MA 6453- PROBABILITY & QUEUEING THEORY
UNIT-1
RANDOM VARIABLES
PART A
1.
If a random variable X takes the values 1,2,3,4 such that
2 P(X=1)=3P(X=2)=P(X=3)=5P(X=4). Find the probability distribution of X
Solution:
Assume P(X=3) = α By the given equation



P( X  1) 
P( X  2)  P( X  4) 
2
3
5
For a probability distribution ( and mass function)  P(x) 1
P(1)+P(2)+P(3)+P(4) =1
 

61
30
    1

 1
 
2 3
5
30
61
15
10
30
6
; P( X  2)  ; P( X  3) 
; P( X  4) 
61
61
61
61
The probability distribution is given by
X
1
2
3
4
P( X  1) 
15 10
30
6
61 61
61
61
Let X be a continuous random variable having the probability density function
p ( x)
2.
 2
,
x 1

f ( x)   3
Find the distribution function of
x
 0, otherwise

x.
Solution:
x
 1 
1
  1
F ( x)   f ( x) dx  
dx  
3
2
2
 x 
1
1x
x
1
3.
A random variable X has the probability density function f(x) given by
x


x  0 . Find the value of c and CDF of X.
f ( x)  cx e ,

 0, otherwise
Solution:
x
x
2

 f ( x) dx  1
0

 cx e
x
dx  1
0

x
x 

c  x e  e   1

0
c1  1
c 1
F x  
x
 f ( x) dx
0
x
  cx e
0
x
  xe
x
dx
x
dx
0
x
x

  x e  e

x
4.
x


0
x
1  x e  e
A continuous random variable X has the probability density function f(x)
given by f ( x)  ce
x

 f ( x) dx  1

 x
ce


x
0

x
2ce
2ce
dx 1
dx 1
dx 1
0

 x 
2c   e   1

0
2c1 1
c
1
2
,  x   . Find the value of c and CDF of X.
Solution:
Case(i ) x  0
x
F x  
 f ( x) dx

x x

ce
dx

x x
c
 e dx

x
 x
 ce 

 
1 x
 e
2
Case(ii ) x  0
F x  

x
 f ( x) dx

x x
ce
dx

c
0 x

x x
e dx  c  e

dx
0
0
x
 x
 x 
 c e 
 c  e 

 

0
 c  ce
x
c
x 

 c 2  e 


x 
1
 2  e 
2

1 x
x0
 2 e ,
F ( x)  
x 
1
  2  e , x  0

 2 
5.
2x

2 e
,
x  0.
If a random variable has the probability density f ( x)  

 0, otherwise
Find the probability that it will take on a value between 1 and 3. Also, find
the probability that it will take on value greater than 0.5.
Solution:
3
3
2x
1

1

 2x
0.5
0.5
P(1  X  3) 
P( X  0.5) 
6.
 f ( x) dx   2e
 f ( x) dx   2e

1
  2x 
dx   e
e


 0.5
Is the function defined as follows a density function?
x2
 0,
1
f ( x)   3  2 x , 2  x  4
18
x 4
 0,
Solution:
4
 3  2 x 2 
1


f
(
x
)
dx

3

2
x
dx


 1

 18
72

 2

2
2
Hence it is density function.
The cumulative distribution function (CDF) of a random variable X is
F ( X )  1  (1  x) e  x , x  0 . Find the probability density function of X.
Solution:
f ( x)  F x 
4
7.
3
2
6
 2x 
dx   e
  e e

1
4

 x 
  x 
 0  1  x   e   1 e 





x
8.
 xe , x  0
The number of hardware failures of a computer system in a week of
operations has the following probability mass function:
No of failures: 0
1
2
3
4
5
6
Probability :0.18
0.28 0.25 0.18 0.06
0.04 0.01
Find the mean of the number of failures in a
week.
Solution:
E( X ) 
 x P( x)
 (0)(0.18)  (1)(0.28)  (2)(0.25)  (3)(0.18) 
(4)(0.06)  (5)(0.04)  (6)(0.01)
9.
 1.92
Given the p.d.f of a continuous r.v X as
0  x 1
6 x(1  x),
follows: f ( x)  
Find the CDF of X.
elsewhere
 0,
Solution:
x
x
x
x
3
2
3
 2
F ( x)   f ( x) dx   6 x(1 _ x) dx   6 x _ 6 x dx  3x  2 x   3x  2 x

0
0
0
0
10.
2
A continuous random variable X has the probability function
f ( x)  k (1  x), 2  x  5 . Find P(X<4).
Solution:
4

f ( x)dx 1
5
 k  1  x  dx  1
2
2
5
 1  x 2 
 k
 1
 2  2
27
k
1
2
2
k 
27
4
2 4
2  1  x 2 
1
16
1  x  dx  
P( X  4)   f ( x) dx 
  25  9 

27
27  2 
25
27
2
2
2
Given the p.d.f of a continuous R.V X as follows:
12.5 x 1.25 0.1  x  0.5
f ( x)  
elsewhere
 0,
Find P(0.2 < X < 0.3)
Solution:
4
11.
P (0.2  X  0.3) 
0 .3
 (12.5 x  1.25) dx
0. 2
0 .3
2


x
 12.5
 1.25 x 


2

 0 .2

12.

 1.25 50.32  0.3  50.2 2  0.2
 0.1875
3
If the MGF of a continuous R.V X is given by M X t  
. Find the
3t
mean and variance of X.
Solution:
1
2
3
t
t t

t
  1    1         ...
t 
3
3  3
 3
1
3
1
E ( X )  coefficient of t  1! 
is the mean
3
2
2
1
2

E ( X )   coefficient of t 2!  2! 
9
9


2
2
2 1 1
Variance  E ( X )  E ( X )   
9 9 9
M X t  
13.
3

3t
1
If the MGF of a discrete R.V X is given by M X t  
distribution of X.
Solution:
4
t
1
1  2e  ,find the
81 

t
1
1 
 t
 t
 t
 t
M X t   1  2e  
1  4C1  2e   4C 2  2e   4C3  2e   4C 4  2e 
81 
81 










1
8 t 24 2t 32 3t 16 4t

 e  e  e  e
81 81
81
81
81
4
2
3
By definition of MGF ,
M X t    e p( x)  p(0)  p(1)e  p(2)e  p(3)e  p(4)e
On comparison with above expansion the probability distribution
X
0
1
2
3
4
tx
is
14.
t
M X t  
8
81
3t
4t
32
16
81
81
1
 , 0  x  10
Find the MGF of the R.V X whose p.d.f is f ( x)  10
.Hence
 0, elsewhere
its mean.
Solution:
p( x)
1
81
2t
10
24
81
1 tx
 10 e dx
0
10
 tx 
1 e 
 
10  t 

0
 10t

1 e
1
 

10 
t


2
3


1 
100t
1000t


1  10t 

 ....  1

10t 
2!
3!



2
1000
 1  5t 
t  .....
31
Mean  coefficient of t  5
15.
Given the probability density function
find k and C.D.F.
Solution:
f ( x) 
k
1 x
2
,   x  ,
4




 f ( x) dx  1
F ( x) 



k

 1  x

 k  tan

 f ( x) dx


dx  1
2
x


k
 1  x
1  
x
1
 
2
dx
x

  1   1

 k   tan     tan      1
 


1  1 
tan x 
 
 

  
 k    1
2 2
1
k
1   1   1  
tan     tan  x  
  
 


1  1
1
1 
  tan x   cot x
2
 

16.
It has been claimed that in 60 % of all solar heat installation the utility bill is
Reduced by atleast one-third . Accordingly what are the probabilities that the
utility bill will be reduced by atleast one-third in atleast four of five installation.
Soln: Given n=5 , p=60 % =0.6 and q=1-p=0.4
p( x  4)  p[ x  4]  p[ x  5]
 5c 4 (0.6) 4 (0.4) 54  5c5 (0.6) 5 (0.4) 55
 0.337
17. The no. of monthly breakdowns of a computer is a r.v. having poisson distbn
with mean 1.8. Find the probability that this computer will function for
a month with only one breakdown.
e  x
Soln: p( X  x) 
, given   1.8
x!
e 1.8 (1.8)1
 0.2975
1!
18. In a company 5 % defective components are produced . What is the probability
That atleast 5 components are to be examined in order to get 3 defectives.
p( x  1) 
Soln: To get 3 defectives ,3 or more components must be examined.
p=5 % =0.05 , q = 1- p=0.95 and k=success=3
p( X  x)  ( x  1)c k 1 p k q x  k , x  k , k  1, k  2,...
p( x  5)  1  p( x  5)
 1   p( x  3)  p( x  4)

1  2c 2 (0.05) 3 (0.95) 0  3c 2 (0.05) 3 (0.95)1

1  0.00048  0.9995
19. A discrete r.v X has mgf
Soln: Given
We know that mgf of poisson is
.Find E(x), var(x) , and p(x=0).
Therefore
λ=2
In poisson E(x) = var(x) = λ
 Mean E ( x)  var ( x)  2
p( X  x) 
e  x
x!
 p( X  0) 
e   0
 e   e 2  0.1353
0!
20. Find the mean and variance of geometric distribution .
Soln: The pmf of Geometric distbn is given by
p( X  x)  p q x1 , x  1,2,3,.....
Mean E ( x)   x p( x)


  x p q x 1  p  x q x 1
x 1
  2q
 p 1  2q  3q
 p 1q
11
 p p  2  p 1 
Mean 
 
x 1
2 1

 3q 31  .....
2

 .....  p1  q 
2
1
p
1
p
E x 2   x 2 p( x)

  x 2 p q x 1
x 1

  x x  1  x  p q x 1
x 1


x 1
x 1
  x( x  1) p q x 1   x p q x 1
1(1  1) p q 11  2(2  1) pq 21  3(3  1) pq 31  ..... 
 2 p  2(3) pq 1  3(4) pq 2  ..... 


 2 p 1  3q  6q 2  ..... 
 2 p 1  q  
3

1
p
1
1
 2 p p 3 
p
p
2
1

2
p
p
Variance  E ( x 2 )  E ( x)
2
1
p
1
p
2
2 1 1
2 1 1
 2      2   2
p  p
p p
p
p
1 1 1 p
q
  2  2
2
p
p
p
p
q
Variance = 2
p

21. Find mgf of geometric distribution.
Soln: The pmf of geometric distribution is given by
Mgf M x (t )  E (e tx )   e tx p( x)


  e tx p q x 1   e tx p q x q 1
x 1
x 1

 
 
x
p
p 
t x
x
e
q

qe t


q x 1
q x 1
2
3
p t

qe  qe t  qe t  ...
q
2
p t

qe 1  qe t  qe t  ...
q



   


 


1
 pe t 1  qe t 1  qe t
 M x (t )  1  qe t
22. Show that for the uniform distribution f ( x) 
origin is
1
,  a  x  a ,the mgf about
2a
sinh at
at
1
, a  x  a
2a
Soln: Given f ( x) 
 
MGF M x (t )  E e tx

a
  e tx f ( x)dx   e tx

a
1
dx
2a
a
1
1  e tx 
tx

e dx 
 
2a a
2a  t   a
1 at
1
sinh at

e  e  at 
2 sinh at 
2at
2at
at
sinh at
M x (t ) 
at
a


23. Define exponential density function and find mean and variance of the same.
Soln: The density function of exponential distribution is given by
f ( x)   e x , x  0

Mean  E x  
 x f ( x) dx



  x e x dx    x e x dx
0
0
 x

 x
  xe
e 

 2 
 0
 

1 

 1  1
  (0  0)   0  2     2  
 

  

1
Mean 


  x
Ex 
2
2
f ( x) dx



0
0
  x 2  e x dx    x 2 e x dx

  x 2 e x 2 xex 2e x 


 3 
2


 0


2 

 2 2
  (0  0  0)   0  0  3     3   2
 

  

2
Variance  E x 2   E ( x)

2
2
1
1
1
   2  2  2
2
 
 

2
24. If X is a normal variate with mean  30 and S.D  5
Find the following P  26  X  40 
Solution:
X
N  30, 52 
   30 &   5
X 
Let Z 
be the standard normal variate

40 30 
 26 30
Z
5
5
 P  0.8  Z  2  P  0.8  Z  0   P  0  Z  2
P  26  X  40  P
 P  0  Z 0.8   0  z  2 
 0.2881  0.4772  0.7653 .
PART B
1.
The density function of a random variable X is given by
f ( x)  kx2  x, 0  x  2 . Find k, mean , variance and rth moment.
Solution:
2
 f x  dx  1
2
0
0
 kx2  x  dx  1
2
2
2

k   2 x  x  dx  1

0
8

k 4    1
3

3
 2
x 

k x 
1

3 

0
3
k
4
2 r
3
 r    x x2  x  dx
4
0

r 2 
3 2  r 1
x
2 x
 dx

4 

0
2
r 3 
 r 2
3 x
x


2

4 r 2 r 3 

0

3 r 3  1
1 
2

 r  2 r  3
4


1
 r
 6 2 
  r  2 r  3
12
24
6
put r = 1,2 1 
 1  2 

34
45 5
Mean = 1 and variance =
6
1
1
1
5
5
The monthly demand for Allwyn watches is known to have the following
probability distribution.
Demand:
1
2
3
4
5
6
7
8
Probability:0.08 0.3k 0.19 0.24 k2
0.1
0.07 0.04
Determine the expected demand for watches. Also, compute the variance.
Solution:
 2  
2.
r 3 
 r 3
3 2
2



4r 2 r 3


2

 P( x) 1
2
(0.08)  (0.3k )  (0.19)  (0.24)  (k )  (0.1)  (0.07))(0.04)  1
2
k  0.3k  0.28  0
E( X ) 
 x P( x)
 k  0.4
 (1)(0.18)  (2)(0.12)  (3)(0.19) 
(4)(0.24)  (5)(0.16)  (6)(0.1)  (7)(0.07)  (8)(0.04)
 4.02 is the mean
2
E( X ) 
2
 x P( x)  (1)(0.18)  (4)(0.12)  (9)(0.19) 
(16)(0.24)  (25)(0.16)  (36)(0.1)  (49)(0.07)  (64)(0.04)
 19.7
2
3.
2
2
Variance  E ( X )  E ( X )  19.07  4.02  3.54
The distribution of a random variable X is given by
x
F ( X )  1  (1  x) e
Solution:
(i)

f ( x)  F  x 
, x  0 .Find the rth moment, mean and variance.

 x 
  x 
 0  1  x   e   1 e 





x
 xe ,
(ii)
x0

r
 r
E  X    r    x f ( x) dx


0

 r
x
xe
x
0
 r 1  x
 x
e
dx
dx
0
 r  1!

(iii) E  X   1  1  1!  2
 2
(iv) E  X    2  2  1!  6 Variance =


2
 2
E X   E ( X ) 2


4.
Suppose that the duration ‘X’ in minutes of long distance calls from your
home, follows exponential law with p.d.f f ( x) 
5), p(3≤X≤6),mean and variance.
Solution:
1
e
5

x
5
, x  0 . Find p(X >
(i) p ( X  5) 


5
5

1
 f ( x) dx   5 e
(ii) p (3  X  6) 
6
6
3
3
x
5

 x


1
dx   e 5   e



5
1
x
5

 f ( x) dx   5 e
6
 x


1.2
 0.5
dx   e 5    e
e



3
(iii)


0
0

x
1
E ( X )   xf ( x) dx   xe 5 dx
5

x
x 




1 
  xe 5 5  e 5 25
5


0
1
 0  25  5
5
2
 2

0
0

x
1 2 5
(iv) E ( X )   x f ( x) dx   5 x e dx

5.
x
x
x






1  2 5
5
5
  x e 5  2 xe 25  2e 125
5


0
1
 0  250   50
5
2
 2
Variance = E  X   E ( X ) 50  25  25


A random variable X has the following probability distribution.
X:
0
1
2
3
4
5
6
7
f(x): 0
k
2k
2k
3k
k2
2k2
7k2+ k
Find (i) the value of k (ii) p(1.5 < X < 4.5 | X > 2) and
1
(iii) the smallest value of λ such that p(X≤λ) > .
2
Solution:

P( x) 1
2
0  k  2k  2k  3k  k  2k
10 k
k
2
 9k  1  0
1
 0.1
10
2
 k   1,
 7k
1
10
2
k 1
A  1.5  X  4.5  2,3,4
B  X  2  3,4,5,6,7
(ii) A  B  3,4
p (1.5  X  4.5 | X  2)  p A | B  
5
5


 10 
2
2
2
2
7
7
2k  3k  k  2k  7k  k 10k  6k
10
(iii)
X
p(X)
F(X)
0
0
0
2
2k = 0.2
0.3
3
2k = 0.2
0.5
4
3k = 0.3
0.8
2
5
k =0.01
0.81
6
2
2k = 0.02
0.83
7
7k2+k = 0.17 1.00
1
From the table for X = 4,5,6,7 p(X) >
and the smallest value is 4
2
Therefore λ = 4.
Find the MGF of triangular distribution whose density function is given by
0  x1
 x,
2  x , 1  x  2

.Hence its mean and variance.
f ( x)  
 0, elsehwere


Solution:
2k  3k
6.
p A  B 
p (3,4)

p B 
p (3,4,5,6,7)
5k

tx
 tX 
M X t   E  e    e f x  dx

 
1 tx
 e
2 tx
x dx   e
0
2  x  dx
1
1
2
tx 
tx
tx 
 tx

e
e
e
e
  2  x 

 x

  1
2 
2 
 t

t
t 0 
t 1

t
t
2t
t
t
e
e
1 e
e
e






2
2
2
2
t
t
t
t
t
t
M X t  
e
2t
t
 2e  1
2
t
expanding the above in powers of t, we get
2
3
4


1  2t  4t  8t  16t  ... 
 

2t
t
2!
3!
4!
e  2e  1 1 

M X t  



2
2
2
3
4

t
t  

t
t
t



 .  1 
 21  t 
2! 3! 4! 
 


3
4
 2

1  2t
6t
14t




 ...

2 2!
3!
4!

t 

2
7.
3
7t
t
 1 t 

 ....
12
4
Mean = E(X) = (coefficient of t) 1! = 1
7
E(X2) = ( coefficient of t2)2! =
6
1
Variance = E(X2) - E(X)2 =
6
Find the MGF of the RV X, whose pdf is given by
1 x
f ( x)  e
,    x   . Hence its mean and variance.
2
Solution:

tx
 tX 
M X t   E  e    e f  x  dx

 


0 tx x
e
e

0 t 1x
e

 tx  x
dx   e e
dx
0
  1 t x
dx   e
dx
0

0
 t 1x 
  1 t x 
e
e





 t  1 
  1  t  

  
0
M X t  
1 1
1 



2 1  t 1  t 
2
1
2
4
 1  t  t  ....
1 t
Mean = E(X) = (coefficient of t) 1! = 0
E(X2) = ( coefficient of t2)2! =2
Variance = E(X2) - E(X)2 = 2
8.
The p.m.f of a RV X, is given by p( X  j ) 
1
2
mean and variance.
Solution:
j
, j  1,2,3... Find MGF,
  e
M X t   E etX 


e
x 0
tx
p( x)
1
2x
 et 
   
x 0  2 

tx
x
2
3
4
 et   et   et   et 
             ....
2 2 2 2
  et   et  2  et 3  et  4

1              ..
  2   2   2   2 



et 1
et


2
et
2  et
1
2
Differentiating twice with respect to t
t  t  t  t 

t
 2  e  e   e   e 
2e






M  t  

et

2
X
t

2  e 


2
t

2  e 


2
2
M X t  
t  t
t 
t 
t

 2  e   2e   2e 2 2  e   e 

 




t 4

2  e 


put t = 0 above E( X )  M X 0  2
E X 2   M X 0  6
 
t

4e  2e
2t
t

2  e 


3
Variance  E X 2  E  X   6  4  2
9.
2
Find MGF of the RV X, whose pdf is given by f x   e
hence find the first four central moments.
Solution:
 x
, x  0 and

 
M X t   E e
 e f x  dx

tX
tx


  e tx e  x dx
0

   e   t x dx
0

 e    t  x 
 

    t   0


  t 
Expanding in powers of t

2
3
1
t t t
M X t  

 1           ...
  t  1   t 
  
 

Taking the coefficient we get the raw moments about origin
1
E  X   coefficient of t 1! 

2
2
2


E  X    coefficient of t 2! 
2

 


  

  

E X 3  coefficient of t 3 3! 
E X 4  coefficient of t 4 4! 
6
3
24
4
and the central moments are
1  0
2
 2   2  2C11 1  1

2

2
2
1

2

1

2
1


2
2
3
 3   3  3C1 2 1  3C 2 1 1  1

6

3
3
2 1

2 
3
1 1


1

2

3

2

3
2
4
4
 4   4  4C1 3 1  4C 2  2 1  4C 3 1  1

24

4
4
6 1

3 
6
2 1
2
 
2
4
1

4

1

4
9


4
1
10.
If the MGF of a (discrete) RV X is
and p ( X = 5 or 6).
Solution:
5  4e
t find the distribution of X
M X t  
1
1

t
5  4e
 4e t
51 
5

1   4e t
 1  
5  5




  4e t
  
  5
2
  4e t
  
  5
3


  ...


By definition
tx
 tX 
M X t   E  e    e p( x)


1  e
t0
t
t2
16
125
p3 
p(0)  e p(1)  e
p(2)  ...
On comparison
p(0) 
1
5
p1 
4
25
p2 
64
625
r
1  4
  , r  0,1,2,3
5 5
p X  5 or 6  p X  5  p X  6
In general p X  r  
5

1  4 1  4
    
5 5 5 5
6
5
1  4  4
   1  
5 5  5

11.
9  4
 
25  5 
5
 3x
If X has the probability density function f x   k e
, x0
Find (i) k (ii) p(0.5 ≤X≤1) (iii) Mean of X.
Solution:
(i)
1
 f ( x) dx
p (0.5  X  1) 
0. 5
1

 3e
3 x
dx
0. 5
1
  e 3 x 
 3

 3  0.5
  e 3  e 1.5
(iii)

Mean  E  X  
 x f ( x) dx
0

  3x e 3 x dx
0

 e 3 x
e 3 x 
 3  x


3
9 0

1 1
 3  
9 3
1
 ,
x2
12.
If a RV X has the pdf f ( x)   4
.
0, otherwise
Obtain (i) p(X <1) (ii) p( X >1) (iii) p( 2X+3 > 5)
(iv) p ( X < 0.5 | X < 1
Solution:
1
(i) p( X < 1) =
1
1
 4 dx  4 x
1
2
2

3
4
1
1
1
 4 dx  4 x
1
2
1
1
Hence p( X >1)= 1 - p( X ≤1) =
2
(iii) p( 2X+3 > 5) = p( X > 1 ) = 1- p( X ≤1)
(ii) p( X ≤1) =
1
3
=
4
4
p ( X  0.5  X  1)
p ( X  1)
p ( 0.5  X  0.5  X  1)
p ( X  1)
p( 0.5  X  0.5 )
p( X  1)
1
0.5
 0.5
1
3
4
3

=
=
=
1
 4 dx x


(iv) p ( X < 0.5 | X < 1)
= 1=
1
1
1
3
13.
If X has the distribution function
x 1
 0,
1
1 x4
 ,
3

1
F x    ,
4 x6
2

 5 , 6  x  10
6
1,
x  10
(1) Probability distribution of X (2) p(2<X<6) (3) Mean (4)
variance
Solution:
(1)As there is no x terms in the distribution function given
is a discrete random variable. Hence the probability distribution is given by
X
1
4
6
10
p( X )
1
3
1
3
1 1

2 3
1

6
(2) p(2<X<6) = p(4) =
1
5
6
1
6
1
6
1
1
1
 1  14
p( x)  1   4   6   10  
3
 3
6
 3
6
1
1
1
 1  95
(4) E(X2) =  x 2 p( x)  1   16   36   100  
3
 3
6
 3
6
95 196 89
Variance = E(X2) – E(X)2 =
=
3
9
9
A continuous random variable X has the distribution function
:
x 1
 0

4
F ( x)  k 1  x  :
1 x  3
 0
:
x 3

Find k, the probability density function f(x) and P(X <2).
Solution:
Since it is a distribution function
F ()  F (3)  1
(3) Mean = E(X) =
14.
5 1

6 2
1
3
x
k 3  14  1
k 
1
16
The density function is
1  x 3 , 1  x  3
1
3
f x   F x  
41  x  
p(X < 2) = F(2) =
16
4
1
2  14  1
16
16
15.
If the cumulative distribution function of a R.V X is given by
4

1  2 , x  2
F x    x
find (i) P(X < 3) (ii) P(4 < X < 5) (iii) P ( X ≥3).
0,
x2
Solution:
(i) P(X < 3) = F(3) = 1 
4
3
2

5
9
4

(ii) P( 4 < X < 5) = F(5) - F(4) = 1  2   5 


1  4   21  3  9

2  25 4 100
 4 
5
4
=
9
9

4 
(iii) P ( X ≥3) = 1- F(3) = 1 - 1 
= 1
2
 3 
16.
Find the recurrence relation for the moments of the Binomial distribution.
Soln: The k th order central moment is given by

 
 k  E  X  X   E  X  n p k
k
n

  x  n p  p( x)
k
x 0
n
  x  n p  n c x p x q n x
k
x 0
k
n
  nc
x 0
x  n p 
k
x

p x q n  x              (1)
Differentiating (1) w.r.to p,we have
d k
dp
k x  n p 
n
 nc
x 0
x
n
 nk  n c
x 0
n
 nk  n c
x 0
x

(n) p x q n  x   x  n p  x. p x 1 q n  x  p x (n  x)q n  x 1 (1)
k
n
x
x  n p k 1 p x q n x   n c x x  n p k p x 1q n x 1 xq  (n  x) p 
x 0
x  n p 
n
 nk k 1   n c
x 0
 nk k 1 
k 1
x
p q
n x
n
 nc
x 0
k
x x  n p 
1 n
nc
pq x 0
 nk k 1 
k 1
x
x
x  n p 
k
p x q n x
x( p  q)  np 
p q
p x q n x
x  np 
p q
x  n p k 1 p x q n x
1
 k 1
pq
, by (1)
 d

  k 1  pq  k  nk k 1 
 dp

This is the recurrence relation for the moments of the Binomial distribution
17. Prove that poisson distribution is the limiting case of Binomial distribution.
(or)
Poisson distribution is a limiting case of Binomial distribution under the
following conditions
(i)
n , the no.of trials is indefinitely large , i.e, n  
(ii)
p, the constant probability of success in each trial is very small ,i.e p  0


np   is inf inite or p  and q  1  , is positive real
(iii)
n
n
Soln: If Xis binomial r.v with parameter n & p ,then

p ( X  x)  n c x p x q n  x , x  0,1,2,....n
n!
p x (1  p ) n  x
(n  x)! x!

n(n  1)( n  2)....( n  ( x  1))( n  x)!      

  1  
(n  x)! x!
n  n
x
n(n  1)( n  2)....( n  ( x  1))
x!


 
n
x
 
1  
 n
n
n x
 
1  
 n
x
x
 1  2
 x  1      
 x n.n 1   n1  .....n1 
 1    1  
n   n  n
n x!  n   n 


n
x 
1   2   x  1   
1   1  .....1 
 1  
x!  n   n  
n   n
Taking limit as n  
p( X  x)  lim
n 
n 



x!
x
x!
 
1  
 n
x
on both sides
lim
x
n
x
lim
n 
x 
1   2   x  1      
1   1  .....1 
 1   1  
x!  n   n  
n   n  n
n
 1   2   x  1
1   1  .....1 

n 
 n  n  
 
(1.1....1) . e  .1 
e  x
x!
Hence the proof.
 p ( X  x) 
e


x
x!
lim
n 
 
1  
 n
n
lim
n 
x
 
1  
 n
, x  0,1,2,.....
, x  0,1,2,..... and it is poisson distbn.
18. Find the recurrence relation for the central moments of the poisson distbn. and
hence find the first three central moments .
Soln: The k th order central moment  k is given by
 k  E X  X   E  X   k
k

   x    p( x)
k
x 0
 k 

 x   
x 0

k
e . x
         (1)
x!
Diff. (1) w.r.to  we have
k 1
 

d k
e   . x  x   
k 1
  k  x    (1)

e  .x x 1  (e  ). x 
d
x!
x!
x 0 


x

x 1


.
.
k 1 e
k e
 k  x   
  x   
(x  )
x!
x!
x 0
x 0


  k k 1    x   
x 0
  k k 1 
1

k 1
e  . x
x !
by (1)
 k 1
 d

  k 1    k  k k 1               (2)
 d


x
which is the recurrence formula for the central moments of the poisson distbn.
since  0  1 and 1  0
put k= 1 in (2)
d

11    1  111 
 d

d

   (0)  1 0 
 d

2  
put k= 2 in (2)
d

 21     2  2  21 
 d

d

   ( )  2  1 
 d

 3   (1  0)

19. Prove that the sum of two independent poisson variates is a poisson variate,
while the difference is not a poisson variate.
Soln: Let X 1 and X 2 be independent r.v.s that follow poisson distbn. with
Parameters 1 and  2 respectively.
Let X  X 1  X 2
p ( X  n)  p ( X 1  X 2  n)
n
  p X 1  r . p X 2  n  r  sin ce X 1 & X 2 are independen t
r 0
nr
e 1 .1 e 2 . 2
.
r!
(n  r ) !
r 0
r
n

e


 1
e
.1 1 n !.
.
2 nr

r 0 r ! n ! (n  r ) !
n
 2
r
e ( 1  2 )
n!
 . r ! (n  r ) ! 
 ( 1  2 )
n
e
n!
n
n !.
r
1
r 0
 . n cr 1 2
r
nr
r 0
2 nr

e ( 1  2 )
(1   2 ) n
n!
This is poisson with parameter ( 1 +  2 )
(ii) Difference is not poisson
Let X  X 1  X 2
E ( X )  EX 1  X 2 
 E( X 1 )  E( X 2 )
 1  2


 E X  X  2 X X 
 E X   E X   2 E X E X
E( X 2 )  E X 1  X 2 
2
2
2
1
1
1
2
2
2
1
2
1
2

 (1  1 )  ( 2   2 )  2(1 2 )
2
2
 (1   2 ) 2  (1   2 )
 (1  2 ) 2  (1   2 )
It is not poisson.
20. If X and Y are two independent poisson variates , show that the conditional
distbn. of X, given the value of X+Y is Binomial.
Soln: Let X and Y follow poisson with parameters 1 and  2 respectively.
pX  r and X  Y  n
p X  Y  n 
pX  r . pX  Y  n

by independent
p X  Y  n 
p X  r X  Y  n  
nr
e 1 .1 e 2 .2
.
r!
(n  r ) !
 ( 1  2 )
e
.(1  2 ) n
n!
r
.
nr
e 1 .1 .e 2 . 2
n!

r !(n  r ) ! e 1e 2 ..(1   2 ) n
r
 n cr
1 r .2 n  r
(1  2 ) r .(1  2 ) n  r
r
 1    2 
 n cr 
 

 1  2   1  2 
where
p
nr
1
1  2
 n cr p r q nr
and q 
2
1  2
This is binomial distbn.
21. It is known that the probability of an item produced by a certain machine will be
defective is 0.05. If the produced items are sent to the market in packets of 20,
find the no. of packets containing atleast ,exactly,atmost 2 defectives in a consignment
of 1000 packets using poisson.
Soln: Give n = 20 , p = 0.05 , N = 1000
Mean   n p  1
Let X denote the no. of defectives.
p X  x  
e   . x e 1 .1x e 1 .


x  0,1,2,...
x!
x!
x!
px  2  1  px  2
1   p( x  0)  p( x  1)
 e 1 e 1 
1
 1 

  1  2 e  0.2642
0
!
1
!


Therefore , out of 1000 packets , the no. of packets containing atleast 2 defectives
 N . px  2  1000 * 0.2642  264 packets
e 1
(ii) px  2 
 0.18395
2!
Out of 1000 packets,=N*p[x=2]=184 packets
(iii)
px  2  p[ x  0]  p[ x  1]  p[ x  2]
e 1 e 1 e 1



 0.91975
0! 1!
2!
For 1000 packets = 1000*0.91975=920 packets approximately.
22. The atoms of radio active element are randomly disintegrating. If every gram of this
element , on average, emits 3.9 alpha particles per second, what is the probability during
the next second the no. of alpha particles emitted from 1 gram is
(i)
atmost 6 (ii) atleast 2 (iii) atleast 3 and atmost 6 ?
Soln: Given   3.9
Let X denote the no. of alpha particles emitted
(i ) p ( x  6)  p ( x  0)  p ( x  1)  p ( x  2)  ......  p ( x  6)

e 3.9 (3.9) 0 e 3.9 (3.9)1 e 3.9 (3.9) 2
e 3.9 (3.9) 6


 ...... 
0!
1!
2!
6!
 0.898
(ii )
p ( x  2)  1  p ( x  2)
 1   p ( x  0)  p ( x  1)
 e 3.9 (3.9) 0 e 3.9 (3.9)1 
1  


0!
1!


 0.901
(iii ) p(3  x  6)  p( x  3)  p( x  4)  p( x  5)  p( x  6)
e 3.9 (3.9) 3 e 3.9 (3.9) 4 e 3.9 (3.9) 5 e 3.9 (3.9) 6




3!
4!
5!
6!
 0.645
23. Establish the memoryless property of geometric distbn.
Soln: If X is a discrete r.v. following a geometric distbn.
 p ( X  x)  pq x 1
, x  1,2,.....

 pq
p( x  k ) 
x 1
x  k 1

 p q k  q k 1  q k  2  ......



 p q k 1  q  q 2  .....  p q k (1  q ) 1
 p q k p 1  q k
Now
px  m  n x  m 

px  m  n and x  m
p x  m 
px  m  n
p x  m 
  qmn
q
 px  m  n x  m  px  n
m
 q n  px  n
24. If X 1 , X 2 be independent r.v. each having geometric distbn pq k , k  0,1,2,....
Show that th conditional distribution of X1 given X 1  X 2 is Uniform distbn.
Soln:
pX 1  r and X 1  X 2  n
p X 1  X 2  n 
p X 1  r X 1  X 2  n  

pX 1  r and X 2  n  r 
n
 p X
s 0

 s and X 2  n  s 
pX 1  r . pX 2  n  r 
n
 p X
s 0

1
1
 s . pX 2  n  s 
qr p qnr p
n
q
s

p qns p
s 0
(i.e) pX 1  r X 1  X 2  n 
qn

n
q
n
by independen t
1
, r  0,1,2,.....n
n 1
s 0
1
, this is uniform distbn
n 1
25. Suppose that a trainee soldier shoots a target in an independent fashion. If the
probability
That the target is shot on any one shot is 0.7.
(i) What is the probability that the target would be hit in 10 th attempt?
(ii) What is the probability that it takes him less than 4 shots?
(iii)What is the probability that it takes him an even no. of shots?
(iv) What is the average no. of shots needed to hit the target?
Soln: Let X denote the no. of shots needed to hit the target and X follows geometric
distribution with pmf pX  x  p q x1 , x  1,2,....
Given p=0.7 , and q=1-p=0.3
101
 0.0000138
(i) px  10  (0.7)(0.3)
(ii)
px  4  p( x  1)  p( x  2)  p( x  3)
 (0.7)(0.3)11  (0.7)(0.3) 21  (0.7)(0.3) 31
 0.973
(iii)
px is an even number  p( x  2)  p( x  4)  ........
 (0.7)(0.3)2 1  (0.7)(0.3)4 1  .....


 (0.7)(0.3) 1  (0.3)2  (0.3)4 .......


    .......
  (0.21) (0.91)
 0.211  (0.3) 2  (0.3) 2

 0.21 1  (0.3) 2
1
2
1
0.21
 0.231
0.91
1
1
(iv) Average no. of shots = E ( X )  
 1.4286
p 0.7

26. The number of personel computer (pc) sold daily at a computer world is uniformly
distributed with a minimum of 2000 pc and a maximum of 5000 pc. Find
(1) The probability that daily sales will fall between 2500 and 3000 pc
(2)What is the probability that the computer world will sell atleast 4000 pc’s?
(3) What is the probability that the computer world will sell exactly 2500 pc’s?
Soln: Let X~U(a , b) , then the pdf is given by
1
f ( x) 
, a xb
ba
1

, 2000  x  5000
5000  2000
1

, 2000  x  5000
3000
(1)
p2500  x  3000 
3000
 f ( x) dx
2500
3000


(2)
1
1
dx 
3000
3000
2500

x
3000
2500
1
3000  2500  0.166
3000
px  4000 
5000
 f ( x) dx
4000
5000

1
1
dx 
3000
3000
4000

x
5000
4000
1
5000  4000  0.333
3000
px  2500  0 , (i.e) it is particular point ,the value is zero.

(3)
27. Starting at 5.00 am every half an hour there is a flight from San Fransisco airport
to Los angles .Suppose that none of three planes is completely sold out and that they
always have room for passengers . A person who wants to fly to Los angles arrive at
a random time between 8.45 am and 9.45 am. Find the probability that she waits
(a) Atmost 10 min
(b) atleast 15 min
Soln: Let X be the uniform r.v. over the interval (0,60)
Then the pdf is given by
1
f ( x) 
, a xb
ba
1

, 0  x  60
60
(a) The passengers will have to wait less than 10 min. if she arrives at the airport
 p(5  x  15)  p(35  x  45)
15
45
1
1
  dx   dx
60
60
5
35
1
60
1

3

x  601 x
15
45
5
35
(b)The probability that she has to wait atleast 15 min.
 p (15  x  30)  p (45  x  60)
30
60
1
1
  dx   dx
60
60
15
45
1
60
1

2

x  601 x
30
60
15
45
28. Establish the memoryless property of exponential distbn.
Soln: If X is exponentially distributed , then
px  s  t x  s  px  t  for any s, t  0
The pdf of exponential distbn is given by
 e x
, x0
f ( x)  
, otherwise
0

p( x  k )   f ( x) dx
k


 e
k
 x

 e  x 
dx   

   k

  0  e k  e x        (1)
px  s  t x  s  

px  s  t and x  s 
px  s 
px  s  t
px  s 
  e  ( s t )
e
 s
 e t  px  t 
 px  s  t x  s  px  t  for any s, t  0
29. The time (in hours) required to repair a machine is exponentially distributed
1
with parameter   .
2
(a)What is the probability that the repair time exceeds 2 hrs ?
(b)What is the conditional probability that a repair takes atleast 11 hrs given
direction exceeds 8 hrs ?
that Its
Soln: If X represents the time to repair the machine, the density function
Of X is given by
f ( x )   e  x

, x0
1 x2
e
2
x 0
(a)


2
2
p ( x  2)   f ( x) dx    e x dx


1
 e
2
2
 x 
1 e 2 
dx  

2  1

 2  2
x
2


  0  e 1  0.3679
px  11 x  8  px  3


3
3
  f ( x) dx    e x dx


1
 e
2
3
x
2
 x 
1 e 2 
dx  

2  1

 2  3
3
3
 


  0  e 2   e 2  0.2231


30. In a certain city the daily consumption of electric power in millions of kilowatt hrs
1
, k  3 . If the power plant has
2
a daily capacity of 12 million kilowatt hours . What is the probability that tha power
supply will be inadequate on any given day.
can be treated as central gamma distbn with  
Soln: Let X be the daily consumption of electric power
Then the density function of X is given by
f ( x) 
k x k 1e  x
k
3
 1  31  2
  x e
2
 

3


2
x e
16
x
 1  2 2
 x e
8
2!
x
x
2
p[ the power supply is inadequate]=p[x>12]



x
x 2e 2
  f ( x) dx  
dx
16
12
12

x

1
  x 2 e 2 dx
16 3

x
x
x


 
1 
  2 x 2 e 2  8 xe 2  16e 2 
16 
 12
 0.0625
31. The daily consumption of milk in a city in excess of 20,000 litres is approximately
1
, k  2 . The city has a daily
distributed as an Gamma distbn with parameter  
10000
stock of 30,000 litres. What is the probability that the stock is insufficient on a particular
day.
Soln: Let X be the daily consumption ,so ,the r.v. Y=X-20000.
Then
k y k 1e  y
fY ( y) 
k
2
y
 1  21 10000
y


 y e
10000
ye
10000 


2
(10000) 2 1!


y
10000
ye
(10000) 2
p[ insufficient stock]=p[X>30000]
=p[Y>10000]

10000

1
(10000) 2

y
10000
ye
dy
2
(
10000
)
10000
 f ( y) dy  
p[Y  10000] 



ye

y
10000
dy
3
y 

,  By substitution method , put t 
10000 

 2e 1
 0.7357
32.. (a). State and prove the additive property of normal distribution.
t2
(b). Prove that “For standard normal distribution N  0,1 , M X  t   e 2 .
Sol: a) Statement:
2
If X 1 , X 2 ,..., X n are n independent normal random variates with mean  1 ,  1  ,
  ,  ,…  ,  then


  ,    .
2
2
n
X 1  X 2  ...  X n also a normal random variable with mean
2
2
n
n
n
2
 i 1
Proof:
i
i 1
i

We know that. M X
1  X 2 ...
MX
1  X 2 ...
Xn
t   e
e
1 t 
t
1
i
 t
But M Xi  t   e
2
 t   M X  t  M X  t  ...M X  t 
Xn
 12
2
e
2
n
t2  i2
2 , i  1, 2....n
2 t 
t 22 2
2
 1   2 ...  n t 

...e
n t 
t 2n 2
2
 12  2 2 ... n 2
t
2
2
n

n

2 2
i
t
i
 t
i1
e
By uniqueness MGF, X 1  X 2  ...  X n follows normal random variable with
i1
2
n
 n

parameter  i ,   i 2  .
 i 1
i 1

This proves the property.
b) Moment generating function of Normal distribution
 M X  t   E  etx 
1

 2


2
e
1  x 
 

tx
2   
e

dx
x 
then dz   dx,    Z  


z
t  z    
1
2
 M X t  
e
dz

2 
Put z 
2
t

e
2


e

 z2

  t z


 2

dz

e
t
2

e

1
 z t 
2

2
 2t 2 
 2 



2t 2
t
2

e
dz 
e e
2


1
 
2
2
dz
z
t


1   12  z t 
dz  1
e
2 
2
the total area under normal curve is unity, we have
Hence M X  t   e
 2t 2
 t 
2
t
For standard normal variable N  0,1
2
M X t   e 2
33. (a). The average percentage of marks of candidates in an examination is 45 will a
standard deviation of 10 the minimum for a pass is 50%.If 1000 candidates appear for the
examination, how many can be expected marks. If it is required, that double that number
should pass, what should be the average percentage of marks?
(b).
Given
that X is
normally
distribution
with
mean
10
and
probability P  X  12   0.1587 . What is the probability that X will fall in the
interval  9,11 .
Solution:
a) Let X be marks of the candidates
Then X N  42,10 2 
X 42
10
P  X  50   P  Z  0.8
Let z 
 0.5  P 0  z  0.8  0.5  0.2881  0.2119
Since 1000 students write the test, nearly 212 students would pass the
examination.
If double that number should pass, then the no of passes should be 424.
We have to find z1 , such that P  Z  z1   0.424
 P  0  z  z1   0.5  0.424  0.076
From tables, z  0.19
50 x1
 z1 
 x 1  50 10 z1  50  1.9  48.1
10
The average mark should be 48 nearly.
b) Given X is normally distributed with mean   10.
x 
be the standard normal variate.
Let z 

12 10
2
z
For X  12, z 


Put z1  2

Then P  X  12   0.1587
P  Z  Z1   0.1587
 0.5  p  0  z  z1   0.1587
 P 0  z  z1   0.3413
From area table P  0  z  1  0.3413
2
1

To find P 9  x  11
 Z1  1 
1
1
For X  9, z   and X  11, z 
2
2
 P 9  X  11  P  0.5  z  0.5  2P  0  z  0.5  2  0.1915  0.3830
34. (a). In a normal distribution,31 % of the items are under 45 and 8% are over
64.Find the mean and standard deviation of the distribution.
(b). For a certain distribution the first moment about 10 is 40 and that the 4th moment
about 50 is 48, what are the parameters of the distribution.
Solution:
a) Let  be the mean and  be the standard deviation.
Then P  X  45  0.31 and P  X  64  0.08
45  
  z1
When X  45 , Z 

 z1 is the value of z corresponding to the area
z1
   z dz  0.19
0
 z1  0.495
45    0.495 ---(1)
64  
 z2
When X  64 , Z 

 z2 is the value of z corresponding to the area
z2
   z dz  0.42
0
 z2  1.405
64    1.405 ---(2)
Solving (1) & (2) We get   10 (approx) &   50 (approx)
b) Let  be mean and  2 the variance then 1  40 about A=10

Mean A  1  10  40 10+40
   50
Also 4  48  3 4  48   2  4
 The parameters are Mean    50
and S.D    2 .