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Statistics

STAT 263

Charles Arthur Molson

Winter

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STAT 263 Formulas (these pages, and tables printed with exam) V 1.7.2a
3(meamedian) p
SK = standard deviation indexp percentile () +1
100
x−µ xx−
IQR = Q3− Q1 z = σ or z = s
s σ
CV =⋅ 100% or =C⋅V 100%
Coefficient of Variation: x µ
1
Chebyshev's Rule: at leak2 of the data fall withink standard deviations of the mean
j
Median for grouped data (similar for any fractile):≈+c
f
Where L is the lower boundary of the class into which the median must fall, f is the frequency of
this class, c is the class interval, and j is the number of values we still lack when we reach L.
= PA )≥ 0 for any eAt ; A ( ) 1 for any vent P; (φ) P0; PA( ) ( ) 1
If A and B are mutually exclusive events, then P(A ∪B)=P(A)+P(B).
If k events are mutually exclusive, then
P(A1∪ A 2 A ∪3…. ∪ A )=k(A )+1(A ) 2…+ P(A ) k
It is always true that: P(A ∪B)=P(A)+P(B) - P(A ∩ B)
If P(B) is not equal to zero, then: P(A) ∩
P(BA ) =
PB()
If A and B are independent: P(A) = P(A|B) and P(A ∩ B) = P(A) P(B)
It is always true that: ( A ∩B ) P ( B = )P( ) P ( A P ) ( )
P(APA ) ( )
Bayes’ Theorem: PA() =
PB()A PBA()A + () ′ () ′
P(at least one success) = 1 – P(zero successes)
If the probabilities of obtaining the amounts a , a , a , … , or a are p , p , p , … , and p ,
1 2 3 k 1 2 3 k
where 1 + 2 + 3 + … + pk= 1, then the mathematical expectation is1 1= a 2 2 a p + k ka p
x n – x 2
Binomial : f(x) n x p (1 – p) for x = 0, 1, 2, …, or n,p, σ =np(1 − p)
Hypergeometric : PX x == f x abxnx − for x = 0, 1, 2, …, or n; x < a, (n – x) < b
ab n
−λ x
) () == f = e λ
Poisson: x! , µ =λ pen () x
Poisson Approx. to the Binomial P(X x )f x ( ) assupmes n 10, 100
x!
Z = x−µ
Normal distribution: σ
Sampling distribution for means – infinite population:
σ σ x − µ x − µ
µ σ = = =σ = = X X Z x
X X X X n n σ x σ
n
Sampling distribution for means – finite population:
σ X Nn −
EX() =µµX X EX () σ X
n N −1
Finding the sample size in estimation of means situations:
2
⎡⎤zα⋅σ
⎢⎥ 2 σ
Formula in text:n =⎢⎥ or use α
E 2 n
⎣⎦
Estimating µ, σ known, (1 − α)·100%, (large sample case , use s for σ if σ not known):
σ σ σ
xZ− α α≤ ≤+xµ µ α equivale=±tto:
2 2n n 2 n
Estimating µ, σ unknown, (1 − α)·100%, (small sample case):
s s s
+ ≤xt ≤ α ±α xt µ α equivalentxto:
n−1,2n−2 n n−, 2n
2 2
2 )1s− − )( 1n2
Estimating

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