Numeracy and Data Analysis

This document provides an analysis of weather data for Kolkata, India from 24th April to 3rd May 2017. It includes tables and charts for temperature, gust, cloud percentage, humidity percentage, rain, and air pressure. The document also explains the steps for calculating mean, median, mode, range, and standard deviation for each parameter.

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Numeracy and Data Analysis

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1. The collected data from www.worldweatheronline.com has been presented in a table
format below in Table 1.
Table 1: Weather Report for Kolkata, India from 24April to 03 May 2017
Weather Parameters 24-Apr 25-Apr 26-Apr 27-Apr 28-Apr 29-Apr 30-Apr 01-May 02-May 03-May
Temp °c 30.50 32.50 28.38 32.00 32.13 32.00 32.00 31.50 30.88 32.00
Gust km/h 27.38 28.75 17.13 19.50 19.38 22.50 22.75 22.63 14.63 16.88
Cloud % 47.13% 9.63% 37.75% 9.13% 11.88% 14.13% 17.50% 19.88% 14.63% 8.50%
Humidity % 72.38% 62.88% 72.63% 63.13% 64.25% 63.38% 62.75% 63.38% 60.00% 60.75%
Rain (mm) 0.0000 0.0000 0.0125 0.0000 0.0000 0.0000 0.0000 1.0000 0.0000 0.0000
Pressure (mb) 1004.63 1003.63 1011.00 1007.38 1006.88 1007.50 1007.25 1009.50 1010.00 1009.75
2. The 10 day weather data has been presented using column chart and line chart.
Figure 1: Column And Line chart for temperature in Kolkata from 24th April to 3rd May 2017
Figure 2: Column And Line chart for gust in Kolkata from 24th April to 3rd May 2017

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Figure 3: Column And Line chart for cloud percentage in Kolkata from 24th April to 3rd May 2017
Figure 4: Column And Line chart for humidity percentage in Kolkata from 24th April to 3rd May 2017
Figure 5: Column And Line chart for rain in Kolkata from 24th April to 3rd May 2017

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Figure 6: Column And Line chart for air pressure in Kolkata from 24th April to 3rd May 2017
3. Steps for the calculations shown and the final values highlighted.
I. Mean
Arithmetic Mean for temperature in degree Celsius =
μT =30 . 50+32 .50+28 . 38+32. 00+32 . 13+ 32. 00+32 . 00+ 31. 50+30 . 88+32. 00
10
= 31.39
Arithmetic Mean for gust in km/h =
μG=27 . 38+28 . 75+ 17 .13+19 . 50+19. 38+22 .50+ 22. 75+22 .63+14 .63+16 . 88
10
=
21.15
Arithmetic Mean for cloud in percentage =
μC=47. 13%+ 9 .63% +37 .75% +9 .13% +11. 88%+ 14 .13%+17 .50% +19 .88% +14 . 63%+8 . 50%
10
=19.01% (Kaliyadan, and Kulkarni, 2019, p.82).

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Arithmetic Mean for humidity in % =
μH =72. 38%+62 . 88%+72 .63%+63 .13%+64 . 25%+63 . 38%+62 . 75%+63 . 38%+60 . 00%+60 . 75%
10
= 64.55%
Arithmetic Mean for rain in millimetre =
μR= 0. 0000+0 . 0000+0 . 0125+0 . 0000+0 . 0000+ 0 .0000+ 0. 0000+1 . 0000+0 .0000+ 0 .0000
10
=0 . 1013
Arithmetic Mean for rain in pressure (mb) =
μP=1004 . 63+ 1003. 63+1011 .00+ 1007. 38+1006 . 88+1007 .50+ 1007. 25+1009 . 50+1010. 00+1009 . 75
10
=1007 . 75
II. Median
Number of observation in the sample is n=10 (even). Hence, the median will be the average
of ( n
2 )
th
and ( n+1
2 )
th
observations.
For temperature: The data is arranged in the ascending order as follows,
Day Temp °c
26-Apr 28.38
24-Apr 30.50
02-May 30.88
01-May 31.50
27-Apr 32.00
29-Apr 32.00
30-Apr 32.00
03-May 32.00
28-Apr 32.13
25-Apr 32.50

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The ( n
2 )
th
observation = 5th observation = 32 and ( n+1
2 )
th
observation = 6th observation = 32
Therefore, median =
1
2 ( 32+ 32 ) = 32 Degree Celsius.
For gust: The data is arranged in the ascending order as follows,
Day Gust km/h
02-
May 14.63
03-
May 16.88
26-Apr 17.13
28-Apr 19.38
27-Apr 19.50
29-Apr 22.50
01-
May 22.63
30-Apr 22.75
24-Apr 27.38
25-Apr 28.75
The ( n
2 )
th
observation = 5th observation = 19.50 and ( n+1
2 )
th
observation = 6th observation =
22.50. Therefore, median =
1
2 ( 19. 50+22 .50 ) =21. 00 km/h.
For cloud %: The data is arranged in the ascending order as follows,
Day Cloud %
03-May 8.50%
27-Apr 9.13%
25-Apr 9.63%
28-Apr 11.88%
29-Apr 14.13%

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02-May 14.63%
30-Apr 17.50%
01-May 19.88%
26-Apr 37.75%
24-Apr 47.13%
The ( n
2 )
th
observation = 5th observation = 14.13% and ( n+1
2 )
th
observation = 6th observation
= 14.63%
Therefore, median=
1
2 ( 14 .13 % +14 . 63 % ) =14 .38 % .
For humidity %: The data is arranged in the ascending order as follows,
Day Humidity %
02-May 60.00%
03-May 60.75%
30-Apr 62.75%
25-Apr 62.88%
27-Apr 63.13%
01-May 63.38%
29-Apr 63.38%
28-Apr 64.25%
24-Apr 72.38%
26-Apr 72.63%
The ( n
2 )
th
observation = 5th observation = 63.13 % and ( n+1
2 )
th
observation = 6th observation
= 63.38%
Therefore, median=
1
2 ( 63. 13 %+ 63 .38 % ) =63 . 25 % .

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For rain (mm): The data is arranged in the ascending order as follows,
Day Rain (mm)
24-Apr 0.0000
25-Apr 0.0000
27-Apr 0.0000
28-Apr 0.0000
29-Apr 0.0000
30-Apr 0.0000
02-
May 0.0000
03-
May 0.0000
26-Apr 0.0125
01-
May 1.0000
The ( n
2 )
th
observation = 5th observation = 0.0000 and ( n+1
2 )
th
observation = 6th observation
= 0.0000
Therefore, median=
1
2 ( 0 .0000+ 0. 0000 ) =0 .0000 mm .
For air pressure (mb): The data is arranged in the ascending order as follows,
Day Pressure (mb)
25-Apr 1003.63
24-Apr 1004.63
28-Apr 1006.88
30-Apr 1007.25
27-Apr 1007.38
29-Apr 1007.50
01-May 1009.50
03-May 1009.75

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02-May 1010.00
26-Apr 1011.00
The ( n
2 )
th
observation = 5th observation = 1007.38 and ( n+1
2 )
th
observation = 6th observation
= 1007.50
Therefore, median=
1
2 ( 1007. 38+1007 . 38 ) =1007 . 44 mm .
III. Mode
Mode of a distribution is the observation with highest frequency, From Table 1 the
frequencies are noted for all the variables.
Temperature: With frequency = 4, 32.0 degree Celsius is found to be the mode.
Gust: No observation has frequency greater than 1. Hence, no mode is found.
Cloud %: No observation has frequency greater than 1. Hence, no mode is found.
Humidity %: With frequency = 2, 63.38% humidity is found to be the mode.
Rain (mm): With frequency = 8, 0.0 mm rain is found to be the mode.
Pressure: with frequency of 4, 32.0 degree Celsius is found to be the mode.
Temperature: No observation has frequency greater than 1. Hence, no mode is found.
IV. Range
Range is considered as the difference between maximum and minimum value.
Range of temperature = Maximum temp – Minimum temp = 32.5 – 28.38 = 4.13 degree
Celsius.
Range of gust = Maximum gust – Minimum gust = 28.75 – 14.63 = 14.13 km/h.

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Range of cloud% = Maximum cloud% – Minimum cloud% = 47.13% – 8.50%= 38.63%
Range of humidity% = Maximum humidity% – Minimum humidity% = 72.63% - 60.00% =
12.63%
Range of rain = Maximum rain – Minimum rain = 1.00 – 0.00 = 1mm.
Range of pressure = Maximum pressure – Minimum pressure = 1011.0 – 1003.63 = 7.38 mb.
V. Standard Deviation
Sample Standard deviation = √ 1
( n−1 ) ( xi−x
¿
)
2
S.D of temperature = √ ( 30 . 5−31 . 9 ) 2+ ( 32 .5−31. 9 ) 2+ ( 28 .38−31. 9 ) 2+.. .+ ( 32 .0−31. 9 ) 2
10−1
=1 .22 Degree Celsius.
S.D of gust = √ ( 27 . 38−21 . 15 ) 2+ ( 28 .75−21 .15 ) 2+ ( 17 .13−21. 15 ) 2+ .. .+ ( 16 .88−21 .15 ) 2
10−1
=4 .55 km/h.
S.D of cloud% = √ ( 47 .13−19. 01 ) 2+ ( 9 .63−19 .01 ) 2+ ( 37 .75−19. 01 ) 2+.. .+ ( 8 .50−19 .01 ) 2
10−1
=13 . 06 %
S.D of humidity% =
√ ( 72 .38−64 . 55 ) 2+ ( 62 . 88−64 . 55 ) 2+ ( 72. 63−64 .55 ) 2+.. .+ ( 60 .75−64 . 55 ) 2
10−1 = 4 . 38 %
S.D of rain =
√ ( 0 . 00−0. 10 ) 2+ ( 0 .00−0 .10 ) 2+ ( 0 . 0125−0 . 10 ) 2 . ..+ ( 1 . 00−0 . 10 ) 2+ ( 0 . 00−0 .10 ) 2+ ( 0 .00−0 . 10 ) 2
10−1
=0 . 32 mm.

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S.D of pressure =
√ ( 1004 . 63−1007 . 75 ) 2+ ( 1003 . 63−1007 . 75 ) 2+ ( 1011 . 00−1007 . 75 ) 2 . ..+ ( 1009 .75−1007 .75 ) 2
10−1
=2 .37 mb (Sarstedt, and Mooi, 2019, pp. 91-150)
4. Here X is the predictor variable, which denotes number of days. Y is the outcome variable.
Let the regression equation be y = mx + c, where “m” is the slope of the regression line and
“c” is the intercept of the equation.
So, the sum of square of errors SSE = ∑ ( Y i−c −mXi )
2
Minimizing the error with respect to the parameters “m” and “c”, it is obtained that
∂SSE
∂ c =−2∑ ( Y i −c−mX i ) =0 ………. (1)
∂SSE
∂ m =−2∑ Xi ( Y i−c−mX i ) =0 ……… (2)
From (1) and (2), the normal equations of the regression model are,
∑ Y i =n∗c +m∗∑ Xi
∑ X i Y i=c∗∑ Xi+ m∗∑ Xi
2
Solving for “m” and “c”, it is obtained
c= ( ∑ Y i ) ( ∑ Xi
2 ) − ( ∑ Xi ) ( ∑ Xi Y i )
n∗( ∑ Xi
2 ) − ( ∑ Xi )
2

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m= n∗( ∑ Xi Y i )− ( ∑ Xi ) ( ∑ Y i )
n∗( ∑ Xi
2 ) − ( ∑ Xi ) 2
Weather Forecast
A. Outcome Variable: Temperature
Day (Xi) Temp °c (Yi) (Xi)^2 Xi*Yi
1 30.50 1.00 30.500
2 32.50 4.00 65.000
3 28.38 9.00 85.125
4 32.00 16.00 128.000
5 32.13 25.00 160.625
6 32.00 36.00 192.000
7 32.00 49.00 224.000
8 31.50 64.00 252.000
9 30.88 81.00 277.875
10 32.00 100.00 320.000
Total 55.00 313.88 385.00 1735.13
Now,
X
¿
=55
10 =5 . 5 and
Y
¿
=313 . 88
10 =31. 39
Normal equations are
∑ Y i =n∗c +m∗∑ Xi => 313.88 = 10 * c + m *55
∑ X i Y i=c∗∑ Xi+ m∗∑ Xi
2
=> 1735.13 = c * 55 + m*385
Solving simultaneously it is obtained,
m=10∗1735. 13−55∗313 . 88
10∗385−55∗55 =0. 1068 c=313 . 88∗385−55∗1735. 13
10∗385−55∗55 =30. 803
So, the regression equation is Y
^¿=0 .1068∗ X
^¿+ 30. 803
¿
¿
Now, for X = 15, 23 the predicted values are found as follows,

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Y =0 . 1068∗15+ 30. 803 =32 . 40 Degree Celsius
Y =0 . 1068∗23+ 30. 803 =33 . 26 Degree Celsius (Dahl, Brun, and Andresen, 2017, pp.455-465)
B. Outcome Variable: Gust
Day (Xi) Gust km/h (Xi)^2 Xi*Yi
1 27.38 1.00 27.375
2 28.75 4.00 57.500
3 17.13 9.00 51.375
4 19.50 16.00 78.000
5 19.38 25.00 96.875
6 22.50 36.00 135.000
7 22.75 49.00 159.250
8 22.63 64.00 181.000
9 14.63 81.00 131.625
10 16.88 100.00 168.750
Total 55.00 211.50 385.00 1086.75
Now,
X
¿
=55
10 =5 . 5 and
Y
¿
=211 .5
10 =21 .15
Normal equations are
∑ Y i =n∗c +m∗∑ Xi => 211.50 = 10 * c + m *55
∑ X i Y i=c∗∑ Xi+ m∗∑ Xi
2
=> 1086.75 = c * 55 + m*385
Solving simultaneously it is obtained,
m=−0 . 927 , c=26 . 252
So, the regression equation is Y
^¿=−0. 927∗ X
^¿+26 . 252
¿
¿
Now, for X = 15, 23 the predicted values are found as follows,
Y
^¿=−0. 927∗15+26 .252
¿ =12 .35 Km/h

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Y
^¿=−0. 927∗23+26 .252
¿ =4 . 93 Km/h (Paras, 2016, pp.161-168)
C. Outcome Variable: Cloud
Day (Xi) Cloud % (Xi)^2 Xi*Yi
1 47.13% 1.00 47.13%
2 9.63% 4.00 19.25%
3 37.75% 9.00 113.25%
4 9.13% 16.00 36.50%
5 11.88% 25.00 59.38%
6 14.13% 36.00 84.75%
7 17.50% 49.00 122.50%
8 19.88% 64.00 159.00%
9 14.63% 81.00 131.63%
10 8.50% 100.00 85.00%
Total 55.00 190% 385.00 858.38%
Now,
X
¿
=55
10 =5 . 5 and
Y
¿
=190
10 =19 .01 %
Normal equations are
∑ Y i =n∗c +m∗∑ Xi => 190.01= 10 * c + m *55
∑ X i Y i=c∗∑ Xi+ m∗∑ Xi
2
=> 858.38 = c * 55 + m*385
Solving simultaneously it is obtained,
m=−2 . 271, c=31 .505
So, the regression equation is Y
^¿=−2 . 271∗ X
^¿+ 31.505
¿
¿
Now, for X = 15, 23 the predicted values are found as follows,
Y
^¿=−2 . 271∗15+31 .505
¿ =0 . 00 % (Negative value avoided, as cloud % cannot be negative)

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Y
^¿=−2 . 271∗23+31 .505
¿ =0 . 00 %
D. Outcome Variable: Humidity
Day (Xi) Humidity % (Xi)^2 Xi*Yi
1 72.38% 1.00 72.38%
2 62.88% 4.00 125.75%
3 72.63% 9.00 217.88%
4 63.13% 16.00 252.50%
5 64.25% 25.00 321.25%
6 63.38% 36.00 380.25%
7 62.75% 49.00 439.25%
8 63.38% 64.00 507.00%
9 60.00% 81.00 540.00%
10 60.75% 100.00 607.50%
Total 55.00 645.50% 385.00 3463.75%
Now,
X
¿
=55
10 =5 . 5 and
Y
¿
=645 . 5
10 =64 .5 %
Normal equations are
∑ Y i =n∗c +m∗∑ Xi => 645.50 = 10 * c + m *55
∑ X i Y i=c∗∑ Xi+ m∗∑ Xi
2
=> 3463.75 = c * 55 + m*385
Solving simultaneously it is obtained,
m=−1 . 049 , c=70 . 323
So, the regression equation is Y
^¿=−1 . 049∗ X
^¿+ 70. 323
¿
¿
Now, for X = 15, 23 the predicted values are found as follows,
Y
^¿=−1 . 049∗15+ 70. 323
¿ =54 . 59 %

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Y
^¿=−1 . 049∗23 +70. 323
¿ =46 .20 %
E. Outcome Variable: Rain
Day (Xi) Rain (mm) (Xi)^2 Xi*Yi
1 0.00 1.00 0.000
2 0.00 4.00 0.000
3 0.01 9.00 0.038
4 0.00 16.00 0.000
5 0.00 25.00 0.000
6 0.00 36.00 0.000
7 0.00 49.00 0.000
8 1.00 64.00 8.000
9 0.00 81.00 0.000
10 0.00 100.00 0.000
Total 55.00 1.01 385.00 8.04
Now,
X
¿
=55
10 =5 . 5 and
Y
¿
= 1. 01
10 =0 . 10
Normal equations are
∑ Y i =n∗c +m∗∑ Xi => 1.01 = 10 * c + m *55
∑ X i Y i=c∗∑ Xi+ m∗∑ Xi
2
=> 8.04 = c * 55 + m*385
Solving simultaneously it is obtained
m=0 . 03 , c=−0 . 064
So, the regression equation is Y
^¿=0 . 03∗ X
^¿−0 . 064
¿
¿
Now, for X = 15, 23 the predicted values are found as follows,
Y
^¿=0 . 03∗15−0 . 064
¿ = 0 .386 mm

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Y
^¿=0 . 03∗23−0 . 064
¿ =0 . 626 mm
F. Outcome Variable: Pressure
Day (Xi) Pressure (mb) (Xi)^2 Xi*Yi
1 1004.63 1.00 1004.625
2 1003.63 4.00 2007.250
3 1011.00 9.00 3033.000
4 1007.38 16.00 4029.500
5 1006.88 25.00 5034.375
6 1007.50 36.00 6045.000
7 1007.25 49.00 7050.750
8 1009.50 64.00 8076.000
9 1010.00 81.00 9090.000
10 1009.75 100.00 10097.500
Total 55.00 10077.50 385.00 55468.00
Now,
X
¿
=55
10 =5 . 5 and
Y
¿
=10077 . 50
10 =1007 . 75
Normal equations are
∑ Y i =n∗c +m∗∑ Xi => 10077.50 = 10 * c + m *55
∑ X i Y i=c∗∑ Xi+ m∗∑ Xi
2
=> 55468.00 = c * 55 + m*385
Solving simultaneously it is obtained
m=0 . 505 , c=1004 . 972
So, the regression equation is Y
^¿=0 .505∗ X
^¿+ 1004 . 972
¿
¿
Now, for X = 15, 23 the predicted values are found as follows,
Y
^¿=0 .505∗15+1004 .972
¿ =1012 .55 mb

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Y
^¿=0 .505∗23+1004 .972
¿ =1016 . 59 mb
References
Dahl, M., Brun, A. and Andresen, G.B., 2017. Using ensemble weather predictions in district
heating operation and load forecasting. Applied energy, 193, pp.455-465.
Kaliyadan, F. and Kulkarni, V., 2019. Types of variables, descriptive statistics, and sample
size. Indian dermatology online journal, 10(1), p.82.
Paras, S.M., 2016. A simple weather forecasting model using mathematical
regression. Indian Research Journal of Extension Education, 12(2), pp.161-168.
Sarstedt, M. and Mooi, E., 2019. Descriptive Statistics. In A Concise Guide to Market
Research (pp. 91-150). Springer, Berlin, Heidelberg.