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STAT 263 (15)
Final

# finalformulaesheet1_7.pdf

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School
Department
Statistics
Course
STAT 263
Professor
Charles Arthur Molson
Semester
Winter

Description
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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