One Fourth Labs
We deliver courseware in AI and related areas
Recap: List of Topics
Descriptive Statistics
Probability Theory
Inferential Statistics
Different types of data
Different types of plots
Measures of centrality and spread
Sample spaces, events, axioms
Discrete and continuous RVs
Bernoulli, Uniform, Normal dist.
Sampling strategies
Interval Estimators
Hypothesis testing (z-test, t-test)
ANOVA, Chi-square test
Linear Regression
What is the effect of transformations on percentiles?
Learning Objectives
What are percentiles?
What are the different measures of spread?
How do you compute the percentile rank of a value in the data?
What are box plots and how to use them to visualise some measures of centrality and spread?
What is the effect of transformations on measures of spread?
What are some frequently used percentiles?
What are percentiles?
Intuition: Percentiles
But ... ...
Suppose you scored 45 out of 100 on a test, how would you rate your performance? Good or bad?
Example
Is it bad? (because you scored less than 50%)
What if the questions were really hard?
What if the time provided was insufficient?
Intuition: Percentiles
Suppose you scored 45 out of 100 on a test. Out of 100 students, only 2 scored greater than 45. How do you rate your performance?
Example
Does it look good now?
Yes, it does ... ...
You can proudly say that you lie in the top 98 percentile of your class (the score of 98% of students was less than or equal to your score)
Percentiles
44, 43, 37, 68, 55, 46, 19, 59, 34, 46, 51, 62, 47, 52, 44, 28, 36, 56, 65, 60, 55, 66, 54, 48, 62
A university conducts a written test for 25 students and decides to call those students for an interview whose score is above the 70th percentile
Example
Can you Identify which students will be called for the interview?
Percentiles
25 students (sorted scores)
\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}_{\textit{70\% of the values in the data}}
70-th percentile
sorted data values
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59, 60, 62, 62, 65, 66, 68
The p percentile of a sample is a value such that p perentage of the values in the data are less than or equal to this value
Percentiles
25 students (sorted scores)
L_p = \frac{p}{100} (n + 1) = \frac{70}{100} (25 + 1) = 18.2
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59, 60, 62, 62, 65, 66, 68
Sort the data
Compute location of the p-th percentile
The 70th percentile lies at location 18.2 !
Percentiles
25 students (sorted scores)
56~~+
0.2 * (59 - 56)
= 56.6
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55
\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}_{\textit{first 17 elements}}
60, 62, 62, 65, 66, 68
\underbrace{~~~~~~~~~~~~~~~~~~~~}_{\textit{last 6 elements}}
\underbrace{}_{\textit{18}}
56
59
\underbrace{}_{\textit{19}}
Where is the position 18.2?
The 70th percentile should be between 56 and 59, greater than 56 but closer to 56
18.2 is between 18 and 19, closer to 18
Percentiles
What is the overall procedure?
L_p = \frac{p}{100} (n + 1) = \frac{70}{100} (25 + 1) = 18.2
Y_p = x_{i_p} + f_p * (x_{i_{p+1}} - x_{i_p})
Sort the data
Compute location of the p-th percentile
integer part of
L_p = i_p
fractional part of
L_p = f_p
18
0.2
Compute p-th percentile as
Percentiles (some more intuition)
Y_p = x_{i_p} + f_p * (x_{i_{p+1}} - x_{i_p})
Y_p = x_{i_p} + f_p * (x_{i_{p+1}}) - f_p * x_{i_p}
Y_p = (1 - f_p) * x_{i_p} + f_p * x_{i_{p+1}}
f_p
if is high then the weightage given to will be lower than that given to and vice versa
f_p
x_{i_p}
x_{i_p + 1}
Percentiles
Y_p = x_{i_p} + f_p * (x_{i_{p+1}} - x_{i_p})
p = 70
L_p = \frac{p}{100} (n + 1) = \frac{70}{100} (25 + 1) = 18.2
i_p = 18, f_p = 0.2
Y_{70} = x_{18} + 0.2 * (x_{19} - x_{18}) = 56.6
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59, 60, 62, 62, 65, 66, 68
Percentiles
Y_{70} = 56.6
A university conducts a written test for 25 students and decides to call those students for an interview whose score is above the 70th percentile
Example
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59, 60, 62, 62, 65, 66, 68
The school will invite only those 7 students whose score was greater than 56.6
Percentiles
L_p = \frac{p}{100} (n + 1) = \frac{80}{100} (25 + 1) = 20.8
i_p = 20, f_p = 0.8
Y_{80} = x_{20} +
0.8~*
(x_{21} - x_{20})
=61.6
Suppose the school changes its decision and now only wants to invite students who scored greater than 80 percentile
Example
(p = 80)
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59
\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}_{\textit{first 19 elements}}
\underbrace{}_{\textit{20}}
60
62
\underbrace{}_{\textit{21}}
62, 65, 66, 68
\underbrace{~~~~~~~~~~~~~~~~~~~~}_{\textit{last 4 elements}}
Percentiles
Suppose the school changes its decision and now only wants to invite students who scored greater than 80 percentile
Example
(p = 80)
L_p = \frac{p}{100} (n + 1) = \frac{80}{100} (25 + 1) = 20.8
i_p = 20, f_p = 0.8
\underbrace{}_{\textit{20}}
60
62
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59
\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}_{\textit{first 19 elements}}
62, 65, 66, 68
\underbrace{~~~~~~~~~~~~~~~~~~~~}_{\textit{last 4 elements}}
\underbrace{}_{\textit{21}}
Y_{80} = x_{20} +
0.8~*
(x_{21} - x_{20})
=61.6
Percentiles
L_p = \frac{p}{100} (n + 1) = \frac{80}{100} (25 + 1) = 20.8
\underbrace{}_{\textit{20}}
60
62
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59
\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}_{\textit{first 19 elements}}
62, 65, 66, 68
\underbrace{~~~~~~~~~~~~~~~~~~~~}_{\textit{last 4 elements}}
\underbrace{}_{\textit{21}}
Why did we have to compute ? Wasn't knowing enough to identify the shortlisted students?
Y_p
L_p
Yes, it was, but the university may also be required to declare the cut-off score, hence we need to compute. also
L_p
Y_{80} = 60 + 0*(62-60)
Percentiles (special case: )
f_p = 0
L_p = \frac{p}{100} (n + 1) = \frac{80}{100} (24 + 1) = 20
i_p = 20, f_p = 0
In such cases the percentile would actually correspond to a value in the data
Suppose there were only 24 students and p=80
Example
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59
\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}_{\textit{first 19 elements}}
\underbrace{}_{\textit{20}}
60
62, 62, 65, 66
\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~}_{\textit{last 4 elements}}
What are some alternative methods found in textbooks?
Alternative 1 (we already saw this)
Sort the data
Compute location of the p-th percentile
L_p = \frac{p}{100} (n + 1)
integer part of
L_p = i_p
If~L_p~is~an~integer:
If~L_p~is~~not~an~integer:
Y_{p} = x_{i_{p}}
Y_p = x_{i_p} + f_p * (x_{i_{p+1}} - x_{i_p})
Alternative 2
L_p = i_p
Sort the data
Compute location of the p-th percentile
L_p = \frac{p}{100} (n)
(note the use of n instead of n+1)
integer part of
If~L_p~is~an~integer:
Y_{p} = \frac{x_{L_p} + x_{L_{p+1}}}{2}
If~L_p~is~~not~an~integer:
Y_{p} = x_{i_{p+1}}
Alternative 2
L_{70} = \frac{p}{100} (n) = \frac{70}{100}*25=17.5
L_{80} = \frac{p}{100} (n) = \frac{80}{100}*25=20
Marks of 25 students and p=70 or p=80
Example
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59, 60, 62, 62, 65, 66, 68
L_{80}~is~an~integer:
Y_{80} = \frac{x_{20} + x_{21}}{2}=\frac{60+62}{2}
L_{70}~is~~not~an~integer:
Y_{70} = x_{18} = 56
Alternative 2: intuition
the p-th percentile is that value in the data such that at least p percentage of the values are less than or equal to it and at least (100-p) percentage of the values are greater than or equal to it
L_{70} = \frac{p}{100} (n) = \frac{70}{100}*25=17.5
Location 18 (i.e., ) is the only location which satisfies both conditions
i_{p}+1
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59, 60, 62, 62, 65, 66, 68
At least 17.5 values should be less than or equal to it (so the location should be 18 or higher)
At least 7.5 values should be greater than or equal to it (so the location should be 18 or lower)
Alternative 2: intuition
the p-th percentile is that value in the data such that at least p percentage of the values are less than or equal to it and at least (100-p) percentage of the values are greater than or equal to it
L_{80} = \frac{p}{100} (n) = \frac{80}{100}*25=20
Both locations 20 and 21 (i.e., ) satisfy the above conditions so just take an average of these two values
L_p~\&~L_{p}+1
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59, 60, 62, 62, 65, 66, 68
Y_{80} = \frac{x_{20} + x_{21}}{2}=\frac{60+62}{2}=61
Alternative 3
Sort the data
Compute location of the p-th percentile
L_p = \frac{p}{100} (n + 1)
(same as alternative 1)
integer part of
L_p = i_p
If~L_p~is~an~integer:
If~L_p~is~~not~an~integer:
Y_{p} = x_{i_{p}}
(same as alternative 1)
Y_p = x_{i_p} + 0.5 * (x_{i_{p+1}} - x_{i_p})
(same as alternative 1 except that use 0.5 instead of )
f_p
Comparison
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59, 60, 62, 62, 65, 66, 68
Alternative 1
Alternative 2
Alternative 3
Y_p = x_{L_p}
Y_p = \dfrac{x_{L_p} + x_{L_{p+1}}}{2}
Y_p = x_{L_p}
Y_p = x_{i_p} + f_p*(x_{i_{p+1}} - x_{i_p})
Y_p = x_{i_{p+1}}
Y_p = x_{i_p} + 0.5*(x_{i_{p+1}} - x_{i_p})
P = 70
P = 80
Case 2
is not integer
L_p
Case 1
is integer
L_p
L_p = \dfrac{p}{100}*(n+1)
L_p = \dfrac{p}{100}*n
L_p = \dfrac{p}{100}*(n+1)
L_{70} = 18.2
Y_{70} = 56.6
L_{80} = 20.8
Y_{80} = 61.6
L_{70} = 17.5
Y_{70} = 56
L_{80} = 20
Y_{80} = 61
L_{70} = 18.2
Y_{70} = 57.5
L_{80} = 20.8
Y_{80} = 61
What are some frequently used percentiles?
Quartiles
19, 28, 34, 36, 37
43, 44, 44, 46, 46
47, 48, 51, 52, 54
55, 55, 56, 59, 60
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{first 25\% data}}
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{third 25\% data}}
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{second 25\% data}}
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{last 25\% data}}
\underbrace{Q_1}_{25\text{-}th~percentile}
\underbrace{Q_2}_{50\text{-}th~percentile}
\underbrace{Q_3}_{75\text{-}th~percentile}
Quartiles divide the data into four equal parts
Quartiles: Example
Shikhar Dhawan T20I scores (50 sorted scores)
0, 1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 5, 5, 5, 5, 6, 6, 8, 9, 10, 10, 11, 13, 14, 15, 16, 23, 23, 24, 26, 29, 30, 30, 32, 33, 35, 41, 42, 43, 46, 47, 51, 55, 60, 72, 74, 76, 80, 90, 92
L_{25} = \frac{25}{100}*(50 + 1) = 12.75
Q_1 = Y_{25} = x_{12} + 0.75 * (x_{13} - x_{12})
L_{50} = \frac{50}{100}*(50 + 1) = 25.5
Q_2 = Y_{50} = x_{25} + 0.5 * (x_{26} - x_{25})
L_{75} = \frac{75}{100}*(50 + 1) = 38.25
Q_3 = Y_{75} = x_{38} + 0.25 * (x_{39} - x_{38})
= 5
= 15.5
= 42.25
Median is same as Q2
median
\underbrace{x_{\frac{n+1}{2}}}_{n~is~odd}~or~ \underbrace{\frac{x_{\frac{n}{2}} + x_{\frac{n}{2} + 1}}{2}}_{n~is~even}
L_{50} = \frac{50}{100}*(n + 1) = \frac{n+1}{2}
Q_2 = Y_{50} = x_{i_p} + 0.5 * (x_{i_p+1} - x_{i_p})
Q_2
Are they the same?
(of course, the are !)
But why do the formulae look different?
Median is same as Q2
L_{50} = \frac{50}{100}*(n + 1) = \frac{n+1}{2}
Case 1: n is odd
L_{50}~will~be~an~integer
(\because n+1~is~even)
i_p=L_p = \frac{n+1}{2}, f_p = 0
Q_2 = Y_{50} = x_{i_p} + 0 * (x_{i_p+1} - x_{i_p}) = x_{\frac{n+1}{2}}
Case 1: n is even
L_{50}= \frac{n}{2} + \frac{1}{2} = i_p + 0.5
i_p= \frac{n}{2}, f_p = 0.5
Q_2 = Y_{50} = x_{i_p} + 0.5 * (x_{i_p+1} - x_{i_p}) = \frac{x_{\frac{n}{2}} + x_{\frac{n}{2}+1} }{2}
Median is same as Q2
Q_2 = Y_{50} = x_{i_p} + 0.5 * (x_{i_p+1} - x_{i_p})
= x_{\frac{n}{2}} + 0.5*(x_{\frac{n}{2}+1} - x_{\frac{n}{2}})~~(\because i_p = \frac{n}{2}) \\~\\
= 0.5*x_{\frac{n}{2}} + 0.5*x_{\frac{n}{2}+1}
= x_{\frac{n}{2}} + 0.5*x_{\frac{n}{2}+1} - 0.5*x_{\frac{n}{2}}
= \frac{x_{\frac{n}{2}} + x_{\frac{n}{2}+1} }{2}
Quintiles
19, 28, 34, 36
37, 43, 44, 44
46, 46, 47, 48
55, 56, 59, 60
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{first 20\% data}}
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{second 20\% data}}
51, 52, 54, 55
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{third 20\% data}}
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{fourth 20\% data}}
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{fifth 20\% data}}
\underbrace{P_1}_{20\text{-}th~percentile}
\underbrace{P_2}_{40\text{-}th~percentile}
\underbrace{P_3}_{60\text{-}th~percentile}
\underbrace{P_4}_{80\text{-}th~percentile}
Quintiles divide the data into five equal parts
Quintiles: Example
Shikhar Dhawan T20I scores (50 sorted scores)
0, 1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 5, 5, 5, 5, 6, 6, 8, 9, 10, 10, 11, 13, 14, 15, 16, 23, 23, 24, 26, 29, 30, 30, 32, 33, 35, 41, 42, 43, 46, 47, 51, 55, 60, 72, 74, 76, 80, 90, 92
L_{60} = \frac{60}{100}*(50 + 1) = 30.6
P_3 = Y_{60} = x_{30} + 0.6 * (x_{31} - x_{30})
= 27.8
Similarly you can compute the other 3 quintiles!
Deciles
19,28 34,36 37,43 44,44 46,46 47,48 51,52 54,55 55,56 59,60
\underbrace{~~~~~~~~~~~~~~~~~~~}
\underbrace{~~~~~~~~~~~~~~~~~~~}
\underbrace{D_1}_{10\text{-}th~percentile}
\underbrace{~~~~~~~~~~~~~~~~~~~}
\underbrace{D_2}_{20\text{-}th~percentile}
\underbrace{~~~~~~~~~~~~~~~~~~~}
\underbrace{D_3}_{30\text{-}th~percentile}
\underbrace{~~~~~~~~~~~~~~~~~~~}
\underbrace{D_4}_{40\text{-}th~percentile}
\underbrace{~~~~~~~~~~~~~~~~~~~}
\underbrace{D_5}_{50\text{-}th~percentile}
\underbrace{~~~~~~~~~~~~~~~~~~~}
\underbrace{D_6}_{60\text{-}th~percentile}
\underbrace{~~~~~~~~~~~~~~~~~~~}
\underbrace{D_7}_{70\text{-}th~percentile}
\underbrace{~~~~~~~~~~~~~~~~~~~}
\underbrace{D_8}_{80\text{-}th~percentile}
\underbrace{~~~~~~~~~~~~~~~~~~~}
\underbrace{D_9}_{90\text{-}th~percentile}
Deciles divide the data into ten equal parts
Deciles: Example
Shikhar Dhawan T20I scores (50 sorted scores)
0, 1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 5, 5, 5, 5, 6, 6, 8, 9, 10, 10, 11, 13, 14, 15, 16, 23, 23, 24, 26, 29, 30, 30, 32, 33, 35, 41, 42, 43, 46, 47, 51, 55, 60, 72, 74, 76, 80, 90, 92
L_{30} = \frac{30}{100}*(50 + 1) = 15.3
D_3 = Y_{30} = x_{15} + 0.3 * (x_{16} - x_{15})
= 5.3
Similarly you can compute the other 8 deciles!
How to compute the percentile rank of a value in the data?
44, 43, 37, 68, 55, 46, 19, 59, 34, 46, 51, 62, 47, 52, 44, 28, 36, 56, 65, 60, 55, 66, 54, 48, 62
Percentile Rank
OR
Compared to other students, how do you rate the performance of the student who scored 44?
What is the percentile rank of the student who scored 44
The percentile rank of a value is the percentage of data values that are less than on equal to it
Percentile Rank: Example 1
c_s = \textit{number of values less than s}
PR_{s} = \frac{c_s+0.5*f_s}{n}*100
PR_s = \textit{percentile rank of the score s}
f_s = \textit{number of values equal to s}
19, 28, 34, 36, 37, 43, 44, 44, 46, 46, 47, 48, 51, 52, 54, 55, 55, 56, 59, 60, 62, 62, 65, 66, 68
PR_{44} = \frac{6+0.5*2}{25}*100 = 28
Percentile Rank: Example 2
PR_{s} = \frac{c_s+0.5*f_s}{n}*100
PR_{s} = \frac{37+0.5*2}{59}*100 = 64.40
Shikhar Dhawan T20I scores (59 sorted scores)
0, 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 8, 9, 10, 10, 11, 13, 14, 15, 16, 19, 23, 23, 23, 24, 26, 29, 30, 30, 31, 32, 32, 33, 35, 36, 40, 41, 41, 42, 43, 46, 47, 51, 52, 55, 60, 72, 74, 76, 80, 90, 92
We typically round it upto the next whole number (65 in this case)
What is the effect of transformations on percentiles?
Transformations
Scaling and Shifting
x_{new} = a*x + c
(a = 5/9, c = -160/9)
[22.46, 23.54, 24.26, 27.86, 30.2, 30.74, 34.52, 35.96, 40.46, 44.06, 52.7, 54.68, 56.66, 57.56, 59.54, 61.52, 62.06, 65.66, 67.46, 70.88, 76.46, 82.4, 83.12, 84.38, 93.02, 94.28, 95.72, 96.44, 108.86, 109.58]
[-5.3, -4.7, -4.3, -2.3, -1.0, -0.7, 1.4, 2.2, 4.7, 6.7, 11.5, 12.6, 13.7, 14.2, 15.3, 16.4, 16.7, 18.7, 19.7, 21.6, 24.7, 28.0, 28.4, 29.1, 33.9, 34.6, 35.4, 35.8, 42.7, 43.1]
\degree F:
\degree C:
Effect of transformations (on percentiles)
x_{new} = a*x + c
L_p^{new} = \frac{p}{100} (n + 1) = L_p
i_p^{new} = i_p,~~~~f_p^{new} = f_p
Y_p^{new} = x_{i_p}^{new} + f_p * (x_{i_{p+1}}^{new} - x_{i_p}^{new})
Y_p^{new} = (a*x_{i_p} + c) + f_p * (a*x_{i_{p+1}} + c - (a * x_{i_p} + c))
= a* (x_{i_p} + f_p*(x_{i_{p+1}} - x_{i_p})) + c
= a*x_{i_p} + c + f_p*a*(x_{i_{p+1}} - x_{i_p})
= a*Y_p + c
Effect of transformations (on percentiles)
Y_p^{new} = a*Y_p + c
L_p = \frac{40}{100}*(n+1) = 12.4
Y_p = x_{12} + 0.4 (x_{13} - x_{12}) = 55.47
Y_p^{new} = \frac{5}{9} * 55.47 + \frac{-160}{9} = 13.03
[22.46, 23.54, 24.26, 27.86, 30.2, 30.74, 34.52, 35.96, 40.46, 44.06, 52.7, 54.68, 56.66, 57.56, 59.54, 61.52, 62.06, 65.66, 67.46, 70.88, 76.46, 82.4, 83.12, 84.38, 93.02, 94.28, 95.72, 96.44, 108.86, 109.58]
[-5.3, -4.7, -4.3, -2.3, -1.0, -0.7, 1.4, 2.2, 4.7, 6.7, 11.5, 12.6, 13.7, 14.2, 15.3, 16.4, 16.7, 18.7, 19.7, 21.6, 24.7, 28.0, 28.4, 29.1, 33.9, 34.6, 35.4, 35.8, 42.7, 43.1]
\degree F:
\degree C:
Summary
\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}_{\textit{p\% of the values in the data}}
p-th percentile
sorted data values
L_p = \frac{p}{100} (n + 1)
Y_p = x_{i_p} + f_p * (x_{i_{p+1}} - x_{i_p})
(alternative methods not recommended)
Percentile (definition)
How to compute percentiles?
Summary
PR_{s} = \frac{c_s+0.5*f_s}{n}*100
x_{new} = a*x + c
Y_p^{new} = a*Y_p + c
Frequently used percentiles
Quartiles
Quintiles
Deciles
How to compute percentile rank of a value?
What is the effect of transformations on percentiles?
What are the measures of spread?
Motivation: Measures of spread
Sample 1: 81, 81, 82, 82, 83, 83, 84, 84, 85, 85 Mean: 83
Median: 83
Sample 1: 31, 41, 61, 72, 83, 83, 94, 105, 125, 135 Mean: 83
Median: 83
All values are very close to the mean & median (low variability in data)
Some values are far from the mean & median (high variability in data)
Measures of centrality don't tell us anything about the spread and variability in the data
Measures of spread (range)
Sample 1: 81, 81, 82, 82, 83, 83, 84, 84, 85, 85 Mean: 83
Median: 83
Sample 1: 31, 41, 61, 72, 83, 83, 94, 105, 125, 135 Mean: 83
Median: 83
Range clearly tells us that the second sample has more variability/spread than the first
Range: (max value - min value) = 85 - 81 = 4
Range: (max value - min value) = 135 - 31 = 104
Measures of spread (range)
Farm yields of wheat (in bushels): 40.1, 40.9, 41.8, 44, 46.8, 47.2, 48.6, 49.3, 49.4, 51.9, 53.8, 55.9, 57.3, 58.1, 60.2, 60.7, 61.1, 61.4, 62.8, 633
Just like the mean, the range is very sensitive to outliers!
Range: (max value - min value) = 633 - 40.1 = 592.9
Note that most values were close to 40.1 : the range however gets exaggerated due to one outlier (633)
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{first 25\% data}}
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{third 25\% data}}
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{second 25\% data}}
\underbrace{~~~~~~~~~~~~~~~~~~~}_{\textit{last 25\% data}}
Inter~Quartile~Range (IQR) = Q_3 - Q_1
Measures of spread (IQR)
40.1, 40.9, 41.8, 44, 46.8
47.2, 48.6, 49.3, 49.4, 51.9
53.8, 55.9, 57.3, 58.1, 60.2
60.7, 61.1, 61.4, 62.8, 633
\underbrace{Q_1}_{25\text{-}th~percentile}
Inter~Quartile~Range
\underbrace{Q_3}_{75\text{-}th~percentile}
IQR = Q_3 - Q_1
L_{75} = \frac{75}{100}*(20 + 1) = 15.75
Q_{3} = Y_{75} = x_{15} + 0.75(x_{16} - x_{15}) = 60.575
L_{25} = \frac{25}{100}*(20 + 1) = 5.25
Q_{1} = Y_{25} = x_{5} + 0.25(x_{6} - x_{5}) = 46.9
= 60.575 - 46.9 = 13.675
Measures of spread (IQR)
Inter~Quartile~Range
40.1, 40.9, 41.8, 44, 46.8
47.2, 48.6, 49.3, 49.4, 51.9
53.8, 55.9, 57.3, 58.1, 60.2
60.7, 61.1, 61.4, 62.8, 633
Measures of spread (IQR)
Clearly not sensitive to outliers (i.e,, will not change drastically if we drop the outlier)
IQR = 13.4
L_{75} = \frac{75}{100}*(19 + 1) = 15
L_{25} = \frac{25}{100}*(19 + 1) = 5
IQR^{new} = x_{15} - x_5 = 13.4
Inter~Quartile~Range
60.7, 61.1, 61.4, 62.8, 633
40.1, 40.9, 41.8, 44, 46.8
47.2, 48.6, 49.3, 49.4, 51.9
53.8, 55.9, 57.3, 58.1, 60.2
Measures of spread (variance)
Question: How different are the values in the data from the typical value (mean) in the data?
Possible Solution: Compute the sum or average deviation of all points from the mean
\sum_{i=1}^{n}(x_{i} - \bar{x})
Issue: We already know that the sum of deviations from the mean is 0
\sum_{i=1}^{n}(x_{i} - \bar{x}) = 0
Motivation: Measures of spread
Sample 1: 81, 81, 82, 82, 83, 83, 84, 84, 85, 85 Mean: 83
Median: 83
Sample 1: 31, 41, 61, 72, 83, 83, 94, 105, 125, 135 Mean: 83
Median: 83
Sum of deviations = 0
The sum of deviations does not tell us anything about the spread of the data
Sum of deviations = 0
Measures of spread (variance)
Observation: We do not care about the sign of the deviation (Both positive and negative deviations contribute to the spread in the data and hence we do want them to cancel each other)
Deviations on left side
Deviations on right side
Mean: 6.0
1 2 3 4 5 6 7 8 9 10
=
Measures of spread (variance)
Issue: The sum of deviations from the mean is 0
\frac{1}{n}\sum_{i=1}^{n}(x_{i} - \bar{x}) = 0
Reason: The positive deviations cancel the negative deviations
Solution1: Use absolute values
Solution1: Use square values
\frac{1}{n}\sum_{i=1}^{n}|x_{i} - \bar{x}|
\frac{1}{n}\sum_{i=1}^{n}(x_{i} - \bar{x})^2
preferred solution
Measures of spread (variance)
Variance:
(if computed from a sample)
(if computed from the entire population)
s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_{i} - \bar{x})^2
\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_{i} - \bar{x})^2
Why is there a difference in the formula?
We will clarify this later once we introduce probability theory
Measures of spread (variance)
x
(x-\bar{x})
(x-\bar{x})^2
| 81 | -2 | 4 |
| 81 | -2 | 4 |
| 82 | -1 | 1 |
| 82 | -1 | 1 |
| 83 | 0 | 0 |
| 83 | 0 | 0 |
| 84 | 1 | 1 |
| 84 | 1 | 1 |
| 85 | 2 | 4 |
| 85 | 2 | 4 |
| 31 | -52 | 2704 |
| 41 | -42 | 1764 |
| 61 | -22 | 484 |
| 72 | -11 | 121 |
| 83 | 0 | 0 |
| 83 | 0 | 0 |
| 94 | 11 | 121 |
| 105 | 22 | 484 |
| 125 | 42 | 1764 |
| 135 | 52 | 2704 |
x
(x-\bar{x})
(x-\bar{x})^2
s^2 = (\frac{1}{10-1})20
s^2 = (\frac{1}{10-1})10146
= 2.22
= 1127.33
\bar{x} = 83
\bar{x} = 83
s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_{i} - \bar{x})^2
Measures of spread (variance)
s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_{i} - \bar{x})^2
Observation: Variance is not measured in the same units as the data
if~the~unit~of~data~is~km
then~the~unit~of~variance~is~km^2
Measures of spread (standard deviation)
s=\sqrt{s^2} = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_{i} - \bar{x})^2}
Observation: Standard deviation is measured in the same units as the data
(if computed from a sample)
(if computed from the entire population)
\sigma = \sqrt{\sigma^2} = \sqrt{\frac{1}{N}\sum_{i=1}^{n}(x_{i} - \bar{x})^2}
Recap of notations
Statistic
Sample (size n)
Population (size N)
Mean
Variance
Standard Deviation
\bar{x} = \dfrac{1}{n}\sum_{i=1}^{n}(x_{i})
\mu = \dfrac{1}{N}\sum_{i=1}^{N}(x_{i})
\sigma^2 = \dfrac{1}{N}\sum_{i=1}^{N}(x_{i} - \mu)^2
s^2 = \dfrac{1}{n-1}\sum_{i=1}^{n}(x_{i} - \bar{x})^2
s = \sqrt{s^2}
= \sqrt{\dfrac{1}{n-1}\sum_{i=1}^{n}(x_{i} - \bar{x})^2}
\sigma = \sqrt{\sigma^2}
= \sqrt{\dfrac{1}{N}\sum_{i=1}^{N}(x_{i} - \mu)^2}
A slight detour ... ...
Why do we square the deviations?
Reason1: The square function has better properties than the absolute function
Observation1: The square function is a smooth function and hence differentiable everywhere
Observation2: The absolute function is not differentiable at
x_i - \bar{x} = 0
Why do we square the deviations?
Reason1: The square function has better properties than the absolute function
Why do we care about differentiability?
In many applications (especially in ML) we need functions which are differentiable
Why do we square the deviations?
Reason2: The square function magnifies the contribution of outliers
Why do we want to magnify the contribution of outliers?
Why do we square the deviations?
Reason2: The square function magnify the contribution of outliers
Why do we want to magnify the contribution of outliers?
\bar{x} = 1.29
s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_{i} - \bar{x})^2 = 9.827
avg.~abs.~dev. = \frac{1}{n}\sum_{i=1}^{n}|x_{i} - \bar{x}| = 1.782
0.1, 0.2, 0.3, 0.3, 0.5, 0.1, 0.4, 0.2, 0.6, 10.2
Example: Toxic content in a fertiliser
What does the variance tell us about the data?
Variance: a measure of consistency
Mean = 39.89
Mean = 41.78
Does the variance capture this?
Observation: Root was more consistent than Cook
Ashes 2017-18 series (runs scored)
Alistair Cook: 2, 7, 7, 10, 14, 16, 37, 39, 244
Joe Root: 1, 9, 14, 15, 51, 58, 61, 67, 83
Variance: a measure of consistency
(x - \bar{x})
(x - \bar{x})^2
| 2 | -39.8 | 1582.3 |
| 7 | -34.8 | 1209.5 |
| 7 | -34.8 | 1209.5 |
| 10 | -31.8 | 1009.8 |
| 14 | -27.8 | 771.6 |
| 16 | -25.8 | 664.5 |
| 37 | -4.8 | 22.8 |
| 39 | -2.8 | 7.7 |
| 244 | 202.2 | 40893.8 |
Cook
(x)
| 1 | -38.9 | 1512.3 |
| 9 | -30.9 | 954.1 |
| 14 | -25.9 | 670.2 |
| 15 | -24.9 | 619.5 |
| 51 | 11.1 | 123.5 |
| 58 | 18.1 | 328.0 |
| 61 | 21.1 | 445.7 |
| 67 | 27.1 | 735.0 |
| 83 | 43.1 | 1858.6 |
Root
(x)
(x - \bar{x})
(x - \bar{x})^2
Variance (s^2) = 5921.41
Standard Deviation (s) = 76.95
Variance (s^2) = 905.86
Standard Deviation (s) = 30.89
Indeed the lower variance shows that Root was more consistent than Cook
Variance: a measure of consistency
The primary objective of manufacturing industries is to ensure that there is little variance in their products
Desirable to have almost 0 variance in
... ... ...
length of sleeves
radius of tyres
weight of dumbbells
What is the effect of transformations on measures of spread
Transformations
Scaling and Shifting
x_{new} = a*x + c
(a = 5/9, c = -160/9)
[22.46, 23.54, 24.26, 27.86, 30.2, 30.74, 34.52, 35.96, 40.46, 44.06, 52.7, 54.68, 56.66, 57.56, 59.54, 61.52, 62.06, 65.66, 67.46, 70.88, 76.46, 82.4, 83.12, 84.38, 93.02, 94.28, 95.72, 96.44, 108.86, 109.58]
\degree F:
[-5.3, -4.7, -4.3, -2.3, -1.0, -0.7, 1.4, 2.2, 4.7, 6.7, 11.5, 12.6, 13.7, 14.2, 15.3, 16.4, 16.7, 18.7, 19.7, 21.6, 24.7, 28.0, 28.4, 29.1, 33.9, 34.6, 35.4, 35.8, 42.7, 43.1]
\degree C:
range_{new} = max_{new} - min_{new}
= \overbrace{(a*max + c)} - \overbrace{(a*min + c)}
= a*(max - min)
= a*range
range = max - min
Effect of transformations (on range)
range = 109.58 - 22.46 = 87.12
range_{new} = \frac{5}{9} * 87.12 = 48.4
Effect of transformations (on range)
[22.46, 23.54, 24.26, 27.86, 30.2, 30.74, 34.52, 35.96, 40.46, 44.06, 52.7, 54.68, 56.66, 57.56, 59.54, 61.52, 62.06, 65.66, 67.46, 70.88, 76.46, 82.4, 83.12, 84.38, 93.02, 94.28, 95.72, 96.44, 108.86, 109.58]
\degree F:
[-5.3, -4.7, -4.3, -2.3, -1.0, -0.7, 1.4, 2.2, 4.7, 6.7, 11.5, 12.6, 13.7, 14.2, 15.3, 16.4, 16.7, 18.7, 19.7, 21.6, 24.7, 28.0, 28.4, 29.1, 33.9, 34.6, 35.4, 35.8, 42.7, 43.1]
\degree C:
Q_1^{new}=Y_{25}^{new} = a*Y_{25} + c = a*Q_1 + c
Q_3^{new}=Y_{75}^{new} = a*Y_{75} + c = a*Q_3 + c
IQR^{new}=Q_3^{new} - Q_1^{new}
IQR^{new}=(a*Q_3 + c) - (a*Q_1 + c)
IQR^{new}=a*(Q_3 - Q_1) = a*IQR
Effect of transformations (on IQR)
Recap:
x_{new} = a*x + c
Y_p^{new} = a*Y_p + c
Effect of transformations (on IQR)
IQR = Q_3 - Q_1 = 47.83
IQR_{new} = \frac{5}{9} * 47.83 = 26.57
Exercise: Compute the first and third quartiles for the transformed data and verify that the new IQR is indeed 26.57
[22.46, 23.54, 24.26, 27.86, 30.2, 30.74, 34.52, 35.96, 40.46, 44.06, 52.7, 54.68, 56.66, 57.56, 59.54, 61.52, 62.06, 65.66, 67.46, 70.88, 76.46, 82.4, 83.12, 84.38, 93.02, 94.28, 95.72, 96.44, 108.86, 109.58]
\degree F:
[-5.3, -4.7, -4.3, -2.3, -1.0, -0.7, 1.4, 2.2, 4.7, 6.7, 11.5, 12.6, 13.7, 14.2, 15.3, 16.4, 16.7, 18.7, 19.7, 21.6, 24.7, 28.0, 28.4, 29.1, 33.9, 34.6, 35.4, 35.8, 42.7, 43.1]
\degree C:
Effect of transformations (on variance)
s^2_{new} = \frac{1}{n-1}\sum_{i=1}^n (x_i^{new} - \bar{x}^{new})^2
=a^2s^2
= \frac{1}{n-1}\sum_{i=1}^n ((\overbrace{a*x_i + c}) - (\overbrace{a*\bar{x} + c}))^2
=\frac{1}{n-1}\sum_{i=1}^n [a(x_i - \bar{x})]^2
= a^2*\underbrace{\frac{1}{n-1}\sum_{i=1}^n (x_i - \bar{x})^2}
Effect of transformations (on std. dev.)
s_{new}=\sqrt{s^2_{new}} = \sqrt{a^2*s^2} = a*s
s^2_{new} = a^2*s^2
Effect of transformations (on variance)
Exercise: Compute the variance and standard deviation for the original and transformed data and verify that:
s^2_{new} = a^2*s^2~~~and~~~s_{new} = a*s
s^2_{new} = a^2*s^2~~~and~~~s_{new} = a*s
[22.46, 23.54, 24.26, 27.86, 30.2, 30.74, 34.52, 35.96, 40.46, 44.06, 52.7, 54.68, 56.66, 57.56, 59.54, 61.52, 62.06, 65.66, 67.46, 70.88, 76.46, 82.4, 83.12, 84.38, 93.02, 94.28, 95.72, 96.44, 108.86, 109.58]
\degree F:
[-5.3, -4.7, -4.3, -2.3, -1.0, -0.7, 1.4, 2.2, 4.7, 6.7, 11.5, 12.6, 13.7, 14.2, 15.3, 16.4, 16.7, 18.7, 19.7, 21.6, 24.7, 28.0, 28.4, 29.1, 33.9, 34.6, 35.4, 35.8, 42.7, 43.1]
\degree C:
Measures of centrality
x_{new} = a*x + c
Measures of spread
Effect of transformations
s_{new} = a*s
s^2_{new} = a^2*s^2
IQR_{new} = a*IQR
range_{new}= a*range
\bar{x}_{new} = a*\bar{x} + c
median_{new} = a*median + c
mode_{new} = a*mode + c
Summary
How do you use mean and variance to standardise data?
Standardising data
Question: How many standard deviations away from the mean is a given value x?
Intuition: Instead of expressing distances in absolute values, express them in units of standard deviations
Mean: 5.38 Std. dev: 3.97
-4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
-1 SD
-2 SD
+1 SD
+2 SD
-3 SD
3 SD
Standardising data
x = \bar{x} + s
x = \bar{x} + 2*s
x = \bar{x} + z*s
If is the mean and s is the std. dev. then find the point which is one std. dev. away from the mean
\bar{x}
what about 2 standard deviations?
what about z standard deviations?
Standardising data
We can express any point in the data as
x_i = \bar{x} + z_i*s
z_i = \frac{x_i - \bar{x}}{s}
where
is called the z-score and tells us the number of standard deviations that the point is away from the mean
z_i
Standardising data
| 0 | -1.36 |
|---|---|
| 2 | -0.85 |
| 4 | -0.35 |
| 5 | -0.1 |
| 4 | -0.35 |
| 13 | 1.92 |
| 6 | 0.15 |
| 9 | 0.91 |
| 0 | -1.36 |
| 5 | -0.1 |
\dots
| 12 | 1.67 |
|---|---|
| 14 | 2.17 |
| 15 | 2.42 |
z_i = \frac{x_i - \bar{x}}{s}
x_i
mean\space (\bar{x}) = 5.38
Std\space (s) = 3.97
Standardising data (Usage in ML)
| 2L | 65 | 5.8 |
| 5L | 60 | 5.5 |
| 10L | 75 | 6 |
| 8L | 70 | 5.3 |
| 4L | 54 | 5.2 |
| 7L | 60 | 5.3 |
| 1L | 50 | 5.3 |
| 20L | 72 | 6.2 |
| 7L | 82 | 6.1 |
| 3L | 67 | 5.9 |
Income (INR)
Height (Feet)
Weight (Kgs)
Range:
20-1 = 19L
82-50 = 32Kgs
6.2-5.2 = 1.0ft
Income
Height
Weight
Health Risk
No Health Risk
ML System
Standardising data (Usage in ML)
| 2L | 65 | 5.8 |
| 5L | 60 | 5.5 |
| 10L | 75 | 6 |
| 8L | 70 | 5.3 |
| 4L | 54 | 5.2 |
| 7L | 60 | 5.3 |
| 1L | 50 | 5.3 |
| 20L | 72 | 6.2 |
| 7L | 82 | 6.1 |
| 3L | 67 | 5.9 |
Income (INR)
Height (Feet)
Weight (Kgs)
Standardised Data
| -0.86 | -0.05 | 0.37 |
| -0.31 | -0.56 | -0.42 |
| 0.6 | 0.97 | 0.89 |
| 0.24 | 0.46 | -0.95 |
| -0.49 | -1.18 | -1.21 |
| 0.05 | -0.56 | -0.95 |
| -1.04 | -1.59 | -0.95 |
| 2.44 | 0.66 | 1.42 |
| 0.05 | 1.69 | 1.16 |
| -0.68 | 0.15 | 0.63 |
Income (INR)
Height (Feet)
Weight (Kgs)
What is the mean and standard deviation of the standardised data?
Standardising data (Usage in ML)
What is the mean and standard deviation of the standardised data?
Is this always the case?
| -0.86 | -0.05 | 0.37 |
| -0.31 | -0.56 | -0.42 |
| 0.6 | 0.97 | 0.89 |
| 0.24 | 0.46 | -0.95 |
| -0.49 | -1.18 | -1.21 |
| 0.05 | -0.56 | -0.95 |
| -1.04 | -1.59 | -0.95 |
| 2.44 | 0.66 | 1.42 |
| 0.05 | 1.69 | 1.16 |
| -0.68 | 0.15 | 0.63 |
Mean: 0
Mean: 0
Mean: 0
Std: 1
Std: 1
Std: 1
Income (INR)
Height (Feet)
Weight (Kgs)
Standardising data (Usage in ML)
Prove that the mean of the standardised data is 0
Proof:
z_i = \frac{x_i - \bar{x}}{s}
\bar{z} = \frac{1}{n}\sum_{i=1}^{n} z_i
\bar{z} = \frac{1}{n}\sum_{i=1}^{n} \frac{x_i - \bar{x}}{s}
\bar{z} = \frac{1}{s}(\frac{1}{n}\sum_{i=1}^{n} x_i - \frac{1}{n}\sum_{i=1}^{n}\bar{x})
\bar{z} = \frac{1}{s}(\bar{x} - \frac{1}{n}*n*\bar{x}) = 0
Standardising data (Usage in ML)
Prove that the standard deviation of the standardised data is 1
Proof:
z_i = \frac{x_i - \bar{x}}{s}
s^2_z = \frac{1}{n-1}\sum_{i=1}^{n} (z_i - \bar{z})^2 = \frac{1}{n-1}\sum_{i=1}^{n} z_i^2
s^2_z = \frac{1}{n-1}\sum_{i=1}^{n} (\frac{x_i - \bar{x}}{s})^2
s^2_z = \frac{1}{s^2}*\frac{1}{n-1}\sum_{i=1}^{n} (x_i - \bar{x})^2
s^2_z = \frac{1}{s^2}*s^2 = 1
Summary (measures of spread)
Measures of spread
range = max - min
IQR = Q_3 - Q_1
s^2 = \frac{1}{n-1}(x_i - \bar{x})^2~~,~~\sigma^2=\frac{1}{N}(x_i - \bar{x})^2
s = \sqrt{\frac{1}{n-1}(x_i - \bar{x})^2}~~,~~\sigma=\sqrt{\frac{1}{N}(x_i - \bar{x})^2}
Except IQR all measures are sensitive to outliers
Measure of consistency in the data
Except variance, all measures have the same unit as the original data
Summary (measures of spread)
Effect of transformations
x_{new} = a*x + c
s_{new} = a*s
s^2_{new} = a^2*s^2
IQR_{new} = a*IQR
range_{new}= a*range
Standardising data
After standardising the data has zero mean and unit variance
z_i = \frac{x_i - \bar{x}}{s}
What are box plots?
Box Plots
x < Q_1 - 1.5 * IQR
x > Q_3 + 1.5* IQR
or if
Box plots are used for visualising spread, median and outliers in the data
We say that x is an outlier if
(a formal definition of outliers)
Box Plots
IQR
Median
Outliers
1.5 * IQR
1.5 * IQR
Box Plots
Shikhar Dhawan T20I scores (59 sorted scores)
0, 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 8, 9, 10, 10, 11, 13, 14, 15, 16, 19, 23, 23, 23, 24, 26, 29, 30, 30, 31, 32, 32, 33, 35, 36, 40, 41, 41, 42, 43, 46, 47, 51, 52, 55, 60, 72, 74, 76, 80, 90, 92
Median = 23
Q1 = 5, Q3 = 41
IQR = 36
Q1 - 1.5 IQR = -49
Q3 + 1.5 IQR = 95
(no outliers on either side)
Box Plots
Median = 23
Q1 = 5, Q3 = 41
IQR = 36
Q1 - 1.5 IQR = -49
Q3 + 1.5 IQR = 95
Box Plots (Variant 1)
Median = 32
Q1 = 27.75, Q3 = 38.25
IQR = 10.5
Q1 - 1.5 IQR = 12
Q3 + 1.5 IQR = 54
max value = 45
min value = 2
[2, 6, 16, 20, 23, 23, 27, 28, 28, 29, 29, 30, 30, 31, 32, 32, 32, 33, 34, 34, 36, 38, 38, 39, 40, 40, 40, 42, 43, 45]
Box Plots (Variant 1)
If there are no outliers move the whisker to the max/min value
Median = 32
Q1 = 27.75, Q3 = 38.25
IQR = 10.5
Q1 - 1.5 IQR = 12
Q3 + 1.5 IQR = 54
max value = 45
min value = 2
0
10
20
30
40
50
max = 45
Box Plots (Variant 2)
Median = 32
Q1 = 27.75, Q3 = 38.25
IQR = 10.5
Q1 - 1.5 IQR = 12
Q3 + 1.5 IQR = 54
max value = 45
min value = 2
[2, 6, 16, 20, 23, 23, 27, 28, 28, 29, 29, 30, 30, 31, 32, 32, 32, 33, 34, 34, 36, 38, 38, 39, 40, 40, 40, 42, 43, 45]
Box Plots (Variant 2)
five number summary: min, max, median, Q1, Q3
Median = 32
Q1 = 27.75, Q3 = 38.25
IQR = 10.5
Q1 - 1.5 IQR = 12
Q3 + 1.5 IQR = 54
max value = 45
min value = 2
0
10
20
30
40
50
Min
Max
Q1
Q3
Median
Box Plots (Variant 3)
max value = 45 (excluding outliers)
min value = 16 (excluding outliers)
Median = 32
Q1 = 27.75, Q3 = 38.25
IQR = 10.5
Q1 - 1.5 IQR = 12
Q3 + 1.5 IQR = 54
[2, 6, 16, 20, 23, 23, 27, 28, 28, 29, 29, 30, 30, 31, 32, 32, 32, 33, 34, 34, 36, 38, 38, 39, 40, 40, 40, 42, 43, 45]
Box Plots (Variant 3)
move the whisker to the max/min value obtained after excluding all outliers
max value = 45 (excluding outliers)
min value = 16 (excluding outliers)
0
10
20
30
40
50
Median = 32
Q1 = 27.75, Q3 = 38.25
IQR = 10.5
Q1 - 1.5 IQR = 12
Q3 + 1.5 IQR = 54
Box Plots: skewness in the data
Right Skewed
Left Skewed
Symmetric
Box Plots (Usage in ML)
Evaluation:
p_1, p_2, p_3, ...., p_{200}
n_1, n_2, n_3, ...., n_{200}
(scores assigned by the system to + reviews)
(scores assigned by the system to - reviews)
Sample Review
ML System
or
Box Plots (Usage in ML)
(No overlap in the IQR of the scores assigned to positive and negative reviews)
Ideal box plot
Box Plots (Usage in ML)
Box plots can be used to visually compare the performance of ML systems
- M1 does poorly, high overlap in IQRs of +ve and -ve
- M2 does poorly, many outliers coinciding with opposite class IQR
- M3 does well in separating the +ve and -ve classes
Box Plots
Median = 23
Q1 = 5, Q3 = 41
IQR = 36
Q1 - 1.5 IQR = -49
Q3 + 1.5 IQR = 95
Measures of Centrality
| Age | Height | Weight | Cholesterol | Sugar level | .... ..... |
|---|---|---|---|---|---|
| 32 | 165 | 75 | 124 | 108 | ... |
| 24 | 172 | 81 | 112 | 98 | ... |
| ... | ... | ... | ... | ... | ... |
| ... | ... | ... | ... | ... | ... |
| ... | ... | ... | ... | ... | ... |
Which is the 7th most grown crop in the country?
Hard to answer
Nominal attributes
Recall
Question: What is the typical value of an attribute in our dataset?
How many runs does Sachin Tendulkar typically score in a match?
How many balls does Sachin Tendulkar typically face in a match?
Measures of Centrality
| Age | Height | Weight | Cholesterol | Sugar level | .... ..... |
|---|---|---|---|---|---|
| 32 | 165 | 75 | 124 | 108 | ... |
| 24 | 172 | 81 | 112 | 98 | ... |
| ... | ... | ... | ... | ... | ... |
| ... | ... | ... | ... | ... | ... |
| ... | ... | ... | ... | ... | ... |
Which is the 7th most grown crop in the country?
Hard to answer
Nominal attributes
Recall
Question: What is the typical value of an attribute in our dataset?
How many runs does Sachin Tendulkar typically score in a match?
How many balls does Sachin Tendulkar typically face in a match?
Motivation: Summarise Big Data
| Age | Height | Weight | Cholesterol | Sugar level | .... ..... |
|---|---|---|---|---|---|
| 32 | 165 | 75 | 124 | 108 | ... |
| 24 | 172 | 81 | 112 | 98 | ... |
| ... | ... | ... | ... | ... | ... |
| ... | ... | ... | ... | ... | ... |
| ... | ... | ... | ... | ... | ... |
Which is the 7th most grown crop in the country?
Hard to answer
Nominal attributes
| 1 | 38.9 | 1512.3 |
| 9 | 30.9 | 954.1 |
| 14 | 25.9 | 670.2 |
| 15 | 24.9 | 619.5 |
| 51 | 11.1 | 123.5 |
| 58 | 18.1 | 328.0 |
| 61 | 21.1 | 445.7 |
| 67 | 27.1 | 735.0 |
| 83 | 43.1 | 1858.6 |
Score
(x)
(x - \bar{x})
(x - \bar{x})^2
Mean: 5.38 Std. dev: 3.97
-4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
-1 SD
-2 SD
+1 SD
+2 SD
-3 SD
3 SD
0
10
20
30
40
50
Left side values
Right side values
Mean: 6.0
Left side values
Right side values
1 2 3 4 5 6 7 8 9 10
Left side values
Right side values
1 2 3 4 5 6 7 8 9 10
Left side values
Right side values
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
Left side values
Right side values
Summarising Data - Part 2
By One Fourth Labs
Summarising Data - Part 2
PadhAI One: FDS Week 3 (MK)
