Poisson approximation to a Binomial distribution

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Poisson approximation to a Binomial distribution
The Poisson distribution is often used as an approximation to the Binomial distribution for large n and small p, as the Poisson probabilities are easier to calculate.  The Poisson approximation to a Binomial distribution may be made provided n is large. In addition p must be quite small, otherwise the two distributions would not exhibit similar amounts of positive skew.

Poisson approximation to a Binomial distribution
n should be quite large n > 50 p should be quite small p < 0.1 np should not be too big np < 5 X ~ Bin(n, p) approximated by Po(np) where  = np.

Example 1 A factory packs bolts in boxes of 500. The probability that a bolt is defective is Find the probability that a box contains 2 defective bolts.

Solution Let X be the r.v. ‘the number of defective bolts in a box’.
This is a Binomial distribution, with n = 500 and p = So X ~ Bin( 500, )

Method 1: Using the Binomial distribution
P(X = x ) = nCx.px.qn-x, x = 0, 1, 2, …….,n. We have n = 500, p = and q = 0.998, so  P(X = 2 ) = 500C2.(0.002)2.(0.998)498 (500  499)/(2) (0.002)2(0.998)498 = (3 s.f.)

Method 2: Since n is large and p is small, we use the Poisson distribution approximation The parameter  = np = 500  = 1 So x ~ Po(1) and P(x = r ) = P(X = 2) = e-11/2! = (3 s.f.) Note: the answer agree to 3 S.F. and the calculations were much easier in method 2.

Example 2 Find the probability that at least two double sixes are obtained when two dice are thrown 90 times. Solution Throw two dice, P( double 6 ) = 1/6 1/6 = 1/36. Let X be the r.v. ‘the number of double sixes obtained when two dice are thrown 90 times’, then X ~ Bin( 90, 1/36 ) and np = 90  1/36 = 2.5

Using the Poisson approximation,
X ~ Po( 2.5 ) and P(X  2) = 1 – P(x  1) = 1- [ e-2.5 ( /1 )]= (3 s.f) Tables

Example 3 In a large town, one person in 80, on the average, has blood of type X. If 200 blood donors are taken at random, find an approximation to the probability that they include at least five persons having blood of type X. How many donors must be taken at random in order that the probability of including at least one donor of type X shall be 0.9 or more?

Solution Let Y be the random variable 'the number of blood donors of type X'. Then Y ~ Bin (n, p) where n = 200 , p = 1/80 We are required to find P (Y  5) and since the criteria for using the Poisson approximation to the binomial distribution are satisfied, this can be approximated to Y ~ Po () where  = np = 2.5

P( Y ≥ 5 ) = 1 – P( Y ≤ 4 ) = 1 – = = ( 3 s.f.).

For the second part, we still have p = 1/80 but it is the value of n (sample size) that is unknown, and we are looking for n such that P (at least one donor has type X blood) is 0.9 or more; i.e. P (X  1)  0.9 using the Poisson approx again  = np = n/80 and so P (Y  1)  is equivalent to

1 – P (Y = 0)  0.9  1 – e–  0.9  1 – e–n/80  0.9 i.e. e–n/80   e n/80  1/0.1  e n/80  10 take logs to base e; log e n/80  loge10 n/80  giving n  (80)(2.30) giving: n  So we need to take at least 185 donors in order that the probability of including at least one donor of type X is 0.9 or more.

Exercises Question1 In a certain manufacturing process the proportion of defective articles being produced is 2%. In a batch of 300 articles, find the probability that: (a) there are fewer than 2 defectives (b) there are exactly 4 defectives.  Answer

Question 2 A medical practice screens a random sample of 250 of its patients for a certain condition which is present in 1.5% of the population. Find the probability that they obtained (a) no patients with the condition (b) at least two patients with the condition. Answer

Question 3 An experiment involving two fair dice is carried out 180 times. The dice are placed in a container, shaken and the number of times a double six is obtained is recorded. Find the probability that the number of times a double six is obtained is: (a) once (b) twice (c) at least three. Answer

Ans1  = np = 300  0.02 = 6 P(X < 2 ) = from tables (b) P(X = 4 ) = P(X4) –P(X3) = – = = (3sf) Go back

Ans2 :  = np = 250  = 3.75 (a)   P(X = 0) = e-3.75= P(X ≥ 2) = 1 – [P(X=0)+P(X=1)] = 1 – [ e-3.75  3.75 ] = 0.888 Go back

Ans3 :  = np = 180  1/36 = 5 (a) P(X=1) = e-5  5 = 0
Ans3 :  = np = 180  1/36 = 5 (a)   P(X=1) = e-5  5 = (b) P(X=2) = e-5  25/2 = (c) P(x ≥ 3 ) = 1 – P(X ≤ 2 ) = 1 – = = (3sf) Go back