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E(X 2 ) = Var (X) = E(X 2 ) – [E(X)] 2 E(X) = The Mean and Variance of a Continuous Random Variable In order to calculate the mean or expected value of.

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Presentation on theme: "E(X 2 ) = Var (X) = E(X 2 ) – [E(X)] 2 E(X) = The Mean and Variance of a Continuous Random Variable In order to calculate the mean or expected value of."— Presentation transcript:

1 E(X 2 ) = Var (X) = E(X 2 ) – [E(X)] 2 E(X) = The Mean and Variance of a Continuous Random Variable In order to calculate the mean or expected value of a continuous random variable, we must multiply the probability density function f(x) with x before we integrate within the limits. To calculate the variance, we need to find E(X 2 ) since

2 Example The continuous random variable X is distributed with probability density function f(x) where f(x) = 6x(1-x) is 0 ≤ x ≤ 1 a) Calculate the mean and variance of X. b) Deduce the mean and variance of (i)Y = 10X – 3 (ii)Z = 2(3 – X) 5 c)Evaluate E(5X 2 – 3X + 1)

3 a)Calculate the mean and variance of X. f(x) = 6x(1-x) = 6x – 6x 2 E(X) =

4 Var (X) = E(X 2 ) – [E(X)] 2 E(X 2 ) = Var (X) =

5 b)Deduce the mean and variance of (i) Y = 10X – 3 (ii) Z = 2(3 – X) 5 (i) E(Y) = E(10X – 3) =10E(X) – 3 =2 6 – 2E(X) = 5 (ii) E(Z) = E 6 – 2X = – 2 x 1 = x 1 – 3 = 2 1 Var(Z) = Var 6 – 2X = x Var (X) = Var(Y) = Var(10X – 3) =10 2 Var(X) =5100 x 1 = x 1 = 5 20

6 c) Evaluate E(5X 2 – 3X + 1) E(5X 2 – 3X + 1) = 5E(X 2 ) – 3E(X) + 1 = 5 x x = Exercise 1.4 Mathematics Statistics Unit S2 - WJEC Homework 11 Homework 12


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