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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.
b
E(X) =
 x. f ( x)dx
a
To calculate the variance, we need to find E(X2) since
Var (X) = E(X2) – [E(X)]2
b
2
x
 . f ( x)dx
E(X2) =
a
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(5X2 – 3X + 1)
a) Calculate the mean and variance of X.
f(x) = 6x(1-x) = 6x – 6x2
E(X) =
1
1
0
0
2
2
3
x
(
6
x

6
x
)
dx

(
6
x

6
x
)dx


1
 6x 6x 



4 0
 3
3
4
31
 21 
2
4
3
1

2
Var (X) = E(X2) – [E(X)]2
E(X2) =
1
1
0
0
2
2
3
4
x
(
6
x

6
x
)
dx

(
6
x

6
x
)dx


1
 6x
6x 



4
5

0
4
5
31 61


2
5
3
2
Var (X) =
3 1
1
  
10  2 
20
4
3

10
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 = 10 x 1 – 3 = 2
2
Var(Y) = Var(10X – 3) = 102 Var(X) = 100 x 1 = 5
20
(ii) E(Z) = E 6 – 2X = 6 – 2E(X) = 6 – 2 x 1 = 1
5 5
5 5
5 5 2
Var(Z) = Var 6 – 2X = 2
5 5
5
2x
Var (X) = 2
5
2x
1 = 1 .
20 125
c) Evaluate E(5X2 – 3X + 1)
E(5X2 – 3X + 1) = 5E(X2) – 3E(X) + 1 =
5x 3 - 3x1 +1= 1
10
2
Exercise 1.4
Mathematics Statistics Unit S2 - WJEC
Homework 11
Homework 12