LECTURES ON PROBABILITY AND STATISTICS COURSE CODE: MTH-301, PROBABILITY & STATISTICS LITAN KUMAR SAHA, ASSISTANT PROFESSOR IN MATHEMATICS

PROBABILITY DISTRIBUTION Random variable: A random variable is a variable which takes specified values with specified probabilities. We shall use capital letters to denote a random variable and the corresponding small letters to represent any specific value of the random variable. On the basis of sample space random variable is generally two types. (i) Discrete random variable and (ii) Continuous random variable Probability Distribution: The set consisting of the values of random variables and their respective probability called probability distribution. Probability distribution of random variable x1 , x 2 , ........., x n

x : x1 ,

x2 ,

x3 ,..............., x n

Px  : Px1 , Px 2 , Px3 ,..........Px n , Probability function: If P  x  is probability function and x is a discrete random variable, then P  x  satisfies the following two properties

i 0  Pxi   1 ii   Pxi   1

Example1: The probability function of a discrete random variable is given below x Px 

-1 .1

0 A

1 2a

Find (i) the value of a ii  Px  1

2 .2

3 3a

4 4a

5 .2

6 5a

iii Px  4 iv P 1  x  2

Solution: (i) Since P  x  is a probability function, then we have 0.5 1  Pxi   1 15a  0.5  1  15a  1  0.5  0.5  a  15  30

ii  Px  1  P0  P 1  0.1  a  0.1 

iii Px  4  P4  P5  P6 

1 2  30 15

4 5 1  0.2   30 30 2

iv P 1  x  2  P0  P1  P2  a  2a  0.2  3a  0.2 

Page 4 of 15

3  0.2  0.3 30

We Create Reality Out of Dreams - Presidency University

LECTURES ON PROBABILITY AND STATISTICS COURSE CODE: MTH-301, PROBABILITY & STATISTICS LITAN KUMAR SAHA, ASSISTANT PROFESSOR IN MATHEMATICS

Example2: The probability function of a discrete random variable is given below x Px 

0 a

1 3a

2 5a

Find (i) the value of a ii  Px  3

3 7a

4 9a

5 11a

6 13a

7 15a

iii Px  4 iv P0  x  5

Example3: The probability function of a discrete random variable is given below

x 1 .1

x Px 

1 x  3 .4

3 x 5 .9

5.5  x 1

Find (i) the probability function of X ii  Px  3

iii Px  4 iv P1.5  x  5.2

Probability density function: If f x  is probability density function and x is a continuous random variable, then f x  satisfies the following two properties 

i  f x  0 ii  f x  dx  1 

b

If B is an event defined by a  x  b then PB   Pa  x  b   f x dx a

Problem4: The probability density function of a continuous random variable x is given by 6 x1  x ; 0  x  1 f x    0 ; otherwise  (i)Show that it is a probability function 1 2 (ii) Find the probability that x lies between and 2 3 Solution: (i) f x  is a probability function if

i  f x  0 ii  f x dx  1 f x   0 Here 1

0

1

Now

 0

1

 x2 x3  f x  dx   6 x1  x  dx 6    1 3 0 2 0 1

Therefore f x  is a probability function.

Page 5 of 15

We Create Reality Out of Dreams - Presidency University

LECTURES ON PROBABILITY AND STATISTICS COURSE CODE: MTH-301, PROBABILITY & STATISTICS LITAN KUMAR SAHA, ASSISTANT PROFESSOR IN MATHEMATICS

1 (ii) P  x  2

2  3

2

3



1

2

3

f x dx   6 x1  x dx  1

2

7 36

2

Problem5: The probability density function of a continuous random variable x is given by 2  x; 1  x  2 f x    3   0 ; otherwise Find i  PX  1.2 (ii) PX  1.2 (iii) P1.2  X  1.6 Problem6: The probability density function of a continuous random variable x is given by f x   ax ; 0  x  10 (i) Find the value of a so that f x  is probability density function (ii) Find the probability that x lies between 2 and 3 Cumulative distribution function: If X is a continuous random variable with density function f x  , then the cumulative distribution function is denoted by F x  and defined by

F x   PX  x  

x

 f x dx



x

Note: (i) If f x  is given then F x  is given by F x  

 f x dx



(ii) If F x  is given then f x  is given by f  x  

dF x  dx

Problem7: Let X is a continuous random variable whose probability density function is given by  k ; 0 x  f x    1  x 2  0 ; otherwise (i)Find the value of k (ii) Find cumulative distribution function of X (ii) Compute P X  1 Solution: (i) Since f x  is probability density function, then we have 





k  1   f x dx 1  0 1  x 2 dx  1 k 1  x  0  1 k  1 therefore f x  

1 ; 0 x 1  x 2 x

1 x  1  dx    1  ; x0 (ii) F x   P X  x    f x dx   2  1 x 1 x  1 x0  0 1  x  x

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x

1

We Create Reality Out of Dreams - Presidency University

LECTURES ON PROBABILITY AND STATISTICS COURSE CODE: MTH-301, PROBABILITY & STATISTICS LITAN KUMAR SAHA, ASSISTANT PROFESSOR IN MATHEMATICS

(iii) P X  1  1  P X  1  1  F 1  1 

1 1  2 2 Problem8: Let X is a continuous random variable cumulative distribution function is given by  0; x  0  F  x    x 2 ;0  x  1  0; x  1  (i) Find probability density function of X Compute ii  P X  1 iii  P X  0.5 iv P X  0.4 v  P0.2  X  0.5

Page 7 of 15

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