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DISCRETE RANDOM VARIABLES Section 2 Choose from the following:

DISCRETE RANDOM VARIABLES Section 2 Choose from the following: Introduction: Car share scheme a success Example 4.3: A discrete random variable Example 4.4: Laura’s Milk Bill End presentation. Car share scheme a success. Car share scheme a success. Expectation.

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DISCRETE RANDOM VARIABLES Section 2 Choose from the following:

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  1. DISCRETE RANDOM VARIABLES Section 2 Choose from the following: Introduction: Car share scheme a success Example 4.3: A discrete random variable Example 4.4: Laura’s Milk Bill End presentation

  2. Car share scheme a success

  3. Car share scheme a success

  4. Expectation Multiply each r value by P(X = r) to form the r P(X = r) column

  5. Expectation Multiply each r value by P(X = r) to form the r P(X = r) column

  6. Expectation Multiply each r value by P(X = r) to form the r P(X = r) column

  7. Expectation Multiply each r value by P(X = r) to form the r P(X = r) column

  8. Expectation Multiply each r value by P(X = r) to form the r P(X = r) column

  9. Expectation Now find the total of the r P(X = r) column

  10. Expectation Expectation = E(X) orm E(X) = m = Sr P(X = r) = 1×0.35 + 2×0.375 + 3×0.205 + 4×0.065 + 5×0.005 = 0.35 + 0.75 + 0.615 + 0.26 + 0.025 = 2

  11. Variance Var(X) = s 2 = Sr 2 P(X = r) – m 2 = 12×0.35 + 22×0.375 + 32×0.205 + 42×0.065 + 52×0.005 – 22 = 0.35 + 1.5 + 1.845 + 1.04 + 0.125 – 4 = 4.86 – 4 = 0.86

  12. Example 4.3 The discrete random variable X has the distribution: • (i) Find E(X). • (ii) Find E(X2). • (iii) Find Var(X) using Var(X)= E(X2) – m2

  13. Example 4.3 : (i) ExpectationE(X) Expectation = E(X) orm E(X) = m = Sr P(X = r) = 0×0.2 + 1×0.3 + 2×0.4 + 3×0.1 = 0 + 0.3 + 0.8 + 0.3 = 1.4

  14. Example 4.3 : (ii) E(X 2) Expectation = E(X2) E(X2) = Sr 2 P(X = r) = 02×0.2 + 12×0.3 + 22×0.4 + 32×0.1 = 0 + 0.3 + 1.6 + 0.9 = 2.8

  15. Example 4.3 : Variance Var(X) = s 2 = Sr 2 P(X = r) – m 2 = 02×0.2 + 12×0.3 + 22×0.4 + 32×0.1 – 1.42 = 0 + 0.3 + 1.6 + 0.9 – 1.42 = 2.8 – 1.96 = 0.84

  16. Example 4.4 : Laura’s Milk Bill

  17. Example 4.4 : Laura’s Milk Bill Laura has one pint of milk on three days out of every four and none on the fourth day. A pint of milk costs 40p. Let X represent her weekly milk bill. (i) Find the probability distribution for her weekly milk bill. (ii) Find the mean (m) and standard deviation (s) of her weekly milk bill. (iii) Find (a) P(X > m+s) and (b) P(X < m−s).

  18. Example 4.4 : (i) Probability distribution Since Laura has milk delivered, it takes four weeks before the delivery pattern starts to repeat.

  19. Example 4.4 : (i) Mean μ or expectationE(X) Expectation = E(X) orm E(X) = m = Sr P(X = r) = 2.00 × 0.75 + 2.40 × 0.25 = 1.50 + 0.60 = 2.10

  20. Example 4.3 : (ii) Standard deviation σ Var(X) = s 2 = Sr 2 P(X = r) – m 2 = 22 × 0.75 + 2.42 × 0.25 – 2.12 = 3.00 + 1.44 – 2.12 = 4.44 – 4.41 = 0.03 Hence s = √0.03 = 0.17 (to 2 d.p.)

  21. Example 4.4 : (iii) Calculating probabilities The probability distribution for Laura’s weekly milk bill: • P(X > μ + σ) = P(X > 2.10 + 0.17) • = P(X > 2.27) • = 0.25 (b) P(X < μ − σ) = P(X < 2.10 − 0.17) = P(X < 1.93) = 0

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