This document presents information about the Poisson distribution. It defines key properties of the Poisson distribution, including that the mean and variance of a Poisson distribution are equal to the parameter λ. It then works through 6 example problems calculating probabilities for various Poisson distributions based on given values of λ. The problems calculate probabilities of certain numbers of events occurring, like errors on a page or accidents on a highway, given the average rate of occurrences.

1. BUSINESS STATISTICS PRESENTATION
– POISSON DISTRIBUTION
Submitted By:
Reshmi Raveendran
212027

2. POISSON DISTRIBUTION
Poisson distribution is a limiting case of Binomial distribution
in which :
- The number of trials are indefinitely large i.e. n → ∞
- Constant probability of success for each trail is very small i.e. p
→0
- If X has a poisson distribution, then mean of X = λ
x = 0, 1,..
- Probability mass function (PMF) = P(x) = e-λ * λx
x!

3. Question 1
Number of errors on a single page has poisson distribution with
average number of errors of 1 per page. Calculate the
probability that there is at least one error on a page?
Sol:
PMF = P(x) = e-λ * λx
x = 0, 1, ……
x!
Where λ is called parameter of the distribution
Here λ = 1 since average number of errors per page is 1
Now P(X>1) = 1 – P(X=0)
= 1 – e-1
= .632

4. Question 2
Number of accidents on an express-way each day is a
poisson random variable with average of 3 accidents per
day. What is the probability that no accidents with occur
today?
Sol:
PMF = P(x) = e-λ * λx
x!
x = 0, 1, …..
Where λ is called parameter of the distribution
Here λ = 3 since average number of accidents per day is 3
Now P(X=0) = e-3 = 0.0498

5. Question 3
A car – hire firm has two cars, which it hires out day by day. The
number of demands for a car on each day is distributed as a poisson
distribution with mean 1.5. Calculate the proportion of days on which
neither car is used and the proportion of days on which some demand
is refused. (e-1.5) = .2231
Sol:
PMF = P(x) = e-λ * λx
x = 0, 1, ….
x!
Here λ = 1.5
Now, proportion of days on which neither car is used :
P(X=0) =
e -1.5(1.5)0
0!
= 0.2231

6. Proportion of the days on which some demand is refused
:
=P(x>2) = 1- P(x<2) = 1- P( x = 0 or x = 1 or x =2 )
= 1 – P(x=0) + P(x=1) + P(x=2)
= 1 – e-1.5(1.5)0 + e-1.5(1.5)1
0!
1!
= 1 – e-1.5 (1 + 1.5 + 2.25/2)
= 1 – (0.2231) (3.625) = 0.1913
+
e-1.5(1.5)2
2!

7. Q. If X has a poisson distribution and P(X=0)=1/2.
What is E(X)?
Sol. Mean of X = E(X) = λ
Since it a poisson distribution, its probability mass function (pmf)
is given by:
P(X) = e- λ λx
X!
1/2 = e- λ λ0
0!
1/2 = e- λ
loge (1/2)= - λ loge e (Taking log on both sides)
loge 1 – loge 2 = - λ (loge e =1)
loge 2 – loge 1 = λ
Hence, λ = log 2 or 0.693

8. Q. If X has a poisson variate such that
P(X=2)= 9 P(X=4)+ 90 P(X=6). Find P(X).
Soln. Using poisson distribution,
P(X) = e- λ λx
X!
e- λ λ2 = 9 e- λ λ4 + 90 e- λ λ6
2!
4!
6!
λ4 + 3 λ2 – 4 = 0
(Solve using quadratic method)
λ2=1
Or λ = 1 (λ cannot be negative)

9. THANK YOU