# Statistics for Rainfall on Sunday – Assignment Example

Case Study I Rainfall on Sunday Sr. No es Frequency x xf Cf 0.0 - 0 44 0.05 2.2 44 2 0 0.2 0 0.15 0 44 3 0.2 - 0.3 5 0.25 25 49 4 0.3- 0.4
1
0.35
0.35
50
5
0.4 - 0.5
1
0.45
0.45
51
6
0.5 - 0.6
0
0.55
0
51
7
0.6 - 0.7
0
0.65
0
51
8
0.7 - 0.8
0
0.75
0
51
9
0.8 - 0.9
0
0.85
0
51
10
0.9 - 1.0
0
0.95
0
51
11
1.0 - 1.1
0
1.05
0
51
12
1.1 - 1.2
0
1.15
0
51
13
1.2 - 1.3
1
1.25
1.25
52
Summation
52

5.5
a) Histogram for the rainfall on Sunday
Histogram for the Rainfall on Thursday
b) For creating a good frequency table, the number of classes should be neither less nor more. With the interval of 0.1, the number of classes is 13, which is neither less nor great. Secondly, the data is being more easily defined because of this class size that is why this size has been selected. In statistics, a proper method has been introduced to select class interval for example, Salkind (2009) argues that a class interval should be such that 10 to 20 number of class cover the entire set of data.
2) The histograms for both parts are almost same however; the number of classes in first histogram is greater than the other one. The highest frequency for class one is same for both histograms. The total frequency of both histograms is also same however, 7 classes in histogram 1 are giving null frequency whereas, only 3 classes in histogram 2 are showing null frequency.
3)
Mean:
Mean = Єxf/Єf
Mean = 0.106
Median:
Median = n/2
Median = 52/2 = 26
Lower limit of median class = 0.0
f = Frequency of median class = 44
cf = Cumulative frequency of preceding class of modal class
n/2 = 26
h = class interval = 0.1
Median = l + {(n/2 – cf)/f} * h
Median = 0.0 + {(26 – 0)/44} * 0.1
Median = 0.059
Mode:
l = lower limit of modal class = 0.0
fo = frequency of modal class = 44
f1 = frequency of preceding class to modal class = 0
f2 = frequency of succeeding class to modal class = 0
Mode = l + {(fo – f1)/(2fo – f1 – f2)} * h
Mode = 0.0 + {(44 – 0)/(88 – 0 – 0)} * 0.1
Mode = 0.05
Other way to present Data
For Rainfall on Sunday
For Rainfall on Thursday
Rainfall on Thursday
Sr. No
Classes
Frequency
x(mid point of class interval)
xf
cf
1
0.0 - 0.1
44
0.05
2.2
44
2
0.1 - 0.2
2
0.15
0.3
46
3
0.2 - 0.3
3
0.25
0.75
59
4
0.3- 0.4
1
0.35
0.35
50
5
0.4 - 0.5
0
0.45
0
50
6
0.5 - 0.6
0
0.55
0
50
7
0.6 - 0.7
0
0.65
0
50
8
0.7 - 0.8
1
0.75
0.75
51
9
0.8 - 0.9
1
0.85
0.85
52
Summation
52

5.2
Mean:
Mean = Єxf/Єf
Mean = 0.1
Median:
Median = n/2
Median = 52/2 = 26
Lower limit of median class = 0.0
f = Frequency of median class = 44
cf = Cumulative frequency of preceding class to modal class = 0
n/2 = 26
h = class interval = 0.1
Median = l + {(n/2 – cf)/f} * h
Median = 0.0 + {(26 – 0)/44} * 0.1
Median = 0.059
Mode:
l = lower limit of modal class = 0.0
fo = frequency of modal class = 44
f1 = frequency of preceding class to modal class = 0
f2 = frequency of succeeding class to modal class = 2
Mode = l + {(fo – f1)/(2fo – f1 – f2)} * h
Mode = 0.0 + {(44 – 0)/(88 – 0 – 2)} * 0.1
Mode = 0.051
4) The report that claims that it rains more on weekends than on other weekdays is reliable because my results also demonstrate that the mean of rainfall on Sunday is 0.106 which is greater than the mean of rainfall on Thursday, which is 0.1. The higher mean of rainfall on Sunday shows more rain on weekends as compared to rain on week days.
6) No
7) In such that there can be many practical errors. The major one is the personal error, which could be because of lack of attention, expertise or ignorance. The instrumental error is another cause of practical error. Moreover, external or environmental conditions also increase the chances of practical error.

Case Study II
1) Range of Each Section
Section 1:
Range = Xm – Xo
Range = 20 – 1
Range = 19
Section 2:
Range = Xm – Xo
Range = 19 – 2
Range = 17
2) Standard Deviation
Section 1:
x
x2
1
1
20
400
20
400
20
400
20
400
20
400
20
400
20
400
20
400
20
400
20
400
Єx=201
Єx2=4001
S.D = (√Єx2/n – (Єx/n) 2)
S.D = (√4001/11 – (201/11) 2)
S.D = (363.727 – 333.89)
S.D = (√29.837)
S.D = 5.46
Section 2:
x
x2
2
4
3
9
4
16
5
25
6
36
14
196
15
225
16
256
17
289
18
324
19
361
Єx=119
Єx2=1741
S.D = (√Єx2/n – (Єx/n) 2)
S.D = (√1741/11 – (119/11) 2)
S.D = (158.27 – 116.94)
S.D = (√41.33)
S.D = 6.42
3) The range of data is the difference between the highest and lowest number of data (StatuTutorialText, n.d.). Therefore, it gives an immediate idea about the variation in data. The data in section 1 is very consistent with very slight variation therefore, its range is 19. On the other hand, relatively many variations have been seen in the data of section 2, therefore, its range is 17.
4) The range is misleading because range is a cluster point and the other observations should fall near it. However, in this case, there is a significant difference between range and all other observations that is why range is misleading.
5) The standard deviation of section 2 is greater than section 1 which means that observation of second data has relatively more variations from its mean value.
Bibliography
Salkind, N. J. (2009) Statistics for People Who (Think They) Hate Statistics: Excel 2007 Edition. [Online]