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Px vs. W P X ð õ r Z l ~ ï Ç X } s ^ u } µ v Á Z î D ð ( o Z Á X h ð rD ð Z Z } o v Æ } Á v s ^ } u } µ v Á Z } o Ç u } v l Á o o l ~ P v } Á X ò. } X U î ï X ì ð X î ì î ì, ^ v l r Z P P ^ } X U î ò X ì ð X î ì î ìd Z , P v r > À ^ } X U í î X ì õ X î ì î í í õ W ï ì h Z í ô W ì ì h Z. F(x) = P(X x) = Zx 1 f(x)dx Therefore f(x) = F(x)0 Compute probabilities using cdf P(a) b Find the cdf c Use the cdf to compute P(X>) d Find the 75 th percentile of the.
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^ x x^ x' x& x x/ x ï ì ó ì õ ð ð ó ô ð ò/ ~ p ^ x x Z v s o W } P v v o Z D P r o W ^K^ ð î U ó ó ð î U ó ò ô ì ì ì ô ì ì í ô ð ñ ï ì ñ ô ñ ì ñ ô ó ô í. ^ x x^ x' x& x x/ x ï ì ó ì õ ð ð ó ô ð ò/ ~ p ^ x x Z v s o W } P v v o Z D P r o /W ^K^ ¨ ò î U õ ñ ¨ ò î U õ ñ ì ì ì ô ì ì í ô ð ò ï ì ñ ô ñ ì ñ ô ó ô í. PX ≤ Y First, let’s consider the denominator PX ≤ Y = X z≥1 PX = z ∩z ≤ Y = X z PX = zPz ≤ Y = X z (1−p)z−1p(1−q)z−1 = X z (1−p)(1−q)z−1p = p X z (1−p−q pq)z−1 = p pq −pq The last step above is again by the identity in Eqn 1 Now we can compute the whole equation EXX ≤ Y = pq −pq p.
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î P X ð î ñ X } v o l Ç } v } Ç U Z Æ ( } í ñ ì r î ì ì Á v í ó ó ñ r } } v s r î ï ï í ó ò ì í ô ì ì W } ( t ( } v í ó ó ó X o Ç } Z v E Á o v r î î õ õ í ó ò ó. (1 1 ) f(x ) L b P (x ) n = 0 n n w h e re { p (x ) } is a p re s c rib e d se q u e n c e of p o ly n o m ia ls , a n d w h e re th e c o e ffic ie n ts b a re n u m b e rs re la te d to f In p a rtic u la r, th e in n u m e ra b le in v e stig a tio n s o n e x p a n sio n s'. P(k) = P(X = k) given by p(1) = p, p(0) = 1−p, p(k) = 0, otherwise Thus X only takes on the values 1 (success) or 0 (failure) A simple computation yields E(X) = p Var(X) = p(1−p) M(s) = pes 1−p Bernoulli rvs arise naturally as the indicator function, X = I{A}, of an event A, where I{A} def= ˆ 1, if the event A occurs;.
O í ô v > Z s v s P D l > µ v l V > } v î ó WZ /^ d D W í í W d u í v ' Ç U ^ Ç v Ç U µ v U < U Z v Ç o î ñ W d u ð v U ' o v v U µ o U > Z U D P ^ , h> &KZ WZ/> í í r í ó. O s v } v s D d } ñ & } µ ' } µ W P ò W } Ç W P ó r í ì µ } u ^ À W } Ç W P í í r í ò. 3 (8 points) Let T 2L(V);V a vector space over R;p(x) a polynomial in R, and a2R Show that the following statement is false ais an eigenvalue of p(T) if and only if a= p( ) for some eigenvalue of T Rather than simply providing a counterexample, prove as strong a statement as possible Solution.
Title Microsoft Word 02_MODULO KIT AMIANTOdocx Author rober Created Date 11/4/19 1019 AM. Ð Ñ Ò Ó Ð Ô Ò Ñ Õ Ö × Ó Ø Ù Ú Û Ü Ý Þ ß à á m q t É n u p o â l Ä Ç Ë ã ä å Ú Û Þ æ à Ý ß ç è q m p v Ì é Ä Ç u n Í È É s t o Î Ê Ð Ñ Ò Ó Ð Ô Ò Ñ Õ ê ë ì Ö × Ñ Ò Ó Ð Ô Ø Ù. ) be the time that the bus arrives We want to compute P(X< Y) = P(X Y.
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2 So we can can write p(x) as a linear combination of p 0;p 1;p 2 and p 3Thus p 0;p 1;p 2 and p 3 span P 3(F)Thus, they form a basis for P 3(F)Therefore, there exists a basis of P 3(F) with no polynomial of degree 2 Exercise 2 Prove or give a counterexample If v.
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