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Further kinematics

Vector methods with projectiles

Variable acceleration in one dimension

Differentiating vectors

Integrating vectors

Application of forces

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Modelling with statics

Friction with static particles

Statics of rigid bodies

Dynamics and inclined planes

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Functions of time: Displacement, velocity, acceleration

Using differentiation to find velocity and acceleration

Maxima and minima problems

Using integration to find velocity and displacement

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Constant acceleration

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Constant acceleration: Deriving formulae

Constant acceleration formulae: SUVAT

Vertical motion under gravity

Modelling in mechanics

Modelling assumptions

Quantities and units in mechanics

Working with vectors

Normal distribution

The normal distribution

The normal distribution: Finding probabilities

The inverse normal distribution function

The standard normal distribution

Finding the mean and standard deviation

Normal approximation to the binomial distribution

Hypothesis testing with the normal distribution

Conditional probability

Set notation

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Probability formulae

Conditional probability: Tree diagrams

Correlation and hypothesis testing

Exponential models

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Hypothesis testing I

Hypothesis testing

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One-tailed tests

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Binomial distribution

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The binomial distribution

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Venn diagrams

Probability: Mutually exclusive and independent events

Probability: Tree diagrams

Representation of data

Outliers

Box plots

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Comparing data sets

Correlation

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Variance and standard deviation

Data collection

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Vectors II

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Vectors in 3D

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Applications to mechanics: 3D vectors

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Numerical methods: Applications to modelling

Integration II

Integrating standard functions

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Parametric differentiation

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Second derivatives: Concave and convex functions

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Sequences revision

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Binomial expansion II

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Binomial expansion of compound expressions

Parametric equations

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The modulus function

y = |f(x)| and y = f(|x|)

Combining transformations

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Proof II

Proof by contradiction

Criticising proofs

Vectors I

Vectors

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Magnitude and direction of a vector

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Solving geometric problems

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Integration I

Integrating x^n

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Differentiation I

Finding the gradient of a curve

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y = e^x

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Pascal's triangle

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y = mx + c

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Proof I

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Explainer Video

Tutor: Labib

Summary

Outliers

​​In a nutshell

Outliers are extreme values outside the trend of a data set. They can be calculated in different ways.


Variable definitions

Variable Name

symbol

Upper quartile​​

​Q3Q_3Q3​​​

Lower quartile​​

​Q1Q_1Q1​​​

​Constant​

kkk​​


​​Equations

Description

Equation

Outliers that are beyond the maximum.

​Q3+k(Q3−Q1)Q_3 + k(Q_3-Q_1)Q3​+k(Q3​−Q1​)​​

Outliers that are below the minimum.

​Q1−k(Q3−Q1)Q_1 - k(Q_3-Q_1)Q1​−k(Q3​−Q1​)​​

The interquartile range. 

Q3−Q1Q_3 -Q_1Q3​−Q1​​​




Finding outliers

Use the given equations to find outliers from a data set. The kkk value will be provided or implied within the question. ​

Q3+k(Q3−Q1)   and   Q1−k(Q3−Q1)\boxed{Q_3 + k(Q_3-Q_1) \ \ \ and \ \ \ Q_1 - k(Q_3-Q_1)}Q3​+k(Q3​−Q1​)   and   Q1​−k(Q3​−Q1​)​​​


Example 1

The mass of 303030 bunnies is recorded. The results in kgkgkg are shown below:​

​​​1.6, 2.3, 2.4, 2.4, 2.6, 2.7, 2.9, 3.1, 3.7, 3.8, 3.8, 3.8, 3.9, 3.9, 3.9,4.1, 4.1, 4.1, 4.3, 4.6, 4.8, 5.0, 5.0, 5.0, 5.2, 5.2, 5.3, 5.4, 5.5, 5.81.6 , \ 2.3, \ 2.4, \ 2.4 ,\ 2.6, \ 2.7, \ 2.9, \ 3.1, \ 3.7, \ 3.8, \ 3.8, \ 3.8, \ 3.9, \ 3.9, \ 3.9, \\ 4.1, \ 4.1, \ 4.1, \ 4.3, \ 4.6, \ 4.8, \ 5.0, \ 5.0, \ 5.0, \ 5.2, \ 5.2, \ 5.3, \ 5.4 , \ 5.5, \ 5.81.6, 2.3, 2.4, 2.4, 2.6, 2.7, 2.9, 3.1, 3.7, 3.8, 3.8, 3.8, 3.9, 3.9, 3.9,4.1, 4.1, 4.1, 4.3, 4.6, 4.8, 5.0, 5.0, 5.0, 5.2, 5.2, 5.3, 5.4, 5.5, 5.8​​


An outlier is observed at 1.5×interquartile range1.5 \times \text{interquartile range}1.5×interquartile range above the upper quartile or 1.5×interquartile range1.5 \times \text{interquartile range}1.5×interquartile range below the lower quartile. Does the data set have any outliers?​


Find the lower quartile which a quarter of the way through the data points:

30×14=7.5→8th term30 \times \dfrac1 4 = 7.5 \rightarrow 8^{th} \ \textit{term}30×41​=7.5→8th term

​​

8th term=3.18^{th} \ \textit{term} = 3.18th term=3.1​​


Find the upper quartile which three quarters of the way through the data points:

30×34=22.5→23rd term30 \times \dfrac34 = 22.5 \rightarrow 23^{rd} \ \textit{term} 30×43​=22.5→23rd term

​​

​23rd term=5.023^{rd} \ \textit{term} = 5.023rd term=5.0


Find the interquartile range:

5.0−3.1=1.95.0 - 3.1 = 1.95.0−3.1=1.9​​


Outliers are values less than:

​3.1−(1.5×1.9)=0.253.1 - (1.5 \times 1.9) = 0.253.1−(1.5×1.9)=0.25


Outliers are values greater than:

5.0+(1.5×1.9)=7.855.0 + (1.5 \times 1.9) = 7.855.0+(1.5×1.9)=7.85​​

​

There are no outliers in the data set. 


Note: When calculating the position of a given quartile always round up.



Cleaning data

Questions may use different definitions for outliers. This may involve the application of different equations like those for mean and standard deviation. Once outliers are identified anomalies can then be removed from a data set with a justifiable reason. This is known as cleaning data.  


​​Example 2

The height of 101010​ tigers are given in the data set below in metres:

1.2, 1.5, 3.1, 3.2, 3.5, 3.7, 4.0, 4.1, 5.3, 8.41.2, \ 1.5, \ 3.1, \ 3.2, \ 3.5, \ 3.7, \ 4.0, \ 4.1, \ 5.3, \ 8.41.2, 1.5, 3.1, 3.2, 3.5, 3.7, 4.0, 4.1, 5.3, 8.4​​


Two sums are also given: 

∑(x)=38∑(x2)=180.94\sum (x) = 38 \quad \quad \sum(x^2) = 180.94∑(x)=38∑(x2)=180.94​​


Outliers are more than a standard deviation away from the mean. Find the outliers and use this information to clean the data set, justified with reasons. 


Calculate the mean:

Mean=x‾=∑xn=3810=3.8\text{Mean} =\overline{x}= \dfrac{\sum x}{n} = \dfrac{38}{10} = 3.8Mean=x=n∑x​=1038​=3.8


Calculate the standard deviation:

Variance=∑x2n−x‾2=180.9410−3.82=3.654\text{Variance} = \dfrac {\sum x^2}{n} - \overline{x}^2 = \dfrac {180.94}{10} - 3.8^2 = 3.654Variance=n∑x2​−x2=10180.94​−3.82=3.654


Variance=Standard deviation=3.654=1.9 (1 d.p.)\sqrt{\text{Variance}} = \text{Standard deviation} = \sqrt{3.654} = 1.9\ (1 \ d.p.)Variance​=Standard deviation=3.654​=1.9 (1 d.p.)


Find the outliers:

smaller than 3.8−1.9=1.93.8 - 1.9 = 1.93.8−1.9=1.9​​

bigger than 3.8+1.9=5.73.8+1.9 = 5.73.8+1.9=5.7​​


Hence the outliers are 1.2, 1.5, 8.4‾\underline{1.2, \ 1.5, \ 8.4}1.2, 1.5, 8.4​.


Identify the anomaly with a reason: 

1.21.21.2 and 1.51.51.5 may not be anomalous as it is possible these are juvenile tigers. 

It is highly unlikely a tiger is 8.4 m8.4 \ m8.4 m because this is a significantly larger than x‾+σ\overline{x}+\sigmax+σ with an almost 3 m3\ m 3 m difference.


Rewrite the cleaned data set: 

1.2, 1.5, 3.1, 3.2, 3.5, 3.7, 4.0, 4.1, 5.3‾\underline{1.2, \ 1.5, \ 3.1, \ 3.2, \ 3.5, \ 3.7, \ 4.0, \ 4.1, \ 5.3}1.2, 1.5, 3.1, 3.2, 3.5, 3.7, 4.0, 4.1, 5.3​​​





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Learn with Basics

Learn the basics with theory units and practise what you learned with exercise sets!
Length:
Measures of location: Quartiles, percentiles, deciles

Unit 1

Measures of location: Quartiles, percentiles, deciles

Measures of spread: Range

Unit 2

Measures of spread: Range

Jump Ahead

Score 80% to jump directly to the final unit.

Optional

This is the current lesson and goal (target) of the path

Outliers

Unit 3

Outliers

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FAQs - Frequently Asked Questions

What is the most common equation for outliers?

There are two common equation which are used. Values with are greater than: Lower quartile - k(Interquartile range). Values which are less than: Upper quartile + k(interquartile range).

What is cleaning data?

Removing anomalies from a data set.

What is an outlier?

Outliers are extreme values outside the trend of a data set.

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