The probability distribution for a Poisson random variable X iS
例一.
A manufacturer of a certain type of LCD-screen finds that the number of “deadspots” on a screen is a Po(0.8) random variable. A screen with more than 3 “deadspots” has to be sold at a discount. Find the proportion of screens that are sold at a discount.
Solution
Let X be the number of “dead spots” on a screen. Then X ~ Po(0.8).
The proportion of screens that are sold at a discount
= P( X > 3)
= 1 – p(0) – p(1) – p(2) – p(3)
= 0.00908
例二.
A radio active source is emitting α-particles at an average of 2.6 particles per minute.Find the probability that the number of particles emitted in one minute is
(a) 0
(b) 1
(c) more than 1
Solution
Let X be the number of particles emitted in one minute.
(c) P(X > 1) = 1 – 0.0743 – 0.193 = 0.7327
Poisson Approximation to Binomial Distribution
A rule of thumb for the conditions under which this approximation may be used is
n ≥ 100, np <>
例一.
A company ships memory chips in boxes of 200. It is known that 0.001 of all thechips are defective. Find the probability that less than 3 chips in a box aredefective.
Solution
Let X be the number of defective chips in a box. Then X ~ Bin(200, 0.001).The average number of defective chips in a box = 200 × 0.001 = 0.2The distribution of X may be approximated by Po(0.2).
Let X be the number of defective chips in a box. Then X ~ Bin(200, 0.001).The average number of defective chips in a box = 200 × 0.001 = 0.2The distribution of X may be approximated by Po(0.2).





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