Essay About Acb P And Symmetric Then P

Discrete Variables (bar Graph)
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Discrete Variables (bar graph)[pic 1]Md – middle value (position = )[pic 2]Continuous Variables (histogram)Each class contains its lower limit but not its upper Area (f or f* ) = Width (X) . Height (density)P ( X = a) = 0 = f*1X1 +…+f*n(Xn)[pic 3]Md and percentiles follow:From F* find class which contains datahow much has been covered up till start of class (a)how much to be covered to reach the mdA= wh i.e. (Md – a)h = ADensity curvesBell – symmetricRight tail Dist. – majority lower values  > md[pic 4]Left tail Dist. – majority higher values  < md[pic 5]Range = max – minInterquartile range = C75 – C25 = Q3 – Q1V(x) =  -  **note =  = f*( i.e. the X value gets squared![pic 6][pic 7][pic 8][pic 9]S(x) = [pic 10]Standard Score = Z = [pic 11]Normal Distribution  [pic 12] -  or md – bell’s center[pic 13][pic 14] – S(x) – width of bell[pic 15]standard normal distribution (z – table) [pic 16]P(z(a)[pic 17]P(z>a) = 1 – (a)[pic 18](-a) = 1-(a)[pic 19][pic 20]P(a(b) – (a)[pic 21][pic 22](a) = 0.8        (or any number)[pic 23](0.84) = 0.7995 (one below)[pic 24](0.85) = 0.8023 (one above)[pic 25][pic 26](a) = 0.33 (any number below 0.5)[pic 27](a) = 1 – 0.33[pic 28](-a) = 0.67[pic 29]ProbabilityAUB – union – all values within A and BAB – intersection – only values included in both[pic 30]AB =  – disjoint or mutually exclusive events [pic 31][pic 32]  repeat N times count how many times A happens n(A)[pic 33]notes:1) 0   3) P(   4) P() = 1- P(A)   5) if AcB P(A)[pic 34][pic 35][pic 36]For any Event!!:P(AUB) = P(A) + P(B) – P(AB)[pic 37]If  is symmetric then P(w) =  and P(A) = [pic 38][pic 39][pic 40]P(A/B) =   and P(AB) = P(A) X P(B/A) = P(B) X P(A/B)[pic 41][pic 42]Independent EventsP(E/F) = P(E)P(F/E) = P(F)Theorem if E & F are Independent then P(EF) = P(E) X P(F)[pic 43]3 events – P(ABC) = P(A) X P(B) X P(C)[pic 44][pic 45]If A & B are disjoint i.e. AB =  then P(A/B) = 0 and not P(A) thus they are necessarily dependent.[pic 46][pic 47]RV’sE(x) = x1p1+…+xnpnif X = a and P(a) = 1 then E(x) = ay = ax then E(y) = E(ax) = aE(X)y=x+b then E(y) = E(x) + bE(w) = E(X+Y) = E(X) + E(Y)V(x) = E(X2) – (E(X))2V(a) = 0y = ax then V(y) = V(ax) = a2v(X)y=x+b then V(y) = V(x) V(w) = V(X+Y) = V(X) + V(Y) +2cov(X,y)Binomial  – finds the probability that K successes will occur in n number of attempts [pic 48]P(x=k) = [pic 49]And  = [pic 50][pic 51]E(x) = n.pV(x) = n.p.q where q = (1-p)At least KP(XK) = 1- P(X[pic 52]P( at least K) = 1 – P ( not k)For example P(X2) = 1- P(X=0) – P(X=1)[pic 53]P(X1) = 1- P(X<1) = 1 – P(X=0) (if its 2 people square it etc.) [pic 54]If  P(X=0) = (1-P)^n        Geometric  - finds the probability that a success will occur for the 1st time on the nth attempt.[pic 55]P(x=k) = [pic 56]P(x>K) = [pic 57]P(Xk)(x=L/x>K) = P(X= L-K)E(X) = [pic 58]V(X) =  if p = 0 then V(X) = [pic 59][pic 60]CorrelationCov(X;Y) = E(XY) – E(X).E(Y)[pic 61]notes:if X= ax +b and Y= cy +d then cov(XY) = acCov(XY) but [pic 62]-1[pic 63]if X and Y are dependent they are not necessarily correlated they may be dependent in a non linear way but if correlated then they are dependent  iff y = ax + b[pic 64][pic 65]if Cov(XY) >0 then V(X+Y) incif Cov(XY) <0 then V(X+Y) dec