Statistics (S1) - How Tos
How To revision cards on S1 Statistics including formulae, step-by-step instructions and examples.
- Mathematics
- Statistics, averages and distributionsNumerical methodsNumerical MeasuresProofProbabilityDistributionsLawsSymbolsMeasures of Spread
- AS
- AQA
- Created by: Betsy_2018
- Created on: 26-12-16 10:23
Population and Sample Standard Deviation
Population
σ = | ∑(x - µ)^2
o.o√ n
- Use the population stadard deviation if it specifies that it is the entire population of data
- Use the population standard deviation if all data is present (e.g number of passengers on a us each day of the week: 102, 353, 401, 485, 209, 314, 175)
Sample
s = | ∑(x - µ)^2
.. . √ n-1
- Use the sample standard deviation when is specifies that it is a sample
- estimates the population based on a sample
Probability: Union and Intersection
- Event = set of outcomes
- Sample space = set of all possible outcomes
- Exhaustive events = set of outcomes cover all possible outcomes of the sample space
- Mutually exclusive events = 2 events can not occur at the same time
- Independent events = occurance of events do not affect others
- Complements = An event and the opposite of that event (e.g A and A')
- Conditional probability = given = [(e.g B|A) = independent if P(A) = P(A|B)]
Union = ∪= 'or', 'both'
- Addition Law = P(A∪ B) = P(A) + P(B) - P(A∩B)
- Mutually Exclusive Law = P(AUB) = P(A) + P(B)
Intersection = ∩ = 'overlap', 'and'
- Multiplication Law = P(A∩B) = P(A) x P(B|A)
- Independent Law = P(A∩B) = P(A) x P(B)
Standard Deviation Summary
- A much more precise measure of spread than just the range or the interquartile range.
- For S1, there are 5 (unfortunately!) formulae that you need to be able to recognise - they will be on your formula sheet - and are each for different types of data
- The standard deviation is the difference between each piece of data and the mean
Symbols
σ = Sigma = population standard deviation
∑x = total of data
x̄ = sample mean
µ = population mean
n = number of pieces of data = ∑ƒ
s = sample standard deviation
Binomial Formula and Cumulative Binomial Tables
P(X = x) = (n r)Px(1-p)(n-x) = n! ÷ (n-X)!X!
Example
20 friends, probability that 5 will go swimming next week? p= 1/7 n= 20 r= 5
(20 5) 1/75 x 6/715 = 0.0914
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Cumulative Binomial Tables
- This is the same concept, but instead of =, it is going to be < > ≤ ≥
- you will need to use the formula above, but add each term's applied formula together
Example
p= 0.25 n= 8 X~B(8, 0.25)
p(X<3) = P(X=0) + P(X=1) + P(X=2)
Binomial Distribution Summary
X = successes = variable = 'pass or fail'
n = number of trials (number of things) taken r at a time = fixed
P = probability of successes
Q = probability of failure (1-P)
X ~ B(n, p) = binomial probability Bi = 2 outcomes
(n r) = nCr
Note: the formula can be used OR the statistics table, each labelled n=[ ] Either way, draw a numberline! No graphs are drawn.
When using the tables, you may be required to do '1-P(X)' if you are trying to find (P≥?)
or P(X) - P(X-1) when trying to find an exact value
or P(X) - P(Y) to find a range
Mean and Variance of Binomial Distribution
Mean
= µ
= np
= (number of trials x probability)
Variance
= σ2
= np(1-p)
= (number of trials x probability)(1 - probability)
Standardising a Normal Variable
z = x - µ / σ
x = µ + zσ
Example
mean = 50
x = 51
σ^2 = 25 (5^2)
Therefore, 51 - 50 / 5 = 1/5
P(Z<0.2) = 0.579 (3sf)
Normal Distribution Types (1)
Finding a probability using table 3, given x
z = x-µ / σ
Finding an x value using table 4, given a probability
x = µ + zσ
Make x the opposite (e.g if you need to find the first 5%, switch to 95%) because p cannot <0.5
Setting up 2 equations with unknown µ and σ, solving simultaneously
Given 2 x values, x = µ + zσ, probability between the 2 x values
Normal Distribution Types (2)
Standard deviations away from the mean (Modelling Normal Distribution)
- 68% area lies in the ranges of µ ± σ (1 standard deviation)
- 95.5% area lies in the ranges of µ ± 2σ (2 standard deviations)
- 99.7% area lies in the ranges of µ ± 3σ (3 standard deviatoons)
Finding the sample mean
sample size n
x̅ distributed by µ and standard deviation σ/√n
z = x - x̅ / σ/√n
x = x̅ + z(σ/√n)
Normal Distribution Summary
- bell-shaped
- draw graphs!
- median = mode = mean = symmetrical
- total area = 1
- high population density close to the mean
- X ~ N(0, 1^2)
X is normally distributed with µ (mean) and σ2 (variance)
Standard Normal Distribution
Mean = µ = 0
Standard Deviation = 1
- Z ~ N(0, 1)2
Z is normally distributed with µ(=0) and σ2(=1)
Normal Distribution Tables
For this, we are using Table 3 and Table 4.
Table 3 is to find the probability that Z (mean = 0 and variance = 1) is normally distributed to euqal less than or equal to z
Example
1.36 on a graph. P(Z≤1.36) = 0.913 on Table 3
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Table 4 is to find the value of z that satisfies P(Z≤z) = p
Example
p = 0.9
P(Z≤z) = 0.9
z = 1.2816 on Table 4
The Critical Limit Theorem
x̅ ~N(µ, (σ/√n)^2)
z = x̅- µ / (σ/√n)^2
x̅ = µ + z(σ/√n)
Example
Weight of pebbles are distributed with mean 48.6g and standard deviation 8.5g. Random sample of 50.
P(x̅<49.0g) n= 50
z = 49.0 - 48.6 / 8.5/√50 = 0.33 (2dp)
P(x̅<49.0g) = P(Z<0.33) = 0.629 (3sf)
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