Mean- Mean is an essential concept in mathematics and statistics. The mean is the average or the most common value in a collection of numbers Types of Mean A. Arithmetic Mean a. Simple Arithmetic Mean b. Weighted Arithmetic Mean B. Geometric Mean C. Harmonic Mean 1.Calculation of Simple Arithmetic Mean a) Direct Method b) Shortcut Method c) Step Deviation Method 2. Calculation of Weighted Arithmetic Mean a) Direct Method b) Shortcut Method Merits and Demerits of Different types of Mean.

1. BIOSTATISTICS AND RESEARCH METHODOLOGY
Unit-1: mean
PRESENTED BY
Himanshu Rasyara
B. Pharmacy IV Year
UNDER THE GUIDANCE OF
Gangu Sreelatha M.Pharm., (Ph.D)
Assistant Professor
CMR College of Pharmacy, Hyderabad.
email: sreelatha1801@gmail.com

2. MEAN
• Mean is an essential concept in mathematics and statistics. The mean is the average or the most
common value in a collection of numbers.
• In statistics, it is a measure of central tendency of a probability distribution along median and
mode.
Mean
Arithmetic
Mean
Simple
Arithmetic
Mean
Weighted
Arithmetic
Mean
Geometric
Mean
Harmonic
Mean

3. A) ARITHMETIC MEAN
• The arithmetic mean is the sum of all observations divided by the total number of observations is known
as simple Arithmetic mean.
1. Calculation of Simple Arithmetic Mean
• Simple Arithmetic Mean can be calculated by 2 methods:
a) Direct Method
b) Shortcut Method
I. Calculation of simple arithmetic mean in a series of individual observations.
a) Direct Method: x
̄ =
𝐱𝟏+𝐱𝟐+𝐱𝟑−−−−−−−−−− +𝐱𝐧
𝐧
𝐨𝐫 x̄=Σx/n
x
̄ = Arithmetic Mean
Σx= Sum of all values of the variable, x
n= Number of observations.
• Example: Calculate the mean incubation period of 12 polio cases from the data given below.
16,20,21,19,17,24,23,20,22,18,18,22
Solution: x̄=Σx/n
where Σx= 240 ; n =12
x
̄ = 240/12= 20 So, the mean incubation period of polio cases was found to be 20.

4. b) Shortcut Method: x
̄ = A+
Σ𝒅
𝐧
where
x
̄ = Arithmetic Mean
A= assumed mean
d= deviation of items from the assumed mean (x-A)
Σd= sum of deviations.
Example: The following example shows the calculation of arithmetic mean from the observations.
Number of mean= 18,20,21,19,28,22,29,30
Solution: Assumed mean=24
x
̄ = A+
Σ𝒅
𝐧
A= 24; 𝛴d = −5; n = 8
x
̄ = 24+
−5
8
= 24-0.625=23.37
Number of Observations
x
Deviations from assumed mean
d= (x-A)
18 -6
20 -5
21 -4
19 -3
28 -2
22 4
29 5
30 6
n= 8 Σ𝑑 = −20 + 15 = −5

5. II. Calculation of Simple Arithmetic Mean in a Discrete Series.
a) Direct Method :x̄ =
𝜮𝐟𝐱
𝒏
or x
̄ =
𝜮𝐟𝐱
𝜮𝒇
Where
x
̄ = arithmetic mean
n= number of observations
Σf= sum of frequencies
Σfx= sum of values of the variable and their corresponding frequencies.
Example: Find the arithmetic mean of a sample of
reported cases of mumps in school children
of the following data.
Solution: n=Σf= 30 and Σfx= 1848
Mean = x
̄ =
𝜮𝐟𝐱
𝜮𝒇
=𝟏𝟖𝟒𝟖
𝟑𝟎
= 𝟔𝟏. 𝟔
Blood
LDL (x)
Frequenc
y (f)
f *x
52 7 364
58 5 290
60 4 240
65 6 390
68 3 204
70 3 210
75 2 150
Total n/Σf=30 Σfx=1848
Blood LDL No. of
Patients
52 7
58 5
60 4
65 6
68 3
70 3
75 2

6. b) Shortcut Method:
x
̄ = 𝐀 +
𝚺𝐟𝐝
𝚺𝐟
where
x
̄ = Arithmetic mean
A= Assumed mean
Σfd= Sum of the deviations from the assumed mean and the respective frequencies.
Σf= Sum of frequencies
• Example: Ten patients were examined for uric acid test. The operation was performed 1050 times and
the frequencies thus obtained for different number of patients (x) are shown in the following table.
Calculate the arithmetic mean by short-cut method.
Solution:
• Let 5 be the assumed mean. i.e., a=5. Let us prepare the following table in order to calculate the
arithmetic mean.
x 0 1 2 3 4 5 6 7 8 9 10
f 2 8 43 133 207 260 213 120 54 9 1

7. f d=x-5 fd
0 2 -5 -10
1 8 -4 -32
2 43 -3 -129
3 133 -2 -266
4 207 -1 -207
5 260 0 0
6 213 1 213
7 120 2 240
8 54 3 162
9 9 4 36
10 1 5 5
Σf= 1050 Σfd= 12
Arithmetic Mean= x
̄ = a+ Σfd/ Σd= 5+17/1050= 5+0.0114= 5.0114 cm.
Hence the average for uric acid is 5.0114

8. III) Calculation of Simple arithmetic Mean in a Continuous series.
a) Direct Method: x
̄ =
𝚺𝐟𝐦
𝚺𝐟
Where x
̄ = arithmetic mean
Σfm = sum of values of mid − points mutiplied by the respective frequency of each class.
Σf= sum of frequency
m= mid-point of various classes.
Mid point (m)=
lower limit+upper limit
2
Example: Compute the arithmetic mean from the following data:
Age
Group
0-10 10-20 20-30 30-40 40-50 50-60
No. of
Patients
5 10 25 30 20 10
Age group No. of Patients (f) Mid – points (m) fm
0-10 5 5 25
10-20 10 15 150
20-30 25 25 625
30-40 30 35 1050
40-50 20 45 900
50-60 10 55 550
Σf= 100 Σfm= 3300
Solution:
x
̄ =
𝚺𝐟𝐦
𝚺𝐟
Σfm= 3300 ;Σf= 100
x
̄ =
𝟑𝟑𝟎𝟎
𝟏𝟎𝟎
= 33

9. b) Shortcut Method: x
̄ =A+
𝚺𝐟𝐝
𝚺𝐟
Where x
̄ = Arithmetic mean
A= Assumed mean
Σf = Sum of frequency
d= Deviation of mid-points from assumed mean(m-a)
f= Frequency of each other
Example: Compute the arithmetic mean from the following data:
Age
Group
0-10 10-20 20-30 30-40 40-50 50-60
No. of
Patients
5 10 25 30 20 10
Age group No. of Patients (f) Mid – points (m) fm
0-10 5 5 25
10-20 10 15 150
20-30 25 25 625
30-40 30 35 1050
40-50 20 45 900
50-60 10 55 550
Σf= 100 Σfm= 3300
Solution:
x
̄ =A+
𝚺𝐟𝐝
𝚺𝐟
A= 35; Σfd= -200; Σf= 100
x
̄ = 35+
(−200)
100
= 35+ (-2)= 33

10. c. Step Deviation Method:
x
̄ = 𝑨 +
𝚺𝐟d1
𝚺𝐟
x i – when class-interval are unequal
x
̄ = 𝐀 +
𝚺𝐟d1
𝚺𝐟
𝐱 𝐜 − when class − interval are unequal
Where x
̄ = assumed mean
d1 = deviations of mid points from assumed mean (m-A)/I or c
Σf= sum of frequencies
i= class- interval
c= common factor
d= deviations from assumed mean
• Example: Calculate by step deviation method the arithmetic mean of the blood test for triglyceride of
384 patients.
Triglyceri
des
5 10 15 20 25 30 35 40 45 50
No. of
Patients
20 43 75 67 72 45 39 9 8 6

11. x f Dx= x-a D’= d/i f x d’
5 20 -25 -5 -100
10 43 -15 -4 -172
15 75 -10 -3 -225
20 67 -5 -2 -134
25 72 0 -1 -72
30 45 5 0 0
35 39 10 1 39
40 9 10 2 18
45 8 15 3 24
50 6 20 4 20
Σdx= 384 Σfds’x’= -598
Solution:
Let a=20 and I= 5
x
̄ = a +
Σfd′x′
Σf
x i
30 −
598
384
x 5 = 30 – 1.56 x 5 = 30 – 7.80= 22.2

12. • Merits of Arithmetic Mean.
• It is easy to understand and easy to calculate.
• It is rigidly defined.
• It is based on all the observations.
• It provides a good basis for comparison.
• It is not affected by fluctuations of sampling.
• Demerits of Arithmetic Mean.
• The mean is unduly affected by the extreme items.
• It cannot be accurately determined even if one of the values is not known.

13. WEIGHTED ARITHMETIC MEAN
• It may be defined as the average whose component items are being multiplied by certain value known
as “Weights” and the aggregate of the multiplied results are being divided by the total sum of their
weights instead of the sum of the items.
1. Direct Method:
x̄w=
𝐱𝟏𝐰𝟏
+
𝐱𝟐𝐰𝟐
+
𝐱𝟑𝐰𝟑
+⋯.+
𝐱𝐧𝐰𝐧
𝐰𝟏
+
𝐰𝟐
+
𝐰𝟑
+⋯…….+
𝐰𝐧
=
𝚺𝐱𝐰
𝚺𝐰
where
x̄w= Weighted arithmetic mean
x1,x2,x3……….. xn= variables
w1,w2,w3…….. wn= weights
Σ𝑥𝑤= sum of the products of the values and their respective weights
Σ𝑤= sum of the weights

14. Example: Suppose that a pharmaceutical company conducts a clinical trial on1,000 patients to
determine the average number of disorders in each patients. The data show a large number of patients has
two or three disorders and a smaller number with one or four. Find the mean number of disorders per
patients.
Number of
disorders per
patients
No. of patients
1 73
2 378
3 459
4 90
Solution: As many of the values in this data set are repeated multiple
times, you can easily compute the sample mean as a weighted mean.
Follow these steps to calculate the weighted arithmetic mean:
Step 1: Assign a weight to each value in the dataset:
Step 2: Compute the numerator of the weighted mean formula.
Multiply each sample by its weight and then add the products together:
= (1)(73)+(2)(378)+(3)(459)+(4)(90)
= 73 + 756 + 1377 +360 =2566
Step 3: Now, compute the denominator of the weighted mean formula by
adding the weights together.
= 73 + 378 + 459 + 90 =1000
Step 4: Divide the numerator by the denominator
The mean number of TVs per household in this sample is 2.566.

15. 2. Shortcut (or) Indirect Method:
x̄w = 𝐀𝐰 +
𝚺𝐰. 𝐝𝐱
𝚺𝐰
Where
x̄w= Weighted arithmetic mean
Aw= Assumed weighted mean
dx= Deviation items from assumed mean
w= Weights of various items.
Example: Suppose that a pharmaceutical company conducts a clinical trial on1,000 patients to
determine the average number of disorders in each patients. The data show a large number of patients
has two or three disorders and a smaller number with one or four. Find the mean number of disorders
per patients.
Number of disorders per patients No. of patients
1 73
2 378
3 459
4 90

16. • Solution:
x̄w = Aw +
Σw. dx
Σw
x̄w= 378+
−11006
1000
= 378- 11.006
x̄w= 366.994 = 367
• Merits of Weighted mean.
i. It is simple to calculate
ii. It is easy to understand.
iii. It is rigidly defined.
iv. It is the most representative measures of central tendency.
v. It is not affected by sampling fluctuations.
• Demerits of Weighted mean.
i. The mean can not be calculated when the frequency distribution has open end classes at both the
ends.
ii. It is simple to calculate and can not be located by inspection like median or mode.
iii. It is unduly affected by extremely high or low values.
X w Xw d(x-A) wd
1 73 73 -305 -22265
2 378 756 0 0
3 459 1377 81 37179
4 90 360 -288 -25920
Σw=
1000
Σwd= -
11006

17. B) GEOMETRIC MEAN
• It is defined as the nth root of the product of n items of a series.
• If the geometric mean of items 5 and 20 is to be calculated, we apply the square root.
• If the geometric mean of items 5, 10 and 20 is to be calculated, we apply the cube root.
1. Calculation of geometric mean in a series of individual observations.
i. The log are found out of the given values (x).
ii. Then the logs are added (Σlog x).
iii. The sum of the logs is divided by number of observations and the antilog of the value is found out.
• G.M. of items 5,10,12 and 15 would be fourth root of the product of these figures
G.M.= x1x2x3 −− − xn
4
5 ∗ 10 ∗ 12 ∗ 15 =
4
9000 = 9.74
Where
G.M.= geometric mean
n= number of items
x= values of the variable
• The value of geometric mean is always less than the value of the arithmetic mean.
2. Calculation of geometric mean in a discrete.
G.M. = Antilog{
𝚺𝐟 𝐥𝐨𝐠 𝐱
𝚺𝐟
}
i. Find logs of the variable x.
ii. Multiply these logs with the respective frequencies and obtain the total Σf log x.

18. Example: The number of Basophils in blood of 40 patients of a hospital and their frequency were
recorded as: 12,15,17,20,24 and frequencies 7,9,11,7,6. Find out the value of geometric mean.
Solution:
G.M. = Antilog{
Σf log x
Σf
}
= antilog{
49.0619
40
}
= antilog (1.2265)
= 16.85
No. of Basophils
x
Frequency
f
Log x f log x
12 7 1.0792 7.5544
15 9 1.1761 10.5849
17 11 1.2304 13.5344
20 7 1.3010 9.1070
24 6 1.3802 8.2812
Total n= 40 Σf log x = 49.0619

19. Merits of Geometric Mean.
• It is based on all the observations.
• It is rigidly defined.
• It is not much affected by the fluctuations.
• It is suitable for averaging ration, rates and percentages.
Demerits of Geometric Mean.
• It is difficult to understand and to calculate.
• It cannot be calculated when there are both negative and positive values.
• If one or more of the values are zero, the geometric mean would also be zero.

20. C) HARMONIC MEAN
• It is a type of statistical average which is a suitable measure of central tendency when the data pertains to speed,
rates and time.
• According to Downie and Heath, “ The harmonic mean is used in averaging rates when the time factor is variable
and the act being performed is constant”.
H.M.= n ÷ Σ
1
X
Calculations of Harmonic Mean
i. Calculation of Harmonic Mean in a series of Individual Observations.
H.M.=n÷ 𝛴
1
x
ii. Calculation of Harmonic Mean in a Discrete Series'
H.M.= Σf÷ 𝛴(fx
1
x
)
iii. Calculation of Harmonic Mean in a continuous series.
H.M.=f÷ 𝛴(
f
m
)
Where
n= number of observations
f= frequency
m= mid-points Relationship among Arithmetic Mean, Geometric Mean and Harmonic Mean
A.M.>G.M.>H.M.

21. Calculation of Harmonic Mean in a series of Individual Observations
Example: Haemoglobin percentage of six patients were recorded as 1,5,10,15,20 and 25. Find out the
value of harmonic mean.
Solution: HM = n / [(1/x1)+(1/x2)+(1/x3)+…+(1/xn)]= n÷ 𝛴
1
x
H.M=
6
437
300
=
1800
437
= 4.19
 Merits of Harmonic Mean.
• It is rigidly defined.
• It is based on all the observations of a series.
• It gives greater weightage to the smaller items.
• It is not much affected by sampling fluctuations.
 Demerits of Harmonic Mean.
• It is not easy to calculate and understand.
• It cannot be calculated if the one value is zero.
• It gives larger weightage to the small items.
• It cannot be calculated if negative and positivity