaceutical
examples
Parametric test: t-test (Sample, Pooled or Unpaired and Paired), ANOVA (One way and Two way), Least
Significance difference.
2

3. 3
Number of class test Marks obtained
1 14
2 18
3 16
4 17
5 20
Total 85
Grouped frequency distribution
Age in years Frequency (Number of persons)
10-20 14
20-30 18
30-40 16
40-50 17
50-60 20
Total 85
MEASURES OF CENTRAL TENDENCY
The measures of central tendency are also usually called the averages.
They give us an idea about the concentration of the values in the
central part of the distribution.
The following are the three measures of central tendency that are
common use.
(1)Mean (or) Averages (or) Arithmetic Mean (A.M)
(2)Median
(3)Mode

4. Arithmetic mean or mean
Definition: Arithmetic mean (or) simply the mean of a variable is defined as the sum of
the observations divided by the number of observations. It is denoted by the symbol 𝑥
PROBLEMS ON MEAN
Problem 1: Calculate mean for the following data.
20,40,35,25,40.
Solution:
Mean = Sum of the observations =
xi
Number of observation n
= 20+ 40+35+ 25+ 40
5
= 160
5
= 32
Mean( x ) = 32
Problem-2: Calculate mean for the following data.
x 1 2 3 4 5
f 3 4 5 6 2
4

5. 5
x f fx
1 3 3
2 4 8
3 5 15
4 6 24
5 2 10
TOTAL 𝑵 = ∑𝒇 = 𝟐𝟎 ∑𝒇𝒙 = 𝟔𝟎
Mean(x) =
fx
=
60
= 3
N 20
Problem-3: Calculate mean for the following data.
CI 0-10 10-20 20-30 30-40 40-50 50-60
No. of
students
12 18 2 10 12 6
SOLUTION:
CIass Interval f 𝑴𝒊𝒅 𝑷𝒐𝒊𝒏𝒕𝒔(𝒎)
𝑳. 𝑳 + 𝑼. 𝑳
=
𝟐
fm
0-10 12 5 60
10-20 18 15 270
20-30 2 25 50
30-40 10 35 350
40-50 12 45 540
50-60 6 55 330
Total 𝑵 = ∑𝒇 = 𝟔𝟎 ∑𝒇𝒎 = 𝟏𝟔𝟎𝟎
Mean(x) =
fm
=
1600
= 26.67
N 60

6. 6
1. Arrange the data in ascending (or) descending order.
2. Count the number of observations (or) number of items is denoted by (n)
n +1 th
3. Median =Value of 2 term
Problem-4: Calculate Median for the following data.
23,24,20,19,19,15,18,14,20.
SOLUTION:
1. Arrange the data in ascending order:
14,15,18,19,19,20,20,23,24
2. Count the number ot
fhobservations (n) = 9 (ODD Observations)
n +1
3.Median =Value of 2 term
9 +1 th
=Value of 2 term
10 th
=Value of
2
term
=Value of (5)th
term
Median =19
Problem-5: Calculate Median for the following data.
21,28,29,20,19,5,8,3,2,22,20,21.
SOLUTION:
1. Arrange the data in ascending order:
2,3,5,8,19,20,20,21,21,22,28,29.
2. Count the number of observations (n) = 12 (EVEN
Observations)

7. 10
term
=Value of 2
13 th
=Value of
2
term
=Value of (6.5)th
term
6th term + 7th term 20+ 20
Median = = = 20
2 2
IN DISCRETE SERIES :
1. Arrange the data in ascending (or) descending order.
2. Find out the Cummulative Frequency (C.F).
N +1 th
3.Median = Size of 2 term.
Problem-6: Obtain the Median for the following frequency distribution.
x 1 2 3 4 5 6 7 8 9
f 8 10 11 16 20 25 15 9 6
SOLUTION:
x f Cummulative
Frequency(C.F)
1 8 8
2 10 18
3 11 29
4 16 45
5 20 65
6 25 90
7 15 105
8 9 114
9 6 120
TOTAL 𝑁 = ∑𝑓 = 120

8. term.
Median = Size of 2
120 +1 th
term
= Size of 2
121 th
= Size of
2
term
= Size of (60.5)th
term
= C.F > 60.5th term
Median = 5
IN CONTINUOUS SERIES:
Problem-7: Obtain the Median for the following frequency distribution.
CI 0-10 10-20 20-30 30-40 40-50
f 150 100 150 200 125
SOLUTION:
CI f CF
0-10 150 150
10-20 100 250(C.F)
(L)20-30 150(f) 400
30-40 200 600
40-50 125 725
TOTAL 𝑵 = ∑𝒇 = 𝟕𝟐𝟓

9. N
th
N o w , w e h a ve to fin d the M e d i a n class = S ize o f ter m
2
7 2 5
th
te rm
= Size of 2
= S ize of (3 6 2 .5 )t h
ter m
= C . F > 362.5 t h t e r m
M ed ian cla ss = 2 0 3 0
Me d i a n = L + 2
N
C .F
h
f
W h ere L = L o w er l i m it of th e M e d ia n class
f = frequency of the Me d i a n class
C . F = C u m m u l a t i v e Fr e q u e n c y of class preceding the m e d i a n class
h = widt h of the class int erval
1 5 0
M e d i a n = 2 0 +
3 6 2 . 5 2 5 0
1
0
= 2 0 +
1 1 2 . 5
10
1 5 0
= 2 0 + 7.5
= 2 7 . 5

10. IV B.PHARMACY (BIO STATISTICS)
Mode
The mode refers to that value in a distribution, which occur most frequently. It is
an actual value, which has the highest concentration of items in and around it. It shows
the Centre of concentration of the frequency in around a given value. Therefore, where the
purpose is to know the point of the highest concentration it is preferred. It is, thus, a positional
measure.
Its importance is very great in agriculture like to find typical height of a crop variety,
maximum source of irrigation in a region, maximum disease prone paddy variety. Thus, the
mode is an important measure in case of qualitative data.
PROBLEMS ON MODE
Individual series: The value of that occurs Maximum times is known as
Mode(Model) and it is denoted by z.
Problem-8: Find the Mode for the following data.
2,5,8,6,6,2,6,1,3,6.
SOLUTION:
Mode(z) = 6 (Uni-model)
Problem-9: Find the Mode for the following data.
2,3,2,4,2,3,8,9,5,3,7,1,3,2,2,3.
SOLUTION:
Mode(z) = 2,3 (Bi-Model)
Discrete Series: The value of the variable(observation) which has the heighest
frequency is known as Mode(Model) and it is denoted by z.
Problem-10:Determine the Mode for the following data.
x 1 2 3 4 5 6 7 8
f 4 9 16 25 22 15 7 3
SOLUTION: Mode(z) = 4

11. IV B.PHARMACY (BIO STATISTICS)
Problem-11: Calculate the Mode age for the given sample of Smokers.
Age 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-
100
No. of
Smokers
8 12 24 31 32 20 9 3
SOLUTION:
Age No.of
Smokers(f)
20-30 8
30-40 12
40-50 24
50-60 31(𝑓1)
60-70 32(𝑓)
70-80 20(𝑓2)
80-90 9
90-100 3

12. 2(32) 31 20
f f
Mode = L + 2 f f 1 f h
1 2
where :
L = Lower lim it of the Mode class
f = frequency of the mod e class
f1 = frequency of the class preceding the mod e class
f2 = frequency of the class succeeding the mod e class
h = width of the mod e class
Mode = L + f f1
– f f h
2 f 1 2
= 60 +
32 31
10