By Brink D.

After interpreting the idea publication approximately statistics it's time to try your wisdom to ensure that you're organized in your exam.This unfastened workout publication follows an analogous constitution because the idea booklet approximately records. solution questions on for instance likelihood thought, random variables, anticipated worth and the legislation of huge numbers. the entire workouts are via their suggestions.

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**Additional info for Statistics - Exercises**

**Sample text**

The distribution function F is the continuous function given by ⎧ ⎪ ⎪ ⎨0 for x < 0 F (x) = x for 0 ≤ x ≤ 1 ⎪ ⎪ ⎩1 for x > 1 Solution of Problem 16. X takes values in the set {2, 3, 4, . . , 12}. com 34 Statistics – Exercises Solutions Y takes values in the set {−5, −4, . . , 4, 5}. The point probabilities are ⎧ ⎪ 1/36 for k = −5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2/36 for k = −4 ⎪ ⎪ ⎪ ⎪ ⎪ 3/36 for k = −3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 4/36 for k = −2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨5/36 for k = −1 P (Y = k) = 6/36 for k = 0 ⎪ ⎪ ⎪ ⎪ 5/36 for k = 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 4/36 for k = 2 ⎪ ⎪ ⎪ ⎪ ⎪ 3/36 for k = 3 ⎪ ⎪ ⎪ ⎪ ⎪ 2/36 for k = 4 ⎪ ⎪ ⎪ ⎪ ⎩1/36 for k = 5 If we know, say, that X takes the value 12, then we can conclude that Y takes the value 0 (since both dice in that case show 6).

The product X · Y thus takes the value (r + s)(r − s) = r2 − s2 . e. 6 6 (r2 − s2 ) = 0. com 36 Statistics – Exercises Solutions We therefore ﬁnd the covariance Cov(X, Y ) = E(X · Y ) − E(X) · E(Y ) = 0 − 7 · 0 = 0. 6). We have shown in 16 that X and Y are not independent. This problem therefore shows that we cannot conclude conversely that the variables are independent when the covariance is 0. Solution of Problem 20. Chebyshev’s Inequality says 1 . n2 In words this means that the probability that X takes values more than n standard deviations away from its expected value is small when n is large.

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