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Discrete Random Variables 2

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. Cumulative Distribution Function.

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Discrete Random Variables 2

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  1. 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

  2. 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

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

  4. Problem 1 - Solution F(x) = (x + k) x = 1,2,3 8 • Find k • F(3) = 1 so 3 + k = 1 therefore 3+k = 8 and k=5 • 8 b) F(2) = 2 + 5 = 7 8 8 F(1) = 1 + 5 = 6 8 8

  5. Problem 1 - Solution F(x) = (x + k) x = 1,2,3 8 c) F(2.6) means P(X≤2.6) = F(2) = 7/8 d)

  6. 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)

  7. Problem 1 A survey of 100 houses is conducted about the number of TV sets • Calculate the mean number of sets • Find the probability distribution where X is the number of TV sets in a house picked at random • Calculate the expected value of X

  8. Problem 1 - Solution A survey of 100 houses is conducted about the number of TV sets • Calculate the mean number of sets • Mean value Σfx = 0 + 75 + 20 + 15 = 110 = 1.1 • Σx 100 100 b) c) Expected value of X = Σxp(x) Σxp(x) = 0 + 0.75 + 0.20 + 0.15 = 1.1 Note that the mean value = E(X) = 1.1 E(X) is sometimes called the mean of X

  9. Problem 2 Given that E(X)=3 write 2 equations involving p and q and solve the find the value of p and q.

  10. 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. 0.1 + p + 0.3 + q + 0.2 = 1 p + q + 0.6 = 1 p + q = 0.4 0.1 + 2p + 0.9 + 4q + 1 = 3 2p + 4q + 2 = 3 2p + 4q = 1

  11. 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. p + q = 0.4 and so p = 0.4 - q Substitute into 2p + 4q = 1 2(0.4 – q) + 4q = 1 0.8 – 2q + 4q = 1 0.8 + 2q = 1 2q = 0.2 and q = 0.1 p = 0.3

  12. Problem 3 – Expected value of X² E(X²) = Σx²p(x) E(Xn) = Σxnp(x) A discrete random variable X has a probability distribution Write down the probability distribution for X² Find E(X²)

  13. Problem 3 – Solution a) E(X²) = Σx²p(X=x²) = 12/25 +24/25 +36/25 +48/25 = 120/25 = 4.8

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

  15. Var(X) = E(X²) – (E(X))² Variance b) Find E(M) E(M) = Σxp(x) =2/16 +6/16 +12/16 +20/16 +18/16 +14/16 +8/16 = 80/16 = 5 Var(X) = E(X²) – (E(X))² =(4/16 +18/16 +48/16 +100/16 +108/16 +98/16 +64/16)-25 = 440/16 – 25 = 2.5

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