Statistics

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regions (AUB)' AUB ven diagrams notations B. A'nBn C A B BO B ل B B A UB A union B everything in а от в от кони A Алв A intersection B Everthing in A and B A' Everything not in A BCA AD B A complement (A MB)¹ B A OO D Алв B¹ (AMB)UC A B is a subset of A A contains B с ANBI A (AUB) B B B с B J A conditional probability in venn diagrams ex ampe es: AMB P (B) ANB P(A) P(AUB) Mutually exclusive Events B P(AUB) PC A/B) P (BIA) A A = P(AUB) = P(A) + P(B) - P(AMB) 3 7 3/5 Yes no x (y) Z PROOF: P(A) + P(B) -(PNB) = P(x+y)+ P(y+z) - p(y) = x+y + y + z -y +2 x +y B 7 = P Using two way tables with probability Girl 4 PLANB) = P(A)XP (BIA)) (P(ANB) = P(A) P(B¹ | A)) Boy 3 5 4 P (AUB 8 P(A/B) = P(BIA) = = 12 % 19 probability laws and Definitions Dependent events: old O15 = mutually venn diagram independent events ↳ unettected PL BIA) = P(B) P(ANB) = P(A) P(B) LD ³2 P(AIB) = ²/1/₂2 1 PC BIA) = ²/3 2 proot that 2 events are independent exclusive cannot occur at the time P(ANB)=0 :: PC AUB) = P(A) + P(B) A W General P(AUB) = P(A) +P(B)-P(ANB) mutually exclusive events two events cannot happen at the same time P(ANB) = O P(AUB) = P(A) +P(B) A B FOO probability Distribution table let X = x . 1 P(X₂)X₁ 4 properties: 2 3 Binomial Distribution X₂ X3 fixed number of trials. · two outcomes, success/failure →CD Binomial trious are independent Probability of success (p) remains constant cumulative Binomial mode 7 A add up to I example P(z ≤6) ▾ CD > P(x ≤7) = P(≤6) distribution P(x≤ x) for X²B(U,P) probability function: P(X+x) = { K (x+2)₂ X=1₁2 1 x = 3,4 P ( X = 1) + P(x =2) +P(x= 3) + P(x=4) = 1 3K 4n e.g P(X=2) Example: P(x² 7) = 1- P(x≤6) P(26) = 1 - P(Z = 6) P(3525) P(x=s) - P(x²2) Binomial Distribution - cumulative Probability tables + + 3к +4к 2 3 % % at least 2 p=0.08 n = IS 3/14 4. 14 к 90% chance of tottees - p= 0.9 bags - и - 20 20 xnB( 20,0.9) PD (x = 20) - 0.122 6(x>18) PD (x = 19) 14 XnB CIS, (x-2) - 0.392 0.08) = 1 = A = /14 1- (x=1) Binomial Distribution - Mean, variance хив (и,р), up mean variance = иря, where ✓ failure Normal Distribution. M x ~ N(№, o) Normal CD (a<x< b) Example: success M 249 250 251.S Normal CD: for xvN (N₁0²) The volume of coffee dispensed from a maschine x mi, is modelled by x^N ( 250, 1.2²) a.) P(249 x 251.5) b.) (x>247.8) B lower : 249 upper : 251.5 0 : 1.2 M : 250 - кр (1,0²) Z = X-N O Inverse Normal Distribution 3 INVERSE NORMAL x^N(N₁0²) P(x<x) X Standart Normal Distribution z ~ N(0,1). 2 BNN (0.1) ↑ > z 2478 250 60 or 999 for upper bound Inverse Area ↓ find z ↓ solve simultanous equations 251) C.) P(x c lower bound -999 2 = x-μ 0 Correlations scatter diagrams r = 0 X Pearson's product moment correlation coefficient r y 1 2 3 S 1 3 6 8 O → no correlation O [= -1 0 x O linear (= -1 positive sa perfect negative whear correlation sa calculater: Statistics mode Menu - 6 - 2 negative a =-0-37 b = 1.77 r. 0.973 y = -0.4 +1.8% T= 1 [=1 → y = a + bx +4 round to I dp. → -close to one no correlation perfect positive Whear correlation sa line of best fit Normal approx to the Binomial Distribution symmetrical 2 ~N[ up, > пря) continvety correction P(Z <16) → • (x < 15.5) no = P(x = 29) → P ( x < 29.5) H PF P(x>28) → P(x> 28.5) A