BITI2233 Statistics and Probability

Chapter 4 :Continuous Random Variables

Continuous Probability Distributions

A continuous random variable has infinite many values, and those values
can be associated with measurements on a continuous scale.

Theorem :
The function f(x) is a probability density function for the continuous random
variable X, defined over the set of real number R, if

1. 0<f(x)< 1, for all x  R.


2.  f (x) dx = 1,

b
3. P(a < X < b) =  f ( x) dx = F(b) – F(a).
a

Theorem :
The cumulative distribution function, F(x) of a continuous random variable
X with density function f(x) is defined by
x
F(x) = P(X < x) =  f (t )

dt

d
f(x) = F ( x)  F ' ( x)
dx

 Example 1.10 :
Suppose that the error in reaction temperature, in C, for a controlled
laboratory experiment is a continuous random variable X having the probability
density function
x2 
  1  x  2
f ( x)   3 

0 otherwise 

(a) Verify whether f(x) is a pdf

(b) Find P(0 < X < 1) using f(x)
(c) Find P(0 < X < 1) using F(x)

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BITI2233 Statistics and Probability

Solution :
2
2
x2
2
 x3  8  1
(a)  f ( x)dx   dx         1
1 1
3  9  1 9  9 
1
1 1
x2  x3 
dx      0 
1 1
(b) P(0 < X < 1) = 
0
f ( x)dx  
0
3  9 0 9 9

x
x
x2  x3 
x
x 3  1  x 3 1
(c) F ( x)   f ( x)dx   dx         
1 1
3  9  1 9  9  9

2 1 1
P(0 < X < 1) = F(1) – F(0) =    
9 9 9

{ F(x) = 0, x < -1
F(x) = (x3 + 1)/9 -1 < x < 2
F(x) = 1 x > 2. }

Definition of Mean

Let X be a continuous variable with probability distribution f(x). The mean or

expected value of X is

  E[ X ]   x  f ( x)dx


Definition of Variance

Let X be a random variable with probability distribution function f(x) and mean  . The
variance of X is


 2   ( x   ) 2 f ( x)dx


 E ( X 2 )  ( E ( X )) 2
 E( X 2 )   2
 
where E ( X )   x  f ( x)dx E ( X 2 )   x 2  f ( x)dx
 

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BITI2233 Statistics and Probability

Definition of Mean of the random variable g(X)

Let X be a random variable with probability distribution function f(x). The mean or
expected value of the random variable g(X) is


 g ( X )  E ( g ( X ))   g ( x) f ( x)d x


Example 1.11
The total number of hours, measured in units of 100 hours, that a family runs a vacuum
cleaner over a period of one year is a continuous random variable X that has the density
function
 x, 0  x 1

f ( x )  2  x , 1  x  2
0,
 elsewhere
Find the probability that over a period of one year, a family runs their vacuum cleaner
(a) less than 120 hours.
(b) between 50 and 100 hours.
(c) Find the cumulative distribution function F(x).
(d) Find the  and  for the probability density.

Solution
1 1.2 1
(a) P( X  1.2)   xdx   (2  x)dx (b) P(0.5  X  1)   xdx
0 1 0.5
1 1.2 1
 x2   x2   x2 
    2 x    
 2 0  2 1  2  0.5
 0.5  0  2.4  0.72  2  0.5 = 0.5 - 0.125
= 0.5 + 1.68 - 1.5 = 0.375
= 0.68

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BITI2233 Statistics and Probability

(c) For X<0

x
F ( x)   0dt  0

For 0  X<1
x
F ( X )   f (t )dt For X>2,

x x
  tdt F ( X )   f (t )dt
0 
x
t 2  1 2
    tdt   (2  t )dt
 2 0 0 1

1 2
x2 t 2   t2 
     2t  
2  2 0  2 1
For1  X<2,
x 1   4  1 
F ( X )   f (t )dt    0    4     2  

2   2  2 
1 x
  tdt   (2  t )dt =1
0 1
1 x
t 2   t2 
    2t  
 2 0  2 1
1   x2   1 

   0    2 x     2  
2   2  2 
1
  x 2  2x  1
2

0, 0x
 2
x , 0  x 1

 F ( x)   2
 1 x 2  2 x  1, 1 x  2
 2
1, x2


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BITI2233 Statistics and Probability


(d) E ( X )     x  f ( x)dx   E( X 2 )   2

1 2 
  x  xdx   x  (2  x)dx E ( X 2 )   x 2 f ( x)dx
0 1 
1 2 1 2
  x dx   2 x  x dx
2 2
  x 2  xdx   x 2 (2  x)dx
0 1 0 1
1 2
 x3   x3  1 2
    x 2     x 3 dx   2 x 2  x 3 dx
 3 0  3 1 0 1

1 2
1   8   1   x4  2 x4 
   0    4    1        x3  
3   3   3   4 0 3 4 1
1 4 2 1   16   2 1 
      0     4     
3 3 3 4   3   3 4 
1 4 5
=1   
4 3 12

1
1
6
  E( X 2 )   2
1
 1  (1) 2
6
= 0.4

Example 1.12
Consider a new random variable g(X), which depend on X. Each value of g(X) is
determined by knowing the values of X.

Let X be a random variable with density function

x2
 ,  1 x  2
f (x )   3
0,
 elsewhere
(a) Find the expected value of g(X) = 4X + 3.
(b) Find the variance and standard deviation of the random variable g(X)=4X+3.

Solution

(a)  g ( X )  E[ g ( X )]   g ( x)  f ( x)dx


(b)  g2( x )  E ( g ( X )   g ( x ) ) 2 
g ( X )   
2 2
2 x 
  (4 x  3)  ( )dx  g ( x)  f ( x)dx
1 3 

1 2 2 x2
  (4 x 3  3x 2 )dx   (4 x  3  8) 2  ( )dx
3 1 1 3

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BITI2233 Statistics and Probability


3

1 4
x  x3 
2
1 
1 2
3 1
16 x 4  40 x 3  25 x 2 dx
2
1 16 25 3 
 16  8  (1  1)
1
  x 5  10 x 4  x
3 3 5 3  1
=8
1  512 200    16 25 
   160    10  
3  5 3   5 3 
1
 10 or 10.2
5
Standard deviation,   10.2

= 3.2

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