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Discrete Random Variables 2 To understand and calculate with cumulative distribution functions To be able to calculate the mean or expected value of a discrete random variable To be able to calculate the variance of a discrete random variable

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Cumulative Distribution Function The probability that X is less than or equal to x is written as F(x) F(x) = P(X x) 2 coins are thrown. X is the number of heads Sample space HH, HT, TH, TT Cumulative distribution function x012 P(X=x)¼½¼ F(x)¼¾1

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Problem 1 F(x) = (x + k) x = 1,2,3 8 a)Find k b)Write down the distribution table for the cumulative distribution function c)Write down F(2.6) d)Find the probability distribution of X

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Problem 1 - Solution F(x) = (x + k) x = 1,2,3 8 a)Find k F(3) = 1 so 3 + k = 1 therefore 3+k = 8 and k=5 8 b) F(2) = = F(1) = = x123 F(x) 6/86/8 7/87/8 1

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Problem 1 - Solution F(x) = (x + k) x = 1,2,3 8 c) F(2.6) means P(X2.6) = F(2) = 7 / 8 d) x123 F(x) 6/86/8 1/81/8 1/81/8

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Mean or Expected value of a DRV NB p(x) is the same as P(X=x) Expected value of X = E(X) = ΣxP(X=x) = Σxp(x) Statistics experiment Collect data Frequency distribution Mean value Σfx Σx Theoretical approach Probability distribution Expected value Σxp(x)

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Problem 1 A survey of 100 houses is conducted about the number of TV sets a)Calculate the mean number of sets b)Find the probability distribution where X is the number of TV sets in a house picked at random c)Calculate the expected value of X No of sets0123 Frequency

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Problem 1 - Solution A survey of 100 houses is conducted about the number of TV sets a)Calculate the mean number of sets Mean value Σfx = = 110 = 1.1 Σx No of sets0123 Frequency x0123 p(x) b) c) Expected value of X = Σxp(x) Σxp(x) = = 1.1 Note that the mean value = E(X) = 1.1 E(X) is sometimes called the mean of X

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Problem 2 Given that E(X)=3 write 2 equations involving p and q and solve the find the value of p and q. x12345 p(x)0.1p0.3q0.2

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Problem 2 - Solution Given that E(X)=3 write 2 equations involving p and q and solve the find the value of p and q. x12345 p(x)0.1p0.3q p q = 1 p + q = 1 p + q = p q + 1 = 3 2p + 4q + 2 = 3 2p + 4q = 1

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Problem 2 - Solution Given that E(X)=3 write 2 equations involving p and q and solve the find the value of p and q. x12345 p(x)0.1p0.3q0.2 p + q = 0.4 and so p = q Substitute into 2p + 4q = 1 2(0.4 – q) + 4q = – 2q + 4q = q = 1 2q = 0.2 and q = 0.1 p = 0.3

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Problem 3 – Expected value of X² E(X²) = Σx²p(x) E(X n ) = Σx n p(x) A discrete random variable X has a probability distribution x1234 P(X=x) 12 / 25 6 / 25 4 / 25 3 / 25 a) Write down the probability distribution for X² b) Find E(X²)

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Problem 3 – Solution a) x1234 x²x²14916 P(X=x) 12 / 25 6 / 25 4 / 25 3 / 25 a) E(X²) = Σx²p(X=x²) = 12 / / / / 25 = 120 / 25 = 4.8

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Variance Var(X) = E(X²) – (E(X))² Example 2 four sided die numbered 1,2,3,4 are spun and their faces are added (X). a)Find the probability distribution of X b)Find E(M) c)Find Var(M) a) x p(x) 1 / 16 2 / 16 3 / 16 4 / 16 3 / 16 2 / 16 1 / 16

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Variance Var(X) = E(X²) – (E(X))² b) Find E(M) x p(x) 1 / 16 2 / 16 3 / 16 4 / 16 3 / 16 2 / 16 1 / 16 E(M) = Σxp(x) = 2 / / / / / / / 16 = 80 / 16 = 5 Var(X) = E(X²) – (E(X))² =( 4 / / / / / / / 16 )-25 = 440 / 16 – 25 = 2.5

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