Statistical Calculations: Drug Analysis and Probability Problems

This assignment report presents a comprehensive analysis of statistical calculations and probability. It begins with an introduction to statistical methods, including descriptive and inferential statistics, and then delves into the analysis of data related to Drug A, calculating mean, median, mode, standard deviation, and quartiles. The report then explores probability calculations related to seed germination, determining the probabilities of various outcomes. The final section uses secondary data to calculate the standard deviation of continuous data. The report concludes by summarizing the application of these statistical methods in investigating scientific problems, particularly within the context of pharmaceutical drug performance and genetic purity of seeds.

STATISTICAL
CALCULATIONS

Table of Contents
INTRODUCTION...........................................................................................................................3
TASK 1............................................................................................................................................3
P1: Carry out Statistical calculations to investigate a scientific problem...................................3
TASK 2............................................................................................................................................6
2.1: From 3 seeds planted all seeds will germinate.....................................................................6
2.2: Probability that from 3 seeds planted none will germinate..................................................7
2.3: Out of total seeds but one will germinate............................................................................7
Solution:.................................................................................................................................7
2.4: ..............................................................................................................................................8
TASK 3............................................................................................................................................8
On the basis of secondary data the following calculations has been done..................................8
CONCLUSION................................................................................................................................9
REFERENCES..............................................................................................................................10

INTRODUCTION
The statistical calculations uses mathematical science which includes methods such as
recording, classifying and summarising of the data in a manner that becomes meaningful to make
easy interpretation and conclusions by using that particular data. Basically the statistics methods
are divided into two detailed categories such as inferential and descriptive statistics.(Acko, and
et. al., 2014)The statistical calculations used in this project report assists in investigating a
scientific problems, by calculating various variables such as measures of central tendencies such
as mean, median and mode. It also calculates variables such as standard deviation, maximum
range and minimum range which comes under measures of dispersion of statistics. The
probability methods are also utilised in this report to come at different solutions. Normally
distributed data is used in the final task to calculate the standard deviations of described
distribution.
TASK 1
P1: Carry out Statistical calculations to investigate a scientific problem
TABLE OF THE
RESULTS
DRUG A (%) cumulative
56 56
0.31305049
23
0.55950915
3
68 124
0.54206354
8
0.73624965
06
71 195
0.92808363
92
0.96337097
69
50 245
0.74081238
58
0.86070458
68
45 290
0.32122152
88
0.56676408
57
67 357
0.99423670
9
0.99711419
06

64 421
0.70958918
95
0.84237117
09
61 482
0.56450745
77
0.75133711
32
43 525
0.01476796
96
0.12152353
52
71 596
0.55878415
39
0.74751866
46
57 653
0.50927689
12
0.71363638
59
56 709
0.76957715
7
0.87725546
85
60 769
0.42213736
75
0.64972099
21
68 837
0.99575054
41 0.99787301
53 890
0.19024340
57
0.43616901
05
58 948
0.48595563
88
0.69710518
49
58 1006
0.51854876
68
0.72010330
29
49 1055
0.73584397
76
0.85781348
65
61 1116
0.42913711
53
0.65508557
86
64 1180
0.98646567
19
0.99320978
24
62 1242
0.61498407
18
0.78420920
16
67 1309 0.82689411 0.90933718

11 23
65 1374
0.75241557
39
0.86741891
49
60 1434
0.51295595
34
0.71620943
4
MEAN 59.75
MEDIAN 60.5
MODE 56
STANDARD
DEVIATION
7.69669607
27
MAXIMUM RANGE 71
MINIMUM RANGE 43
UPPER QUARTILE 65.5
LOWER QUARTILE 56
INTERQUARTILE 9.5
STANDARD ERROR
1.57108150
69

From the above calculations of data related to Drug A, the mean has been calculated as 59.75 ,
median as 60.5 and the modal value of the data is 56. ( Gelman, and Carlin, 2014)The measures
of dispersion such as standard deviation, maximum range and minimum range is calculated as
1.5710815069, 71 and 43 respectively. The upper quartile , lower quartile and interquartile of the
data series 65.5, 56 and 9.5 respectively.
DRUG B(%)
20
84
62
16
19
26
96
79
26
89
78
95
89
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Column C

67
45
39
MEAN 58.125
MEDIAN 64.5
MODE 20
STANDARD
DEVIATION
30.1725592
772
MAXIMUM RANGE 96
MINIMUM RANGE 16
INTERQUARTILE 64.5
STANDARD ERROR
7.54313981
93
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
0
20
40
60
80
100
120
DRUG B(%)

TASK 2
2.1: From 3 seeds planted all seeds will germinate
Total number of events: 3
Results
Multiple Event Probability Value (decimal) Value (percent)
P(A) 0.667 66.667%
P(A') 0.333 33.333%
P(B) 0 0%
P(B') 1 100%
P(A ∩ B) 0 0%
P(A ∪ B) 0.667 66.667%
P(A | B) - -
P(B | A) 0 0%
2.2: Probability that from 3 seeds planted none will germinate
Total number of events: 3
Results
Multiple Event Probability Value (decimal) Value (percent)
P(A) 0 0%
P(A') 1 100%
P(B) 0.667 66.667%
P(B') 0.333 33.333%
P(A ∩ B) 0 0%
P(A ∪ B) 0.667 66.667%
P(A | B) 0 0%
P(B | A) - -
2.3: Out of total seeds but one will germinate
Here, we are required to calculate number of combinations of a data:
the number of combinations (without repetition) is 10

Solution:
We have n = 10 and r = 1
Combination Formula (no repetition is allowed)
C(n, r) = n! / (r! * (n - r)!)
Find n!
10! = 1×2×3×4×5×6×7×8×9×10
= 3628800
Find r!
1! = 1
= 1
Find (n - r)!
(n - r)! = (10 - 1)! = 9!
9! = 1×2×3×4×5×6×7×8×9
= 362880
Result
C(n, r) = n! / (r! * (n - r)!)
C(n, r) = 3628800 / (1 * 362880)
C(n, r) = 10
It has been calculated from the above combinations that the total number of combinations that
can be made from the above data is calculated as 3628800, by taking n factorial as 10. The result
has come out to be 362880 when we considered (n-r)!. And finally the C(n,r) has been
calculated as 10.
2.4:
The company which is claiming the genetic purity of the seeds states that their seeds are
purely breaded with the guaranteed genotypes although when the farmers cross breaded these

pea plants with the other ones they showed an 75% of the spherical seeds and the other 25% as
the wrinkled seeds. By performing the genetic process the following outcomes of probability
came.( Howell, 2012)
Probability that event A does not occur, P(A'):0.25
Probability that event B does not occur, P(B'):0.75
Probability that event A and/or event B occurs, P(A B): 0.8125∪
Probability that event A and event B both occur, P(A∩B): 0.1875
Probability that either A or event B occurs, but not both: 0.625
TASK 3
On the basis of secondary data the following calculations has been done
Continuous data:
Timings(In seconds)
10
11
11
12
12
12
12
13
13
14
STANDARD
DEVIATION
1.15470053
84
MEAN 12
Formula through which standard deviation has been calculated for the above data: