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K ;> $ "DNA ð a'79 ù G DNA ð a ;Z b a f X(x) dx For f X(x) to be a proper distribution, it must satisfy the following two conditions 1The PDF f X(x) is positivevalued;The mean µ and that attains its maximum value of √1 2πσ ' 0399 σ at x = µ as represented in Figure 11 for µ = 2 and σ 2= 15 The Gaussian pdf N(µ,σ2)is completely characterized by the two parameters µ and σ2, the first and second order moments, respectively, obtainable from the pdf as µ = EX = Z ∞ −∞ xf(x)dx, (12

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Of the book, we can find numbers a,b such that, for any Q ∈ (0,1), there is P(a ≤ z ≤ b)=Q The interval a,b is called a Q × 100% confidence interval for z We can minimise the length of the interval by disposing it symmetrically about the expected value E(z) = 0, since z ∼ N(0,1) is symmetrically distributed about its mean of zeroLet X be the normal random variable with parameters µ = 71 and σ2 = 625, and Z be the standard normal random variable Note that 1 inch equals to 12 feets, σ = √ 625 = 25 Then the first problem is to compute P(X > 6·122) P(X > 74) = P X − 71 25 > 74− 71 25 For the second problem, we need to compute P(X > 6·125X > 6·12) = P(X > 77X > 72) Hence, P(X > 77X >@ A z Ÿ • ˙ \ T † G ƒ š ‹ µ s, H H ˇ Ù † 8 Ñ 8 Ñ 5 Ò 4 š € Ö × Š ⁄ I 5 Ö ˙ \ 6 > ˇ ł c ˇ 0 Ý Þ 6 > m v ß à 7 > ˇ ˘ µ ˙ \ 6 Ö Ù 7 ˇ p ¢ 7 J k J — › K Ý) # ˇ ‰ E » ‹ Š > ˙ \ 6 ¿ ˘ µ Ô Õ 6 É 8 6 " l š ˝ Š < ‹ „ Ö Y L A ^ no2 pqr˜struvwxyR z{ }~ • † ‡ # — n o k

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Y Z E P T G M k h dad c m µ n p f a z y Exponent (basis 10) of decimals E n = 10 n Factor European name American name 10 3 thousand thousand 10 6 million million 10 9 milliard billion • • 10 12 billion • • trillion • • • 10 15 billiard quadrillion • • • • 10 18 trillion • • • quintillion • • • • • 10 21 trilliard sextillion 10 24Be used to construct a random variable Y = g(W) such that Y is uniformly distributed on {0,1,2} 3 Let W have the density function f given by f(w) = 2/w3 for w > 1 and f(w) = 0 for w ≤ 1 Set Y = α βW, where β > 0 In terms of α and β, determine (a) the distribution function of Y;= i E(X i) 3 We often write = E(X) 2 4 ˙2 = Var(X) = E((X )2) is the Variance 5 Var(X) = E(X2) 2 6

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The μopioid receptors (MOR) are a class of opioid receptors with a high affinity for enkephalins and betaendorphin, but a low affinity for dynorphinsThey are also referred to as μ(mu)opioid peptide (MOP) receptorsThe prototypical μopioid receptor agonist is morphine, the primary psychoactive alkaloid in opiumIt is an inhibitory Gprotein coupled receptor that activates the G i alphaOPTION B INSIDE THE FRAME INSTALLATION Fully Inside the Master Frame TOP VIEW INSTALLATION TYPE OPTION A Inside the Frame OPTION B Fully Inside the Frame OPTION C Outside the Frame RECORD MEASUREMENTS THRESHOLD TRANSITION KITS When ordering your screen you can also add a Threshold Transition Kit (see the table below)Each of the kits• Expectation of the sum of a random number of random variables If X = PN i=1 Xi, N is a random variable independent of Xi'sXi's have common mean µThen EX = ENµ • Example Suppose that the expected number of acci

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Y = a e^(b x) where a and b are constants The curve that we use to fit data sets is in this form so it is important to understand what happens when a and b are changed Recall that any number or variable when raised to the 0 power is 1 In this case if b or x is 0 then, e^0 = 1 So at the yintercept or x = 0, the function becomes y = a * 1 or y = a Therefore, the constant a is the yX = E(X), µ Y = E(Y) 1 cov(X,Y) will be positive if large values of X tend to occur with large values of Y, and small values of X tend to occur with small values of Y For example, if X is height and Y is weight of a randomly selected person, we would expect cov(X,Y) to be positive 50 2 cov(X,Y) will be negative if large values of X tend to occur with small values of Y, and small valuesX2 y2 z2, (e) f(x,y) = 4y (x2 1), (f) f(x,y,z) = sin(x)ey ln(z) Section 3 Directional Derivatives 7 3 Directional Derivatives To interpret the gradient of a scalar field ∇f(x,y,z) = ∂f ∂x i ∂f ∂y j ∂f ∂z k, note that its component in the i direction is the partial derivative of f with respect to x This is the rate of change of f in the x direction since y and z are

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More generally, Eg(X)h(Y) = Eg(X)Eh(Y) holds for any function g and h That is, the independence of two random variables implies that both the covariance and correlation are zero But, the converse is not true Interestingly, it turns out that this result helps us prove a more general result, which is that the functions of two independent random variables are also independentR y , ~ a \ d y r z u O r b Q V } ± ¶ z h i y ° ç y ² T ) ß & ± µ Z í ­ µ å Á Õ 7 b q y r µ O j ² T ) ß & y ² è r d } l y = b W O ^ q ` T b b j } k b O Ö y ¦ r \ j * d u = b y ¦ r è o V n j d ^ s z l n s Î W V V ^ s r d } r h yTitle Microsoft Word AI_questionnaire_ finaldocx Author 5003 Created Date PM

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Z ∞ −∞ G(x −y′,t)f(y′)dy′ =u(x,t) In the last line, we used that G is an even function of its first arguement Smooth even functions have zero slope at x =0, ie, ux(0,t)=0 So we solve our semiinfinite domain problem by extending the initial data to −∞Sa> E a pgRNA > y pÄ "cDNA > ã â 8'pgRNA ÃÞ ;(d) the mean of Y;

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ß î È b Y w í 2 b M g!K Z C T I 8 0ò(ý c É ß î È b v 1 Â i g!K Z C T I 8 í c b5 c 8 _ c @ g } ^ 8&ì Ø K Z C T I 8 b É ß î È > g ¡ ­ § Ü b Y w í í Ý M \ G M ö @ 6 b K ^ 8 C T I 8 d V0° b Q#Ý ^ c c /æ*( / > g É ß î È _ P M * ö b s 0É M v b 6 ~ Y w í í ,æ Ý 0É M v b c 6 ~ r O Y w íThus we can relabel the indices according to x′ = y, y′ = zm and z′ = x Thus the eigenstates of y′ are just the eigenstates of z, in terms of the primed indices we have Sx′ = Sy = 0 −i i 0 (15) Sy′ = Sz = 1 0 0 −1 (16) Sz′ = Sx = 0 1 1 0 (17) Note that the eigenstates of Sy are only defined up to a phasefactor, differentZ b a jf(t) g(t)jdt where the integral is in the sense of Riemann (Ca,b,d 1) is a metric space Note that (Ca,b,d 1) and (Ca,b,d ¥) are two distinct metric spaces Definition 22 A sequence in a metric space (X,d)is a mapping from the natural numbers N into X, usually denoted by (xn) (xn) is said to be a convergent sequence to x 2X if 8e > 0, 9N = N(e) 2N j n N =)d(xn, x) e

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H i a c d Y j k a a e ~ i w Y d u f ~ ` g f a g a e ~ i w Y d u f g \ g h l f c k l j k Y f g a k a a e ~ i w Y d u f ~ c d ~ r ~ f Y k a j f l k a ~ h g ~ f l c f g h c l Title INBTESTUApdf Author kkasprzak Created Date PMThat is, √ n(X¯ n −µ)/σ has a limiting standard normal distribution The proof is almost identical to that of Theorem 5514, except that characteristic functions are used instead of mgfs Example (Normal approximation to the negative binomial) Suppose X1,, are a random sample from a negative binomial(r,p) distribution Recall that EX = r(1N ¯ Ñ Ý ¬ O N C I J ² ¹ & J O N ¸ È O t N î Í S O * # % " % J ' B ½ Ó Ú æ C ~ ¶ y k 4 k * Ý J ¼ s T e ¥ } Z ¥ k a Ä ­

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(e) the variance of Y 4 LetLet g(x) be such a function Then E(y g(X)) 2 is minimized when g(X) = EYjX Lecture 26 Examples I Toss 100 coins What's the conditional expectation of the number of heads given the number of heads among the rst fty tosses?40/36 1 5/36 9 4/36 36/36 4 16/36 10 3/36 30/36 9 27/36 11 2/36 22/36 16 32/36 12 1/36 12/36 25 25/36 eg E(X Y Z) = E(X) E(Y) E(Z)) 9 If X and Y are independent, E(XY) = E(X)E(Y) (This rule extends as you would expect it to for more than 2 random variables, eg E(XYZ)=E(X)E(Y)E(Z)) 10 COV(X,Y) = E(X E(X)) * (Y E(Y) = E(XY) E(X)E(Y) Question

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The mean or expected value of g(X) is E(g(X)) = Z g(x)dF(x) = Z g(x)dP(x) = (R 1 1 g(x)p(x)dx if Xis continuous P j g(x j)p(x j) if Xis discrete Recall that 1 Linearity of Expectations E P k j=1 c jg j(X) = P k j=1 c jE(g j(X)) 2 If X 1;;X n are independent then E Yn i=1 X i!

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