The NESA Maths Reference Sheet is a superb useful resource… if you understand how to make use of it! Navigate statistics and combinatorics with our Final NESA Maths Reference Sheet Information.
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All the important thing Maths formulation it is advisable to revise, in a single foldable cheatsheet.
Click on on the next formulation to see what they imply and apply them to a observe query!
Statistical Evaluation 
(z=frac{xµ}{σ})
(textual content{An outlier is a rating:})
Regular distribution
(E(X)=μ=sum xP(X=x)) (Var(X)=E[(Xμ)^2]=E(X^2)μ^2)
Likelihood (P(A∩B)=P(A)P(B)) (P(A∪B)=P(A)+P(B)P(A∩B)) (P(AB)=frac{P(A∩B)}{P(B)}, P(B)≠0)
Steady random variables (P(X≤r) = int _a^r f(x) dx) (P(a < X < b) = int _a^b f(x) dx)
Binomial distribution (P(X=r) = ^n C_r p^r (1p)^{nr}) (X textual content{~ Bin} (n,p)) (⇒ P(X=x) = binom{n}{x} p^x (1p)^{nx}, x=0, 1, . . . , n) (E(X)=np) (textual content{Var} (X) = np(1p)) 
Combinatorics 
(^n P_r = frac{n!}{(nr)!})
(binom{n}{r} = ^n C_r = frac{n!}{r!(nr)!}) ((x+a)^n = x^n +binom{n}{1} x^{n1} a + … + binom{n}{r} x ^{nr}a^r + … + a^n) 
Statistical Evaluation
Use  Formulation  Rationalization 
zscore  (z=frac{xµ}{σ})  (z) = zscore
(x) = rating (µ) = imply/common (σ) = commonplace deviation 
Outliers  start{align*} textual content{An outlier is a rating:} &textual content{lower than } Q_1 – 1.5 instances IQR &textual content{or} &textual content{greater than } Q_3 +1.5 instances IQR finish{align*} 
Outlier: a quantity/worth that doesn’t appropriately signify the info set
Q_{2}: Median of the entire set = the center time period Q_{1}: First quartile = the median of all of the numbers smaller than the median of the entire set Q_{3}: Third quartile = the median of all of the numbers higher than the median of the entire set IQR: interquartile vary = Q_{3} – Q_{1} 
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Subsequent part: Combinatorics
Instance 24:
Think about the ages of a gaggle of people:
1, 2, 4, 5, 13, 14, 24, 70
a) Checklist out all of the outlier worth(s).
b) After eradicating all outliers, calculate the zscore of the third youngest particular person (nearest 2 decimal locations).
Resolution 24 a):
start{align*} Q_1 &= frac{2+4}{2} &= 3 Q_3 &= frac{14+24}{2} &= 19 1.5 instances IQR &= 1.5 instances (Q_3 – Q_1) &= 1.5 instances (19 – 3) &= 24 Q_1 – 1.5 instances IQR &= 3 – 24 &= 21 Q_3 + 1.5 instances IQR &= 19 + 24 &= 43 finish{align*}There aren’t any values smaller than (Q_1) — 1.5 IQR = 21, however there’s a worth higher than (Q_3) + 1.5 IQR = 43 < 70.Subsequently, the outlier worth is 70. 
Resolution 24 b):
start{align*} µ &= frac{1 + 2+ 4+ 5+ 13+ 14+ 24}{7} &= 9 σ &= 7.746 z &=frac{xµ}{σ} &= frac{49}{7.746} ∴ zscore &= – 0.65 textual content{ (nearest 2 decimal locations)} finish{align*} 
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Regular distribution

Use  Formulation  Variables 
Anticipated worth  (E(X)=μ)  E(X): anticipated worth of a regular distribution
μ: imply/common of all of the values within the regular distribution 
Variance  (Var(X)=E[(Xμ)^2]=E(X^2)μ^2)  Var(X): variance of a regular distribution
E(Y): imply of outlined Y perform μ: imply/common of all of the values within the regular distribution 
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Likelihood
Use  Formulation  Rationalization 
Likelihood of each occasion A and occasion B occurring  (P(A∩B)=P(A)P(B))  ∩: or
∪: and : given A and B are occasions that might be outlined as something, e.g. touchdown heads on a coin, profitable a sports activities match or raining the subsequent day. 
Likelihood of occasion A or occasion B occurring (this consists of when each occasions happen)  (P(A∪B)=P(A)+P(B)P(A∩B))  
Conditional chance
Likelihood of occasion A occurring, given occasion B has occurred 
(P(AB)=frac{P(A∩B)}{P(B)}, P(B)≠0) 
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Steady random variables
Formulation  Rationalization 
(P(X≤r) = int _a^r f(x) dx)  (a) and (b) are constants ∈ (mathbb{R})
Need assistance with integration? See our record of guides right here. 
(P(a < X < b) = int _a^b f(x) dx) 
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Binomial distribution
Use  Formulation  Rationalization 
Likelihood of a binomial occasion occurring  (P(X=r) = ^n C_r p^r (1p)^{nr})  Confused? This Newbie’s Information explains the basics of binomials. 
Notation for a binomial distribution  (X textual content{~ Bin} (n,p))  
Binomial enlargement  (P(X=x) = binom{n}{x} p^x (1p)^{nx}, x=0, 1, . . . , n)  
Anticipated worth  (E(X)=np)  
Variance  (textual content{Var} (X) = np(1p)) 
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Combinatorics
Use  Formulation  Rationalization 
Permutations  (^n P_r = frac{n!}{(nr)!})  (n, r ∈ mathbb{Z}^+) (constructive integers)
(^n P_r): The variety of other ways you may order r gadgets from a collection of n gadgets. There are two issues that matter: which gadgets you decide and the order through which you decide them in. 
Combos  (binom{n}{r} = ^n C_r = frac{n!}{r!(nr)!})  (n, r ∈ mathbb{Z})
(^n C_r): The variety of other ways you may decide r gadgets from a collection of n gadgets. The order of the gadgets does NOT matter! Solely which gadgets you decide issues 
Observe: the exclamation mark (!) denotes a factorial — the product of all constructive integers lower than or equal to given quantity.
e.g. start{align*} 
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Instance 25:
There are 5 cyclists collaborating in a race and their names are Anna, Barry, Celia, David and Edgar. The spectators are taking turns guessing who’s going to come back first, second and third. What number of completely different guesses are potential?
Resolution 25:
For simplicity’s sake, let’s check with the cyclists by their first preliminary: A, B, C, D and E. If we have been to select the order of the highest 3, selecting out A for 1st, B for 2nd and C for third wouldn’t be the identical as selecting out B for 1st, A for 2nd and C for third.
See how order issues on this choice? Such an ordered choice known as a permutation, and we’d use the notation (^5 P_3) to point what number of completely different permutations of the highest 3 we are able to make from the 5 individuals. We will discover what number of completely different permutations are potential by merely itemizing out all of the completely different potentialities: ABC, ACB, BAC, BCA, CAB, CBA, BCD, BDC, CDB, CBD, DCB, DBC, CDE, CED, DEC, DCE, ECD, EDC, DEA, DAE, EAD, EDA, AED, ADE, … the record goes on! You possibly can see why itemizing out all of the permutations isn’t all the time the best solution to learn how many there are. Happily, we are able to use the method (^n P_r = frac{n!}{(nr)!}) to mathematically calculate what number of permutations there are. Substitute (n=5) and (r=3) into (^n P_r = frac{n!}{(nr)!}): start{align*} 
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Instance 26:
The prefect elections are underway at Matrix Excessive College. In a cohort of 120 college students, there are 68 feminine college students and 52 male college students. If the prefect physique should include 30 college students and an equal variety of ladies and boys, what number of completely different outcomes for the prefect elections are potential?
Write your reply in scientific notation to the closest 2 decimal locations.
Resolution 26:
On this state of affairs, we are able to see that the order of the scholars picked to be prefects is irrelevant. The one factor that issues is the prefect physique of 30 has an equal variety of ladies and boys.
That’s, 15 feminine college students are chosen to be prefects out of the 68 in whole ((^{68} C_{15})) and that 15 male college students are chosen out of the 52 in whole ((^{52} C_{15})). Therefore, the variety of potential prefect physique combos: start{align*} 
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Use  Formulation  Rationalization 
Binomal theorem  ((x+a)^n = x^n +binom{n}{1} x^{n1} a + … + binom{n}{r} x ^{nr}a^r + … + a^n)  Use this theorem to develop the powers of any expression consisting of the sum of two phrases, i.e. ((x+a)^n), the place (x) and (a) could be any fixed or variable, and (n) is a constructive integer worth. 
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Instance 27:
Discover the coefficient of (x^3) within the expression ((3+2x)^4).
Resolution 27:
start{align*} (32x)^4 &= 3^4 + binom{4}{1} 3^3 instances (2x) + binom{4}{2} 3^2 instances (2x)^2+ binom{4}{2} 3 instances (2x)^3 + (2x)^4 textual content{ (Binomial Theorem)} ∴ textual content{Coefficient of }x^3 &= frac{binom{4}{2} 3 instances (2x)^3}{x^3} &= binom{4}{2} 3 instances (2)^3 &= ^4 C _2 instances 3 instances (8) &= 144 finish{align*} 
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