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BIOSTATISTICS
STANDARD DEVIATION ,VARIANCE
RANGE,
COEFFEICIENT OF SD AND VARIANCE
By Mr Payaam Vohra
NIPER AIR 11
Gold Medalist in MU
MET AIR 07
ICT MTECH SCORE RANK 01
CUET-PG AIR 01
IIT BHU AIR 08
GPAT AIR 43
IV B.PHARMACY (BIO STATISTICS)
IV B.PHARMACY (BIO STATISTICS)
IV B.PHARMACY (BIO STATISTICS)
Q.NO 12: Solution
Given that 20,35,25,30,15
Smallest value(S) =15
L argest value(L) =35
Range = L S = 35 15 = 20
Coefficient of Range =
L S
=
35 15
=
20
=
0
.
4
L + S 35 +15 50
Q.NO 13: Solution
Given that10,20,30,40,50,60,70
Smallest value(S) =10
L argest value(L) =70
Range = L S = 70 10 = 60
Coefficient of Range =
L S
=
70 10
=
60
=
0.75
L + S 70 +10 80
Q.NO 14: Solution
Smallest value(S) = 5
L argest value(L) =30
Range = L S = 30 5 = 25
Coefficient of Range =
L S
=
30 5
=
25
=
0.7143
L + S 30 +5 35
IV B.PHARMACY (BIO STATISTICS)
STANDARD ERROR OF MEAN (𝑆. 𝐸[π‘₯]):
The standard error of a statistic is the standard deviation of its sampling
distribution or an estimate of that standard deviation. If the statistic is the
sample mean, it is called the standard error of the mean.
Example: Standard error of sample mean 𝑆. 𝐸[π‘₯] =
𝜎
βˆšπ‘›
20
IV B.PHARMACY (BIO STATISTICS)
PROBLEMS ON STANDARD DEVIATION (S.D), VARIANCE &
COEFFICIENTS OF VARIATION (C.V) & Standard Error (S.E)
Problem-15: Find the Mean, S.D, Variance, coefficients of variations (C.V)
and Coefficients of standard deviation, Standard Error (S.E) of mean of the
following data on random blood sugar(mg) of 10 individuals recorded in private
hospitals.
110 115 165 90 135 125 105 135 165 155
SOLUTION:
(i)Mean(x)=
xi
= 110 +115+165+ 90 +135+125+105+135+165+155
n 10
10
Mean(x) =130
Calculate for S.D
= 1300
=130
x (x x) (x x)2
110 -20 400
115 -15 225
165 35 1225
90 -40 1600
135 5 25
125 -5 25
105 -25 625
135 5 25
165 35 1225
155 25 625
IV B.PHARMACY (BIO STATISTICS)
(x x)
2
n 10
n 10
(x x)2
= 5975
(ii)S tandard Deviation ( ) = =
5975
= 24.44
(iii)Variance ( 2
)= (24.44)2
= 597.5
(iv) Coefficients of S tan dard Deviation = =
24.44
= 0.19
(v) Coefficients of Variation(CV ) =
x 130
100 =
24.44
100 =19
x 130
(vi) S tan dard Error of Mean S.E(x) = =
24.44
= 7.73
Problems-16: Calculate the S.D, Vaiance, Coefficient of SD, CV, S.E of mean
from the following data.
x 1 2 3 4 5 6 7
f 5 9 12 17 14 19 9
SOLUTION:
Calculate for Mean &SD:
x f fx (x x) (x x)2
f (x x)2
1 5 5 -3.4 11.56 57.8
2 9 18 -2.4 5.76 51.84
3 12 36 -1.4 1.96 23.52
4 17 68 -0.4 0.16 2.72
5 14 70 0.6 0.36 5.04
6 19 114 1.6 2.56 48.64
7 9 63 2.6 6.76 60.84
N = f =85 fx =374 f (x x)2
= 250.6
f (x x)
2
N 85
N 85
IV B.PHARMACY (BIO STATISTICS)
(i)Mean(x) =
fx
=
374
= 4.4
N 85
Mean(x) = 4.4
(ii)S tan dard Deviation ( ) = =
250.6
=1.72
(iii)Variance ( 2
)= (1.72)2
= 2.95
(iv) Coefficients of S tan dard Deviation = =
1.72
= 0.39
x 4.4
(v) Coefficients of Variation(CV ) = 100 =
1.72
100 = 39
x 4.4
(vi) S tan dard Error of Mean S.E(x) = =
1.72
= 0.19
Problem-17: Calculate the mean, S.D and coefficients of variations from the
following data.
Age 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40
No. of
persons
2 5 7 13 21 16 8 3
SOLUTION:
Age No of
persons(f)
Mid
points(x)
π’‡π’Ž (π’Ž
β€” 𝒙)
(π’Ž
β€” 𝒙)𝟐
𝒇(π’Ž βˆ’ 𝒙)𝟐
0.-5 2 2.5 5 -19.4 376.36 752.72
5-10 5 7.5 37.5 -14.4 207.36 1036.8
10-15 7 12.5 87.5 -9.4 88.36 618.52
15-20 13 17.5 227.5 -4.4 19.36 251.68
20-25 21 22.5 472.5 0.6 0.36 7.56
25-30 16 27.5 440.0 5.6 31.36 501.76
30-35 8 32.5 260.0 10.6 112.36 898.88
35-40 3 37.5 112.5 15.6 243.36 730.08
βˆ‘π’‡ = 𝑡
= πŸ•πŸ“
βˆ‘π’‡π’Ž
= πŸπŸ”πŸ’πŸ. πŸ“
βˆ‘π’‡(π’Ž βˆ’ 𝒙)𝟐
= πŸ’πŸ•πŸ—πŸ–
IV B.PHARMACY (BIO STATISTICS)
βˆ‘π‘“π‘š 1642.2
π‘€π‘’π‘Žπ‘› = π‘₯ = = = 21.9
𝑁 75
𝑆tanπ‘‘π‘Žπ‘Ÿπ‘‘ π·π‘’π‘£π‘–π‘Žπ‘‘π‘–π‘œπ‘›(𝑆. 𝐷) = βˆšβˆ‘π‘“(π‘šβˆ’π‘₯)2
= √ = 7.9983
4798
𝑁 75
πΆπ‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ 𝑆tanπ‘‘π‘Žπ‘Ÿπ‘‘ π·π‘’π‘£π‘–π‘Žπ‘‘π‘–π‘œπ‘› =
𝜎
=
7.9983
= 0.3652
π‘₯ 21.9
πΆπ‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘π‘  π‘œπ‘“ π‘‰π‘Žπ‘Ÿπ‘–π‘Žπ‘‘π‘–π‘œπ‘›(𝐢. 𝑉) = 𝜎
Γ— 100 =
7.9983
Γ—
π‘₯ 21.9
100 = 36.521
Problem-18: During the 10 weeks of a session, the marks obtained by two
candidates, Ramesh &Suresh taking the computer programming course are
given below:
Ramesh 58 59 60 54 65 66 52 75 69 52
Suresh 87 89 78 71 73 84 65 66 56 46
(i) Who is better scorer?
(ii) Who is more consistent?
Solution:
Mean of Ramesh = x = 58+ 59+ 60 +54 + 65+ 66 + 52+ 75+ 69+ 52 = 610 = 61
10 10
Mean of Suresh = y = 87 +89+ 78+ 71+ 73+84+ 65+ 66 + 56+ 46 = 715 = 71.5
10 10
(i) π‘šπ‘’π‘Žπ‘› π‘œπ‘“ π‘ π‘’π‘Ÿπ‘’π‘ β„Ž > π‘šπ‘’π‘Žπ‘› π‘œπ‘“ π‘Ÿπ‘Žπ‘šπ‘’π‘ β„Ž
𝑖. 𝑒. , y > x
𝑖. 𝑒. ,71.5 > 61
Therefore, Suresh is better scorer.
(ii) 𝑆. 𝐷 π‘œπ‘“ π‘…π‘Žπ‘šπ‘’π‘ β„Ž = 𝜎1
𝑆. 𝐷 π‘œπ‘“ π‘†π‘’π‘Ÿπ‘’π‘ β„Ž = 𝜎2
= βˆšβˆ‘(π‘₯βˆ’π‘₯)2
𝑛
= βˆšβˆ‘(π‘¦βˆ’π‘¦)2
𝑛
IV B.PHARMACY (BIO STATISTICS)
Ramesh(x) (𝒙 βˆ’ 𝒙) (𝒙 βˆ’ 𝒙)𝟐
Suresh(y) (π’š βˆ’ π’š) (π’š βˆ’ π’š)𝟐
58 -3 9 87 15.5 240.25
59 -2 4 89 17.5 306.25
60 -1 1 78 6.5 42.25
54 -7 49 71 -0.5 0.25
65 4 16 73 1.5 2.25
66 5 25 84 12.5 156.25
52 -9 81 65 -6.5 42.25
75 14 196 66 -5.5 30.25
69 8 64 56 -15.5 240.25
52 -9 81 46 -25.5 650.25
TOTAL 526 1710.5
𝑆. π·π‘œπ‘“π‘…π‘Žπ‘šπ‘’π‘ β„Ž = 𝜎1
526
= √ = 7.25
10
𝑆. π·π‘œπ‘“π‘†π‘’π‘Ÿπ‘’π‘ β„Ž = 𝜎2
1710.5
= √ = 13.08
10
𝜎1 7.25
𝐢. 𝑉 π‘œπ‘“ π‘…π‘Žπ‘šπ‘’π‘ β„Ž = = Γ— 100 = 11.89
π‘₯ 61
𝑦 71.5
𝜎2 13.08
𝐢. 𝑉 π‘œπ‘“ π‘†π‘’π‘Ÿπ‘’π‘ β„Ž = = Γ— 100 = 18.29
C.V of Ramesh is less than C.V of Suresh
𝑖. 𝑒. , 11.89 < 18.29
Therefore, Ramesh is more consistent.
IV B.PHARMACY (BIO STATISTICS)
Problem-19: The mean and SD of two brands of lights bulbs are given below:
Brand-1 Brand-2
Mean 800 hours 770 hours
SD 100 hours 60 hours
Calculate a measure of relative dispersion for the two brands interpret the
results? (H.W)
Problem-20: The runs scored by two batsmen Sehwag & Sachin in ten innings
are as follows:
By
sehwag
10 115 5 73 7 120 36 84 29 19
By
sachin
45 12 76 42 4 50 37 48 13 0
Who is better run? Who is more consistent?
(Homework)
MEASURES OF CORRELATION
Definition: In a bivariate distribution the relationship between two variables is
called correlation (OR) If the change in one variable affects a change in the
other variable, then the variables are said to be Correlated.
For example: 1) The Price & Demand.
2)The Income & Expenditure.
3) The Heights & Weights of a group of persons.
Types of correlation: There are 5 types of correlation. They are as follows:
1. Positive correlation: In bivariate distribution one variable increase (or)
decrease, the other variable also increase(or) decrease, then the relation is
called positive correlation. If r > 0.
Example: Price & Quality
2. Negative correlation: In bivariate distribution one variable increase (or)
decrease, the other variable also decrease (or) increase, then the relation is
called negative correlation. If r < 0.
Example: Price & Sales
3. Non-sense(or)Uncorrelation: The correlation which gives no meaning is
called Non-sense correlation. If r = 0.
Example: Beauty & Honesty

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  • 1. BIOSTATISTICS STANDARD DEVIATION ,VARIANCE RANGE, COEFFEICIENT OF SD AND VARIANCE By Mr Payaam Vohra NIPER AIR 11 Gold Medalist in MU MET AIR 07 ICT MTECH SCORE RANK 01 CUET-PG AIR 01 IIT BHU AIR 08 GPAT AIR 43
  • 2. IV B.PHARMACY (BIO STATISTICS)
  • 3. IV B.PHARMACY (BIO STATISTICS)
  • 4. IV B.PHARMACY (BIO STATISTICS) Q.NO 12: Solution Given that 20,35,25,30,15 Smallest value(S) =15 L argest value(L) =35 Range = L S = 35 15 = 20 Coefficient of Range = L S = 35 15 = 20 = 0 . 4 L + S 35 +15 50 Q.NO 13: Solution Given that10,20,30,40,50,60,70 Smallest value(S) =10 L argest value(L) =70 Range = L S = 70 10 = 60 Coefficient of Range = L S = 70 10 = 60 = 0.75 L + S 70 +10 80 Q.NO 14: Solution Smallest value(S) = 5 L argest value(L) =30 Range = L S = 30 5 = 25 Coefficient of Range = L S = 30 5 = 25 = 0.7143 L + S 30 +5 35
  • 5. IV B.PHARMACY (BIO STATISTICS) STANDARD ERROR OF MEAN (𝑆. 𝐸[π‘₯]): The standard error of a statistic is the standard deviation of its sampling distribution or an estimate of that standard deviation. If the statistic is the sample mean, it is called the standard error of the mean. Example: Standard error of sample mean 𝑆. 𝐸[π‘₯] = 𝜎 βˆšπ‘› 20
  • 6. IV B.PHARMACY (BIO STATISTICS) PROBLEMS ON STANDARD DEVIATION (S.D), VARIANCE & COEFFICIENTS OF VARIATION (C.V) & Standard Error (S.E) Problem-15: Find the Mean, S.D, Variance, coefficients of variations (C.V) and Coefficients of standard deviation, Standard Error (S.E) of mean of the following data on random blood sugar(mg) of 10 individuals recorded in private hospitals. 110 115 165 90 135 125 105 135 165 155 SOLUTION: (i)Mean(x)= xi = 110 +115+165+ 90 +135+125+105+135+165+155 n 10 10 Mean(x) =130 Calculate for S.D = 1300 =130 x (x x) (x x)2 110 -20 400 115 -15 225 165 35 1225 90 -40 1600 135 5 25 125 -5 25 105 -25 625 135 5 25 165 35 1225 155 25 625
  • 7. IV B.PHARMACY (BIO STATISTICS) (x x) 2 n 10 n 10 (x x)2 = 5975 (ii)S tandard Deviation ( ) = = 5975 = 24.44 (iii)Variance ( 2 )= (24.44)2 = 597.5 (iv) Coefficients of S tan dard Deviation = = 24.44 = 0.19 (v) Coefficients of Variation(CV ) = x 130 100 = 24.44 100 =19 x 130 (vi) S tan dard Error of Mean S.E(x) = = 24.44 = 7.73 Problems-16: Calculate the S.D, Vaiance, Coefficient of SD, CV, S.E of mean from the following data. x 1 2 3 4 5 6 7 f 5 9 12 17 14 19 9 SOLUTION: Calculate for Mean &SD: x f fx (x x) (x x)2 f (x x)2 1 5 5 -3.4 11.56 57.8 2 9 18 -2.4 5.76 51.84 3 12 36 -1.4 1.96 23.52 4 17 68 -0.4 0.16 2.72 5 14 70 0.6 0.36 5.04 6 19 114 1.6 2.56 48.64 7 9 63 2.6 6.76 60.84 N = f =85 fx =374 f (x x)2 = 250.6
  • 8. f (x x) 2 N 85 N 85 IV B.PHARMACY (BIO STATISTICS) (i)Mean(x) = fx = 374 = 4.4 N 85 Mean(x) = 4.4 (ii)S tan dard Deviation ( ) = = 250.6 =1.72 (iii)Variance ( 2 )= (1.72)2 = 2.95 (iv) Coefficients of S tan dard Deviation = = 1.72 = 0.39 x 4.4 (v) Coefficients of Variation(CV ) = 100 = 1.72 100 = 39 x 4.4 (vi) S tan dard Error of Mean S.E(x) = = 1.72 = 0.19 Problem-17: Calculate the mean, S.D and coefficients of variations from the following data. Age 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40 No. of persons 2 5 7 13 21 16 8 3 SOLUTION: Age No of persons(f) Mid points(x) π’‡π’Ž (π’Ž β€” 𝒙) (π’Ž β€” 𝒙)𝟐 𝒇(π’Ž βˆ’ 𝒙)𝟐 0.-5 2 2.5 5 -19.4 376.36 752.72 5-10 5 7.5 37.5 -14.4 207.36 1036.8 10-15 7 12.5 87.5 -9.4 88.36 618.52 15-20 13 17.5 227.5 -4.4 19.36 251.68 20-25 21 22.5 472.5 0.6 0.36 7.56 25-30 16 27.5 440.0 5.6 31.36 501.76 30-35 8 32.5 260.0 10.6 112.36 898.88 35-40 3 37.5 112.5 15.6 243.36 730.08 βˆ‘π’‡ = 𝑡 = πŸ•πŸ“ βˆ‘π’‡π’Ž = πŸπŸ”πŸ’πŸ. πŸ“ βˆ‘π’‡(π’Ž βˆ’ 𝒙)𝟐 = πŸ’πŸ•πŸ—πŸ–
  • 9. IV B.PHARMACY (BIO STATISTICS) βˆ‘π‘“π‘š 1642.2 π‘€π‘’π‘Žπ‘› = π‘₯ = = = 21.9 𝑁 75 𝑆tanπ‘‘π‘Žπ‘Ÿπ‘‘ π·π‘’π‘£π‘–π‘Žπ‘‘π‘–π‘œπ‘›(𝑆. 𝐷) = βˆšβˆ‘π‘“(π‘šβˆ’π‘₯)2 = √ = 7.9983 4798 𝑁 75 πΆπ‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ 𝑆tanπ‘‘π‘Žπ‘Ÿπ‘‘ π·π‘’π‘£π‘–π‘Žπ‘‘π‘–π‘œπ‘› = 𝜎 = 7.9983 = 0.3652 π‘₯ 21.9 πΆπ‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘π‘  π‘œπ‘“ π‘‰π‘Žπ‘Ÿπ‘–π‘Žπ‘‘π‘–π‘œπ‘›(𝐢. 𝑉) = 𝜎 Γ— 100 = 7.9983 Γ— π‘₯ 21.9 100 = 36.521 Problem-18: During the 10 weeks of a session, the marks obtained by two candidates, Ramesh &Suresh taking the computer programming course are given below: Ramesh 58 59 60 54 65 66 52 75 69 52 Suresh 87 89 78 71 73 84 65 66 56 46 (i) Who is better scorer? (ii) Who is more consistent? Solution: Mean of Ramesh = x = 58+ 59+ 60 +54 + 65+ 66 + 52+ 75+ 69+ 52 = 610 = 61 10 10 Mean of Suresh = y = 87 +89+ 78+ 71+ 73+84+ 65+ 66 + 56+ 46 = 715 = 71.5 10 10 (i) π‘šπ‘’π‘Žπ‘› π‘œπ‘“ π‘ π‘’π‘Ÿπ‘’π‘ β„Ž > π‘šπ‘’π‘Žπ‘› π‘œπ‘“ π‘Ÿπ‘Žπ‘šπ‘’π‘ β„Ž 𝑖. 𝑒. , y > x 𝑖. 𝑒. ,71.5 > 61 Therefore, Suresh is better scorer. (ii) 𝑆. 𝐷 π‘œπ‘“ π‘…π‘Žπ‘šπ‘’π‘ β„Ž = 𝜎1 𝑆. 𝐷 π‘œπ‘“ π‘†π‘’π‘Ÿπ‘’π‘ β„Ž = 𝜎2 = βˆšβˆ‘(π‘₯βˆ’π‘₯)2 𝑛 = βˆšβˆ‘(π‘¦βˆ’π‘¦)2 𝑛
  • 10. IV B.PHARMACY (BIO STATISTICS) Ramesh(x) (𝒙 βˆ’ 𝒙) (𝒙 βˆ’ 𝒙)𝟐 Suresh(y) (π’š βˆ’ π’š) (π’š βˆ’ π’š)𝟐 58 -3 9 87 15.5 240.25 59 -2 4 89 17.5 306.25 60 -1 1 78 6.5 42.25 54 -7 49 71 -0.5 0.25 65 4 16 73 1.5 2.25 66 5 25 84 12.5 156.25 52 -9 81 65 -6.5 42.25 75 14 196 66 -5.5 30.25 69 8 64 56 -15.5 240.25 52 -9 81 46 -25.5 650.25 TOTAL 526 1710.5 𝑆. π·π‘œπ‘“π‘…π‘Žπ‘šπ‘’π‘ β„Ž = 𝜎1 526 = √ = 7.25 10 𝑆. π·π‘œπ‘“π‘†π‘’π‘Ÿπ‘’π‘ β„Ž = 𝜎2 1710.5 = √ = 13.08 10 𝜎1 7.25 𝐢. 𝑉 π‘œπ‘“ π‘…π‘Žπ‘šπ‘’π‘ β„Ž = = Γ— 100 = 11.89 π‘₯ 61 𝑦 71.5 𝜎2 13.08 𝐢. 𝑉 π‘œπ‘“ π‘†π‘’π‘Ÿπ‘’π‘ β„Ž = = Γ— 100 = 18.29 C.V of Ramesh is less than C.V of Suresh 𝑖. 𝑒. , 11.89 < 18.29 Therefore, Ramesh is more consistent.
  • 11. IV B.PHARMACY (BIO STATISTICS) Problem-19: The mean and SD of two brands of lights bulbs are given below: Brand-1 Brand-2 Mean 800 hours 770 hours SD 100 hours 60 hours Calculate a measure of relative dispersion for the two brands interpret the results? (H.W) Problem-20: The runs scored by two batsmen Sehwag & Sachin in ten innings are as follows: By sehwag 10 115 5 73 7 120 36 84 29 19 By sachin 45 12 76 42 4 50 37 48 13 0 Who is better run? Who is more consistent? (Homework) MEASURES OF CORRELATION Definition: In a bivariate distribution the relationship between two variables is called correlation (OR) If the change in one variable affects a change in the other variable, then the variables are said to be Correlated. For example: 1) The Price & Demand. 2)The Income & Expenditure. 3) The Heights & Weights of a group of persons. Types of correlation: There are 5 types of correlation. They are as follows: 1. Positive correlation: In bivariate distribution one variable increase (or) decrease, the other variable also increase(or) decrease, then the relation is called positive correlation. If r > 0. Example: Price & Quality 2. Negative correlation: In bivariate distribution one variable increase (or) decrease, the other variable also decrease (or) increase, then the relation is called negative correlation. If r < 0. Example: Price & Sales 3. Non-sense(or)Uncorrelation: The correlation which gives no meaning is called Non-sense correlation. If r = 0. Example: Beauty & Honesty