Advanced Level Mathematics: Statistics 1 by Steve Dobbs PDF

By Steve Dobbs

ISBN-10: 052153013X

ISBN-13: 9780521530132

Written to check the contents of the Cambridge syllabus. information 1 corresponds to unit S1. It covers illustration of information, variations and mixtures, chance, discrete random variables and the traditional distribution.

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Extra info for Advanced Level Mathematics: Statistics 1

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See [dF] for a thorough discussion of this problem. In the following we introduce some of the most common discrete distributions. 2 Bernoulli Scheme A simple and useful model from which some discrete distributions can be derived is the Bernoulli scheme. It can be thought of as a potentially infinite sequence of trials, each of them with two possible outcomes called success and failure. Each trial is performed in the same known conditions and we assume that there is no influence between different trials.

Indeed, for 0 < H < N for every pair i, j with i = j, we have: cov(E i , E j ) = P(E i E j ) − P(E i )P(E j ) = as P(E i E j ) = H H−N <0 N2 N − 1 N −2 N −2 H (H − 1)Dn−2 H (H − 1) Dn−2 H (H − 1) . 12); possible cases are sequences with no repetition of length n from a set of N elements, whereas in counting favorable cases we first select two different white balls for the ith and the jth drawings and then the remaining n − 2 balls from a set of N − 2 elements. The variance of X is then obtained by means of the formula for the variance of the sum of n random numbers: n σ 2 (X ) = σ 2 (E i ) + i=1 =n cov(E i , E j ) i, j i=j H H H−N N −n H H H (1 − ) + n(n − 1) 2 =n (1 − ) , N N N N −1 N −1 N N where n(n − 1) is the number of ordered pairs i, j, with i = j, which can be chosen out of {1, .

M and j = 1, . . , n. Given ψ : R2 −→ R, the expectation of the random number Z = ψ(X, Y ) can be obtained from the joint distribution of X, Y : m n ψ(xi , y j ) p(xi , y j ) . 3) i=1 j=1 The proof is completely analogous to that one in the case of a single random number. For example, we can compute P(X Y ): m n P(X Y ) = xi y j p(X = xi , Y = y j ). i=1 j=1 If X and Y are stochastically independent and φ1 , φ2 are two real functions φi : R −→ R with i = 1, 2, we have that P(φ1 (X )φ2 (Y )) = P(φ1 (X ))P(φ2 (Y )).

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Advanced Level Mathematics: Statistics 1 by Steve Dobbs


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