Introduction:
Example 1: If a publisher of technical books takes a great pain to ensure that his books are free of typological errors, so that the probability of any given page containing atleast one such error is 0.005 and errors are independent from page to page
(i) what is the probability of its 400 page novels will contain exactly one page with error.
(ii) atmost three pages with errors.
[e−2 = 0.1353 ; e−0.2. = 0.819].
Solution :
n = 400 , p = 0.005
np = 2 = λ
(i) P(one page with error) = P(X = 1) = e−λ λ1/1! = e-2 21/1!
= 0.1363 × 2 = 0.2726
(ii) P(atmost 3 pages with error) = P(X ≤ 3)
= Σ e−λ λx / x! [limits 0 to 3]
= `sum` e−2 (2)x / x!
= e2 [1 +2/1! + 22/2! + 23/3!]
= e−2 (19/3 )= 0.8569
Hope you liked the above example. Please leave your comments, if you have any doubts.
Definition: A random variable X is a Poisson distribution if the probability mass function of X is P(X = x) =e−λ λx / x!, x = 0,1,2, …for some λ > 0
The mean of Poisson Distribution denoted by λ, and the variance is denoted by λ.
The parameter of Poisson distribution is λ.
The mean of Poisson Distribution denoted by λ, and the variance is denoted by λ.
The parameter of Poisson distribution is λ.
The Poisson distribution is a restrictive case of binomial distribution table under the following conditions.
(i) Number of trials(n) is indefinitely huge(large), that is n → ∞.
(ii) The constant probability(p) of success in each trial is very less.
(i) Number of trials(n) is indefinitely huge(large), that is n → ∞.
(ii) The constant probability(p) of success in each trial is very less.
ie., p → 0.
(iii) np = λ is finite where λ is a positive real number. When an event occurs rarely, the distribution of such event may beassumed to follow a Poisson distribution. The poisson distribution examples is give below..
(iii) np = λ is finite where λ is a positive real number. When an event occurs rarely, the distribution of such event may beassumed to follow a Poisson distribution. The poisson distribution examples is give below..
(i) what is the probability of its 400 page novels will contain exactly one page with error.
(ii) atmost three pages with errors.
[e−2 = 0.1353 ; e−0.2. = 0.819].
Solution :
n = 400 , p = 0.005
np = 2 = λ
(i) P(one page with error) = P(X = 1) = e−λ λ1/1! = e-2 21/1!
= 0.1363 × 2 = 0.2726
(ii) P(atmost 3 pages with error) = P(X ≤ 3)
= Σ e−λ λx / x! [limits 0 to 3]
= `sum` e−2 (2)x / x!
= e2 [1 +2/1! + 22/2! + 23/3!]
= e−2 (19/3 )= 0.8569
Hope you liked the above example. Please leave your comments, if you have any doubts.
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