Probability Distribution
Probability Distribution
Probability Distribution
Poisson Distribution
By
Govinda Gyawali
Assistant Professor
Butwal Multple Campus
Butwal
Poisson Distribution:
Let X be a discrete random variable follows Poisson distribution with parameter
‘λ’ and its probability mass function is
𝒆−𝝀 𝝀𝒙
P(X=x) = p(x) =
𝒙!
; x = 0, 1, 2, ………….∞ Mean = Variance = λ
Where, x = no. of success , e = 2.7183 , base of natural logarithm
λ = mean no. of success ( Average rate)
Under following Conditions Binomial distribution tends to Poisson
distribution
- When the no. of trials ‘n’ is very large, n→ ∞.
- When probability of getting Points to remember
success ‘p’ is very small,p→ 𝟎. n< 𝟐𝟎, p>0.05 Binomial Distribution
- np = 𝝀 is finite. n≥20, p≤ 0.05 Poisson Distribution
Source: Richard I. Levin 'et.al.’ ,Statistics for Management, 7th edition(pg.243)
Application of Poisson distribution
The Poisson Distribution is used in those situations where the
probability of happening of an event is small ,i.e. the event rarely
occurs.
➢No. of twin birth in the hospital
➢No. of suicides reported in a city
➢No. of printing mistake per page in a book
➢No. of accidents in a busy road
➢No. of wrong telephone calls per minute in a switch board
➢No. of customer per minutes in a grocery shop.
➢No. of deaths from cancer, heart attack, snake bites etc.
Numerical Example:
Assume a Poisson variate X with λ =5.0. What is the probability that (a) X=1
(b) X≤ 3 (c) X>2. Also, determine the mean and standard deviation of the
random variable X.
Solution:
Here, λ = 5
Let X~P(λ) then
𝒆−𝝀 𝝀𝒙
P(X=x) = p(x) = (-) 5
𝒙! Shift ln
=
0.0067379
𝒆−𝟓 𝟓𝒙
= ; x = 0, 1, 2, ………….∞ 3 Shift 7 =6
𝒙!
𝒆−𝟓 𝟓𝟏 𝒆−𝟓 ×𝟓
a) P(X=1) = = = 0.006737×5 =0.0336897
𝟏! 𝟏
x = 0, 1, 2, 3, 4, 5,……∞
b) P(X≤ 𝟑) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
𝒆−𝟓 𝟓𝟎 𝒆−𝟓 𝟓𝟏 𝒆−𝟓 𝟓𝟐 𝒆−𝟓 𝟓𝟑
= + + +
𝟎! 𝟏! 𝟐! 𝟑!
50 51 52 53
= 𝑒 −5 + + +
0! 1! 2! 3!
5 25 125
= 0.006737× 1 + + +
1 2 6
𝒆−𝟑 𝟑𝟎 𝒆−𝟑 ×𝟏
P(X=0)= = = 𝒆−𝟑 = 0.0498
𝟎! 𝟏
𝒆−𝟔 × 𝟔𝒙
= ; x = 0, 1, 2, ………….∞
𝒙!
The probability that exactly ten cheques bounced is
𝒆−𝟔 × 𝟔𝟏𝟎
P( X= 10) =
𝟏𝟎!
−𝟔 𝟔𝟎𝟒𝟔𝟔𝟏𝟕𝟔
=𝒆 × = 𝒆−𝟔 × 16.662857
𝟑𝟔𝟐𝟖𝟖𝟎𝟎
= 0.002478752 ×16.662857
= 0.041303093
Numerical Example:
A manufacturer of pins know on an average 3% of its production is
defective. He sells pins in boxes of 100 and guarantees that not more
than two pins will be defective. What is the probability that a box
randomly selected (i) will meet the guaranteed quality (ii) will not meet
the guaranteed quality.
Solution:
Here, number of pins (n) = 100
probability of defective pin (p) = 3% =0.03
∴ λ = np = 100× 0.03 = 3
Let X~P(λ) then
𝒆−𝝀 𝝀𝒙
P(X=x) = p(x) =
𝒙!
𝒆−𝟑 × 𝟑𝒙
= ; x = 0, 1, 2, ………….∞
𝒙!
x = 0, 1, 2, 3, 4, 5,……∞
a)The probability that the box will meet the guaranteed quality is
P( not more than 2 pin defective) = P(X≤ 𝟐)
= [P(X=0) + P(X=1) + P(X=2)
𝒆−𝟑 𝟑𝟎 𝒆−𝟑 𝟑𝟏 𝒆−𝟑 𝟑𝟐
= + + ]
𝟎! 𝟏! 𝟐!
−3 30 31 32
= 𝑒 + +
0! 1! 2!
3 9
= 0.0498× 1 + +
1 2
= 0.0498 × 1 + 3 + 4.5
= 0.0498 × 8.5
= 0.4233
Contd…
b) The probability that the box will not meet guaranteed quality is
P( will not meet guaranteed quality) = P ( more than 2 pins defective)
= P(X>2)
= 1 – P(X≤ 2)
= 1 – 0.4233
= 0.5767
Numerical Example:
An actuary from the Nepal life insurance company has determined that
0.0001 of the elderly population incurs a rare disease each year. A random
sample of 10000 Medicare patient records is to be evaluated, what is the
probability that
a) none of these Medicare patients will have incurred the rare disease.
b) at least two of the Medicare patients will have incurred the rare disease
Solution:
Here, no. of Medicare patients (n) = 10000
Probability of the elderly population incurs a rare disease ( p) = 0.0001
∴ λ = np = 10000× 0.0001 = 1
Let X~P(λ) then
𝒆−𝝀 𝝀𝒙
P(X=x) = p(x) =
𝒙!
𝒆 × 𝟏𝒙
−𝟏
= ; x = 0, 1, 2, ………….∞
𝒙!
a) The probability that none of Medicare patients will have
incurred the rare disease is
𝒆−𝟏 × 𝟏𝟎
P(X=0) = = 𝒆−𝟏 = 0.3679
𝟎!
x = 0, 1, 2, 3, 4, 5,……∞
b)The probability that at least two of the Medicare patients will have
incurred the disease is
P( X≥ 𝟐) = 1 - P(X< 𝟐)
= 1 - [P(X=0) + P(X=1)]
𝒆−𝟏 𝟏𝟎 𝒆−𝟏 𝟏𝟏
=1- [ + ]
𝟎! 𝟏!
10 11
=1- 𝑒 −1 +
0! 1!
1
= 1 - 0.3679× 1 +
1
= 1 - 0.3679 × 1 + 1
= 1 - 0.3679 × 2
= 1- 0.7358
= 0.2642
Numerical example
The number of accidents in a year attributed to taxi drivers in a city
follows Poisson distribution with mean 3. Out of 1000 taxi drivers, find
the approximately number of drivers with (i) no accident in a year (ii)
more than 2 accidents in a year.
Solution:
Here, total number of taxi drivers (N) = 1000
no. of accidents in a year (i.e mean), λ = 3
Let X~P(λ) then
𝒆−𝝀 𝝀𝒙
P(X=x) = p(x) =
𝒙!
𝒆−𝟑 × 𝟑𝒙
= ; x = 0, 1, 2, ………….∞
𝒙!
a)Probability of no accident in a year is
𝒆−𝟑 𝟑𝟎 𝒆−𝟑 ×𝟏
P(X=0) = = = 𝒆−𝟑 = 0.0498
𝟎! 𝟏
Expected no. of drivers with no accident = N × P(X=0)
= 1000 × 0.0498
= 49.8 ≈ 50
b) The probability of more than 2 accidents in a year is
P(X> 𝟐) = 1 - P(X≤ 𝟐)
x = 0, 1, 2, 3, 4, 5,……∞
= 1- [P(X=0) + P(X=1) + P(X=2)]
𝒆−𝟑 𝟑𝟎 𝒆−𝟑 𝟑𝟏 𝒆−𝟑 𝟑𝟐
= 1- [ + + ]
𝟎! 𝟏! 𝟐!
30 31 32 3 9
= 1- 𝑒 −3 + + = 1 - 0.0498× 1 + +
0! 1! 2! 1 2
= 1 - 0.0498 × 1 + 3 + 4.5
= 1- 0.0498 × 8.5
= 1- 0.4233
= 0.5767
The expected number of drivers with more than 2 accidents in a year is
= N × P(X> 𝟐)
= 1000 × 0.5767
= 576.7 ≈ 𝟓𝟕𝟕
Numerical Example
A lottery has very large number of tickets, one in every 500 of which
entitles the purchaser to prize. Find the minimum number of tickets the
agent must sell to have 95% chance of selling at least one prize winning
ticket.
Solution:
1
Here, probability of winning prize (p) = = 0.002
500
number of tickets (n) = ?
Let X~P(λ) then
𝒆−𝝀 𝝀𝒙
P(X=x) = p(x) = ; x = 0, 1, 2, ………….∞
𝒙!
x = 0 , 1 , 2 , 3 , ……..∞
According to question,
P(X≥ 𝟏) = 0.95
1 – P(X<1) = 0.95
1 – P(X=0) = 0.95
P(X=0) = 1-0.95 =0.05
𝒆−λ × λ𝟎
= 0.05
𝟎!
𝒆−λ = 0.05
Taking natural logarithm ln on both sides, we get np = 2.9957
−λ ln e= ln (0.05) (∵ ln e = 1)
n × 0.002 = 2.9957
2.9957
−λ × 1 = −2.9957 𝑛=
λ = 2.9957
0.002
∴ 𝒏 = 1497.86 ≈ 1498
Numerical Example
Fit a suitable distribution to the following data.
X 0 1 2 3 4 Total
f 109 65 22 3 1 200
𝒆−𝟎.𝟔𝟏 (𝟎.𝟔𝟏)𝟎
x=0, f(0) = 200 × = 200 × 𝟎. 𝟓𝟒𝟑𝟒 ×1 = 108.67 ≈ 𝟏09
𝟎!
𝒆−𝟎.𝟔𝟏 (𝟎.𝟔𝟏)𝟏
x=1 , f(1) = 200 × = 200 × 𝟎. 𝟓𝟒𝟑𝟒 ×0.61 =66.288 ≈ 𝟔𝟔
𝟏!
𝒆−𝟎.𝟔𝟏 (𝟎.𝟔𝟏)𝟐
x= 2 , f(2) = 200 × = 200 × 𝟎. 𝟓𝟒𝟑𝟒 ×0.1860 = 20.22 ≈ 𝟐𝟎
𝟐!
𝒆−𝟎.𝟔𝟏 (𝟎.𝟔𝟏)𝟑
x =3 , f(3) = 200 × = 200 × 𝟎. 𝟓𝟒𝟑𝟒 × 𝟎. 𝟎𝟑𝟕𝟖𝟑 = 4.111 ≈ 𝟒
𝟑!
𝒆−𝟎.𝟔𝟏 (𝟎.𝟔𝟏)𝟒
x = 4 ,f(4) = 200 × = 200 × 𝟎. 𝟓𝟒𝟑𝟒 × 𝟎. 𝟎𝟎𝟓𝟕𝟕 = 0.6269 ≈ 𝟏
𝟒!
Hence, required expected frequency distribution is
X 0 1 2 3 4 Total
Expected Frequency 109 66 20 4 1 200