stat

1. (0.8)

0.8 0
(0.8) 0.8
(5.8)

by: Sudheer pai
Page 1 of 5
Session 06
Exercise: The number of accidents occurring in a city in a day is a Poisson variate with mean
0.8. Find the probability that on a randomly selected day
(i)There are no accidents
(ii)There are accidents
Solutions:
Let X: number of accidents per day.
Then, X is P(λ=0.8).
The p.m.f. is –
, 0,1,2,3,....
!
( )
0.8
 x

x
e
p x
x
(i) Probability that ob a particular day three are no accidents is ------------
P[no accidents] = P[X=0]=p(0)
= 0.449
0!
  

e
e
(ii) P[accidents occur] = 1-P[no accidents]
= 1-p(0) = 1-0.449 = 0.551
Exercise: The number of persons joining a cinema queue in a minute has Poisson distribution
with parameter 5.8. Find the probability that (i) no one joins the queue in a particular minute
(ii)2 or more persons join the queue in the minute.
Solution:
Let X : number of persons joining the queue in a minute Then, is P(λ=5.8).
The p.m.f is -------------------
, 0,1,2,3,....
!
( )
5.8
 x

x
e
p x
x
(i) P[no one joints the queue] = P[X=0] = p(0)
=
e5.8 (5.8)0
0!
= e5.8 =0.003
(ii) P[two or more join] =

[ 2]
 
1 P [ X
2]
  
1 { p (0) p
(1)}
5.8 0 5.8 1
(5.8)
 
  
1 1 5.8
  
1 0.003 6.8
1 0.0204 0.9796
(5.8)
1!
0!
1
5.8
  
  
  
  

 
e
e e
P X

Exercise: The average number of telephone calls booked at an exchange between 10-
00 A.M. and 10-10 A.M. is Find the probability that on a randomly selected day 2 or
more calls are booked between 10-00 A.M. and 10-10 A.M. On how many days of a
year, would you expect booking of 2 or more calls during that times gap.

2. 4

4

by: Sudheer pai
Page 2 of 5
Solutions:
Let X : number of telephone calls booked at the exchange during 10-00 A.M. to 10-
10 A.M. Then, X is P(λ=4).
The p.m.f is ---
, 0,1,2,3,...
!
( )
4
 x

x
e
p x
x
P[2 or more calls] = 1-P[less than 2 calls]
= 1 – [p(0) +p(1)]
= 1 – e-4[(40)/0! +(41)/1!]
= 1 – 0.0183[1+4]
= 1 – 0.0915 = 0.9085
An year has 365 days. Out of these N = 365 days, the number of days on which there
will be 2 or more calls is ---
N *P[2 or more calls] = 365 * 0.9085 = 332
Exercise: 2 percent of the fuses manufactured by a firm are expected to be defective,
Find the probability that a box containing 200 fuses contains
(i) defective fuses
(ii) 3 or more defective fuse
Solutions:
2 percent of the fuses are defective. Therefore, probability that a fuses is defective is
p = 2/100 = 0.02
Let X denote the number of defective fuses in the box of 200 fuses. Then, X is B(n =
200, p = 0.02)
Let X denote the number of defective fuses in the box of 200 fuses. Then, X is B (n =
200, p = 0.02)
Here, p is very small and n is very large. Therefore, X can be treated as Poisson
variate with parameter λ=np = 200 * 0.02 = 4.
The p.m.f. is ----
, 0,1,2,3,...
!
( )
4
 x

x
e
p x
x
P[box has defective fuses] = 1-P[no defective fuses]
= 1 – p(0)
= 1 – e-4[(40)/0!
= 1 – 0.0183 =0.9817
P[3 or more defective fuses] = 1-P[less than 3 defective fuses]
= 1 – [p(0)+p(1) +p(2)]
= 1 – e-4[1+4+8]
= 1 – 0.0183*13
= 1 – 0.2379 =0.7621
Exercise: The probability that a razor blade manufactured by a firm is defective is
1/500. Blades are supplied in packets of 5 each. In a lot of 10,000 packets, how many
packets would
(i)Be free defective blades?
(ii) Contains exactly one defective blade?(e-0.01=0.99)

3. Solution:
Let X be the number of defective blades in a packet of 5 blades. Then, X is B (n = 5, p
= 1/500)
Since p is very small and n is sufficiently large, X is treated as Poisson variate
(0.01)

3


3 0
3 3
   
by: Sudheer pai
Page 3 of 5
with parameter λ=np = 5*(1/500) = 0.01
, 0,1,2,3,...
!
( )
0.01
 x

x
e
p x
x
(i) P[ no defective blades] = p(0)
= e-0.01(0.01)0/0! = 0.99
The number of packets which will be free of defective blades is -----
N * P[no defective blades] = 10000*0.99 = 9900
(ii) P[one defective blade] = p(1)
The probability that a razor blade manufactured by a firm is defective is 1/500. Blades
are supplied in packets of 5 each. In a lot of 10,000 packets, how many packets would
(i)Be free defective blades?
(ii) Contains exactly one defective blade?(e-0.01=0.99)
(ii) P[one defective blade] = p(1)
= e-0.01(0.01)1/1! = 0.0099
The number pf packets which will have one defectives blade is ----
N * P[one defective blade] = 10000*0.0099 = 99
Exercise:On an average, a typist mistakes while typing one page. What is the
probability that a randomly observed page in free of mistakes? Among 200 pages, in
how many pages would you expect mistakes?
Solutions:
Let X: number of mistakes in a page.
Then, X is P(λ=3).
The p.m.f. is ---
, 0,1,2,3,....
!
( )
3
 x

x
e
p x
x
P[page is free of mistakes] = p(0)
0.0498
0!
e
e
P[page has mistakes] = 1-P[Page has no mistakes]
= 1-0.0498=0.9502
Among 200 pages, the expected number of pages containing mistakes is ---
N*P[page has mistakes] = 200*0.9502=190
Exercise: In a Poisson distribution P[X=2] = P[X=3]. Find P[X=4].
Solution:
Let λ be the parameter
Here, P[X = 2] = P[X=3]

4.   
3

0.0498 81 
e  e 
  
  
          
156 0 16 1 5 2 2 3 1 4 0 5
36

by: Sudheer pai
Page 4 of 5
3
1
2! 3!
3
2 3

 
 
e e
And so, λ=3
The p.m.f. is ----
, 0,1,2,....
!
( )
3
 x

x
e
p x
x
P[X=4] = p(4) =
e332
4!
= 0.1681
24

Exercise: For a Poisson variables 3 * P[X=2] = P[X=4]. Find standard deviation.
Solution:
Here, 3*P[X = 2] = P[X=4]
3 4
3
2! 4!
3
2
2 4

 
 

And so, λ2=36
That is , λ=3
Thus, the parameter is λ=6
The standard deviation S.D.(X) =   6  2.449
Exercise: The following data relates to the number of mistakes in each page of a book
containing 180 pages.
No of mistakes per page: 0 1 2 3 4 5 or more Total
No. of Pages 156 16 5 2 1 0 180
Fit a Poisson distribution to the data. Obtain the theoretical frequencies
Solution:
Let X denotes the number of mistakes per page. Then, X is a Poisson variate. The
parameter is ---
0.2
180
180
 
N
fx
 x
The p.m.f is ---

5. 0.2


 180
 
 180

0

e
  
by: Sudheer pai
Page 5 of 5
, 0,1,2,3,....
!
( )
0.2
 x

x
e
p x
x
The frequency function is---
0.2
!
0.2
0.2 0
0.2
0!
, 0,1,2,3,....
180 0.8187 147.37
T
x
x
e
T
x
X
NORMAL DISTRIBUTION
A probability distribution which has the following probability density functions(p.d.f)
is called Normal distribution
Here, the variable X is continuous and it is called Normal variate.
Note 1: The distribution has two parameters, namely, μ and σ.(Here, Π=3.14 and
e=2.718.
Note 2 : This normal distribution has Mean E(X) = μ and Variance = V(X) = σ2.
S.D.(X)=σ.
Note 3: A normal variate with parameters μ and σ is denoted by N(μ,σ2)
Note 4: The normal p.d.f. can also be written as-
EXAMPLES FOR NORMAL VARIATE
Many of the variables which occur in nature have normal distribution. Some
examples are –
1.Height of students of a college
2.Weight of apples grown in an orchard
3.I.Q(Intelligence Quotient) of a large group of children.
4.Marks scored by students in an examination