OBJECTIVES a Understand the concept of mathematical expectation
OBJECTIVES a) Understand the concept of mathematical expectation b) Use: i) E(X) = ii) E[g(X)] = iii) Var(X) = iv) Var(X) = E(X)2 – [E(X)]2 v) Var(a. X b) = a 2 Var(X)
Expectation of X 1. The expected value of a random variable X is the mean of X and denoted by E(X). Note: is the mean of a sample is the mean of a population
If X is a discrete random variable with probability distribution function P(X=x) then, the expected value is the sum of the probability of each possible outcome of the experiment multiplied by the outcome value.
If g(x) is a function of a discrete random variable X then ,
Variance of A Discrete Random Variable denoted by Var(X) or or Var(X) = E(X 2) – [E(X)]2
Properties a and b are constants
Example 1 The probability distribution of a discrete random variable X is given as follows, x 1 2 3 P(X=x) Calculate a) E(X) c) Var(X) b) E(X 2) d) Standard deviation of X
c) Var(X) = E(X 2) – [E(X)]2 d) Standard deviation of X =
Example 2 A discrete random variable X has the following probability distribution: x P(X=x) 1 2 t 0. 1 0. 2 0. 7 a)Find the value of t if E(X) = 4. b)Find i) E(3 X+2) ii) Var (2 X-1) iii) Var ( 1 -3 X)
x Solution P(X=x) a) E(X) = 4 1(0. 1) + 2(0. 2) + t(0. 7) = 4 0. 1 + 0. 4 +0. 7 t = 4 0. 7 t = 3. 5 t=5 b)i)E(3 X+2) = 3 E(X)+2 = 12+2 = 14 1 2 t 0. 1 0. 2 0. 7
x 1 P(X=x) 2 5 0. 1 0. 2 0. 7 ii) E(X 2) = 1(0. 1)+4(0. 2)+25(0. 7) = 0. 1 + 0. 8 +17. 5 = 18. 4 Var (X) = E(X 2)-(E(X))2 = 18. 4 -16 =2. 4
Var (2 X-1) = 4 Var(X) =4(2. 4) = 9. 6 iii) Var (1 -3 X)= Var (X) =9(2. 4) =21. 6
Example 3 The probability distribution of a discrete random variable X is given as follows, x 0 1 2 P(X=x) a b c If E(X) = 1. 1 and Var (X) = 0. 54, find the constants a, b and c
x Solution P(X=x) a + b +c =1……(i) E(X) = 1. 1 b+2 c =1. 1…. (ii) Var(X)=0. 54 E(X 2)-(E(X))2 = 0. 54 b+4 c – (1. 1) = 0. 54 b+4 c =1. 75…. (iii) 0 1 2 a b c (iii)-(ii) 2 c = 0. 65 c = 0. 325 b = 1. 1 - 2(0. 325) = 0. 45 From (i) a =1 -0. 45 -0. 325 = 0. 225
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