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Unformatted text preview: Chapter 12 Chapter Poisson Approximation to Binomial Distribution Binomial
A. Continuous Base, e Suppose you deposit $1 with the bank. The bank is very generous. It offers you 100% interest, i.e. $1, per annum. By the end of year 1, you have $2 altogether. You ask the bank, “Can I have the interest calculated halfyearly?” and the bank agrees. Then you will get (1+0.50)² =2.25 dollars after 1 year. dollars Suppose you keep asking, “Could you please calculate the interest monthly? daily? by seconds?.....?” and the bank keeps agreeing, you might think you are making a huge fortune. Is that true? Let us do the calculations: You see, there is a limit to the amount receivable. You can never get more than $2.7183. The number The is denoted e, iin honour of the great n mathematician Leonhard Euler (17071783) It is the called the continuous growth base. growth It 1n lim (1 + n ) = 2.718281828.... n →∞ Now suppose the bank offers you Now $x (instead of $1) p.a. Then on a continuous growth, you will get: will
xn 1n 1 x.x (1 + ) = lim (1 + ) = lim (1 + ) = ex lim n n→∞ n n n n →∞ →∞ x x x
n dollars after one year. x The function is called the exponential e function. A similar argument follows for similar negative interest, i.e. if your investment of $1 keeps shrinking. investment Then, Example 1(a) A certain sum of money, $A, is invested in a fund. What will it become after 19 years if : (a) it increases by 0.01% per day? Solution: 0.01%=0.0001; 19 years=6,935 days Amount = Ae 0.0001 x 6935 = Ae 0.6935 ≅ 2A Example 1(b) A certain sum of money, $A, is invested in a fund. What will it become after 19 years if : (b) it reduces by 0.01% per day? Solution: Amount = Ae
− 0.0001 x 6935 = Ae − 0.6935 1 ≅A 2 B. Poisson approximation to Binomial distribution
Let X ~ Bin (n, p) be a binomial variable. Then P( X = x) = ( )p q
n x x n− x , x = 0,1,....., n When n is large, the computation of P(X=x) may be tedious. There are two cases: Case 1: 0.1<p<0.9. Here, p is NOT too far from 0.5. We used the Normal approximation to the Binomial distribution. This was discussed in Ch11. Case 2: p≤0.1 or p≥0.9. ≤0.1 Here p is small (alternatively, is big, which is equivalent to q being small) , a very common situation, especially when p represents the proportion of defectives in a batch of items. Then we need to use another approximation method, to be discussed below: In this case, we can write as P(r ) = ( n ) p r q n − r
r where r =0,1,2,………
The formula (2) is called the Poisson probability formula. It relates to the probability of rare events, such as accidents, deaths. Therefore, it is useful in actuarial science. We shall only look at it as an approximation to the binomial. Note: Although we did not mention it explicitly, we have implicitly assumed in the above argument that as n→∞, p also tends to 0 in such a way that λ=np remains constant. Poisson Approximating Binomial Poisson Theorem (PABT) Theorem
For X ~ Bin(n, p) where n is large and p is small (p≤0.1), we have λx e − λ P( X = x) = , x = 0,1,2,......... x!
where λ=np Example 2:
It is known that 5% of the books bound at a certain bindery have defective pages. Find the probability that 2 out of 100 books bound by this bindery will have defective pages, using (a) The exact method (i.e. the binomial formula) (b) The PABT, and by direct calculation (i.e. without using the Poisson tables) Solution:
Let X=(counting) the number of books with defective pages X ~ Bin (n=100, p=0.05) Find P(X=2)=? Using Binomial formula:
(a) P( X = x) = ( n ) p x q n − x x P ( X = 2) = ( 100 )(0.05) 2 (0.95) 98 2 = 4950 × 0.0025 × 0.00656 = 0.0812 X ~ Bin (n=100, p=0.05) Using Poisson approximation X approx ~Poisson(λ=np=5) =5)
(a) Solution: P( X = x) = x λ e −λ x! 5 2 e −5 25 ×e −5 P ( X = 2) = = =0.0842 2! 2 (from Poisson tables with λ= 5 and r = 2) OR With calculator: OR
x ON SHIFT e ON ( −) 2 5 ) x5 x ÷ 2 EXE 0.0842 Example 3:
A fire insurance company has 3780 policy holders. The probability is 1/1200 that any one of the policyholders will file a claim in any given year. Find the probability that at most 2 of the policy holders will do so next year. Solution:
Let X = (counting) the number of policyholders who make a claim in a particular year. X~Bin(n=3780, p=1/1200) Using Poisson Approx: X approx~Poisson(λ= np =3.15) =3.15) Cannot use Poisson tables here, so must Cannot use calculator: use We need to find P(X≤2)=?
Using Poisson formula:
λx e − λ (3.15) x e −3.15 P( X = x) = = x! x! (3.15) 0 e −3.15 (3.15)1 e −3.15 (3.15) 2 e −3.15 P( X ≤ 2) = + + 0! 1! 2! = 0.04285 + 0.13498 + 0.21260 = 0.3904 Compare with the exact answer:
Using Binomial formula:
P( X = x) = ( n ) p x q n − x x 1 0 1199 3780 3780 1 1 1199 3779 3780 1 2 1199 3778 P( X ≤ 2) = ( 3780 )( )( ) + ( 1 )( )( ) + ( 2 )( )( ) 0 1200 1200 1200 1200 1200 1200 = 0.042795889 + 0.134919483 + 0.212619152 = 0.3903 This is so horrible! Example 4:
The proportion of defectives of a large batch of items is 7.5%. A random sample of 80 items is drawn. If not more than five defective items are found in the sample , then the batch is accepted. What is the probability that the batch is accepted? Solution:
Let X=(counting) the number of defective items X ~ Bin (n=80, p=0.075) Find P(X≤5)=? To give the exact answer we would use the binomial formula, which will take a long time to calculate since we cannot use Binomial tables here. ( )p q P ( X ≤ 5) = ( )(0.075) (0.925) + ( )(0.075) (0.925) + ( )(0.075) (0.925) + ( )(0.075) (0.925) + ( )(0.075) (0.925) + ( )(0.075) (0.925)
P( X = x) =
80 2 80 4 n x x n− x 80 0 0 80 80 1 1 2 4 78 76 80 3 80 5 3 77 5 75 79 So let’s try the Poisson So approximation instead. approximation
X ~ Bin (n=80, p=0.075) X approx~Poisson(λ= np = 6) 6) Find P(X≤5) =P(0)+P(1)+P(2)+P(3)+P(4)+P(5) =0.0025+0.0149+0.0446+0.0892+0.1339+0.1606 =0.4457 from Poisson tables with λ= 6 and r = 0,1,2,3,4,5 This is so much easier!!! Example 5: 92% of the students of a language school can pass its professional test. If 50 students take the test, what is the probability that at least 48 will pass? Solution: Let Y=(counting) the number of students who pass the test Y ~Bin (n=50, p=0.92) We need to find P(Y≥48)=? We could use the Binomial formula We to calculate the exact answer: to
( )p q P( X ≥ 48) = ( )(0.92)
P( X = x) =
n x x n− x 50 48 48 (0.08) 2 + ( )(0.92)
50 49 49 (0.08)1 + ( )(0.92)
50 50 50 (0.08) 0 Which is OK, BUT can we find a quicker way to get the answer? So, let’s try the Poisson approximation…….. Let Y= (counting) the number of students Let who PASS the test who Y ~ Bin (n=50, p=0.92) We need to find P(Y≥48)=? n iis large here, BUT p is NOT small so a Poisson approx s won’t work! But wait a minute! Let’s try the question from another point of view…….. Let’s Let X= (counting) the number of students who FAIL the test X ~ Bin (n=50, p=0.08) X ~ approx Poisson(λ= 4) Now we need to find Now P(X≤2) = 0.0183 + 0.0733 + 0.1465 P(X≤2) = 0.2381 from Poisson tables with λ= 4 and r = 0, 1, 2 and which is so fast!!! which More Examples on Normal approximation More and Poisson approximation to Binomial distribution distribution (Compare Ex 6, 7 and 8) (Compare
Example 6: A lorry load of potatoes has, on the average, one rotten potato in 6. A green grocer tests a random sample of 90 potatoes and decides to turn away the lorry if he finds more than 10 rotten potatoes in the sample. Find the probability that he accepts the consignment. (Use a suitable approximation method) Solution for Example 6: Let X = the number of rotten potatoes X ~ Bin ( n = 90, p = 0.167= ) Reject consignment if P[ X >10 ] Accept if P[X ≤10] It is too difficult to use the Binomial formula here, so do a NORMAL APPROXIMATION TO THE BINOMIAL using a continuity correction factor. When n is large and p is close to 0.5 then do Normal approx to the Binomial distribution. (That is, if 0.1<p<0.9) 1 E(X) = np = 90 x = 15 6 1 5 Var(X) = np(1 p) = 90 x x = 12.5 6 6 ⇒ 1 6 1 ∴ X ~ Bin( n = 90, p = ) ~ approx N( µ =15 , σ² =12.5) 6 10.5 − 15 P[ X ≤ 10] ≅ P[ Z ≤ ] = P[ Z < 1.27 ] 12.5 = 0.50 0.398 = 0.102 Example 7:
The proportion of defectives in a very large batch of items is 0.03. A random sample of size 100 is selected. The batch is accepted if no more than five defectives are found in the sample. What is the probability of accepting the batch? Solution for Example 7: POISSON APPROXIMATION to the BINOMIAL POISSON DISTRIBUTION DISTRIBUTION
Let Y = the number of defectives Y ~ Bin(n=100, p=0.03) When n is large and p ≤ 0.1 then do Poisson approx to Binomial 0.1 then do Poisson approx to Binomial Y ~ approx Poisson (λ=np=100x0.03=3) λ y e − λ 3 y e −3 P[Y = y ] = = ; y = 0,1,2,........ y! y!
P(Y ≤ 5) = P (Y = 0) + P(Y = 1) + P (Y = 2) + P(Y = 3) + P(Y = 4) + P(Y = 5) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 = 0.9160 from the POISSON TABLES distribution. Example 8:
A book containing 500 pages has 60 misprints. Its Chapter 2 has 40 pages. Find the probability that it contains no more than 3 misprints. (Assume misprints occur randomly and independently on each page.) Solution A for Example 8
Let X = the number of misprints in chapter 2 60 X ~ Bin (n=40, p= =0.12) 500 Use the binomial formula to find the probabilities: P[ X = x ] = n p x (1 − p ) n −x = 40 (0.12) x (0.88) 40 −x x x () () (You MUST use your calculator here because p=0.12 is NOT in the binomial tables) ( P ( X = ) =( 1 P ( X =2) =( P ( X =3) =(
P ( X =0) = 40 0
40 1 40 2 40 3 )(0.12) (0.88) )(0.12) (0.88) )(0.12) (0.88) )(0.12) (0.88)
0
1 2 40
39 38 =0.006015986
=0.03281447 =0.08725667 3 37 =0.150716066 P ( X ≤ 3) = P( X = 0) + P ( X = 1) + P( X = 2) + P( X = 3) = 0.276803193
Comment: This is the EXACT answer BUT it did take quite a long time to do! Solution B for Example 8 Let’s try another method which may save time: Let’s
The Normal approximation to the binomial The Normal approximation to the binomial (since n is large and 0.1 < p < 0.9) Don’t forget the continuity correction!
E(X) = np = 40 x 0.12 = 4.8 Var(X) = np(1 p) = 40 x 0.12 x 0.88 = 4.224 ∴ X ~ Bin( n = 40, p =0.12) ~ approx N( µ = 4.8, σ²= 4.224) 3.5 − 4.8
P[ X ≤ 3] ≅ P[ Z ≤ 4.224 = P[ Z < − 0.63] = 0.5 − 0.2357 = 0.2643 Comment: The Normal approximation method gives quite a good approximation to the exact answer and is certainly faster than the exact method. Try the Poisson approximation to the Binomial: Try Solution C for Example 8 Can we find an even faster method? Can Note: n = 40 is large and although p is NOT in between 0.1 and 0.9, p = 0.12 is on the border line of whether to perform a Normal approximation or a Poisson approximation. Also, I feel that this is the perfect situation for a Poisson distribution since you are counting the number of misprints per unit (PER chapter 2 in this case) ∴ X ~ Bin(n = 40, p = 0.12) ~ approx Poisson(λ=μ=np=4.8) P( X ≤ 3) = P( X = 0) + P( X = 1) + P( X = 2) + P ( X = 3) = 0.0082 + 0.0395 + 0.0948 + 0.1517 = 0.2942 from the POISSON TABLES
Comment: This method is just as good as the Normal approx method BUT IS MUCH, MUCH FASTER!!! Note: This course STAT 0301 is presenting you with many methods in which to obtain solutions. It is much better for the student who is able to select the method which is the most efficient. Speed in answering questions is important – the faster you answer this question the more time you have to answer the next question and thus finish the test paper! Summary of Poisson distribution
POISSON
Let X =the number of successes per UNIT X ~ Poisson (λ=μ) x −λ λe µe P[ X = x] = = for x=0,1,2,………….. x! x!
Note: Mean(X)=μ=E(X) = Var(X) =λ x −µ Poisson distribution Example 9: Example
The yearly number X of accidents at a typical traffic intersection in a large city has a mean of μ=2.2. The traffic division of the city keeps records of the yearly number of accidents at each intersection in the city and wishes to be alerted when the number of accidents is unusually large at any intersection. They decide to target an intersection for safety improvement when the number of accidents is 4 or greater. What is the probability that a typical intersection will have 4 or more accidents in a year? Solution for Example 9
Let X=the number of accidents at an intersection PER YEAR X~ Poisson(λ=μ=2.2) 2 x e −2.2 λx e − λ 2. for x=0,1,2,………….. P[ X = x] = = x! x!
P( X ≥ 4) = 1 − P( X ≤ 3) = 1 − {P( X = 0) + P( X = 1) + P( X = 2) + P ( X = 3)} = 1 − {0.1108 + 0.2438 + 0.2681 + 0.1966} = 1 − 0.8193 = 0.1807 (Look up λ = 2.2 and r = 0, 1, 2, 3) from the Poisson tables Poisson distribution Example 10:
Telephone calls enter a college switchboard on the average of two every 3 minutes. What is the probability of five or more calls arriving in a 9 minute period? Let X = number of calls college switchboard receives per 3 MIN λX X ~ Poisson( )= 2 Let Y = number of calls the switchboard receives per 9 MIN λY Y ~ Poisson( )= 6 y −6 λy e −λ 6 e for y=0,1,2,………….. P[Y = y ] = = y! y!
P(Y ≥ 5) = 1 − P(Y ≤ 4) = 1 − {P(Y = 0) + P(Y = 1) + P (Y = 2) + P (Y = 3) + P (Y = 4)} = 1 − {0.0025 + 0.0149 + 0.0446 + 0.0892 + 0.1339} = 1 − 0.2851 = 0.7149 Look up Poisson tables λ = 6 and r = 0, 1, 2, 3, 4 Solution for Example 10 Poisson distribution Example 11:
The probability that a tornado will occur in a particular area of the country is considered to be a constant 0.000444 square miles per year. City A has an area of 40 square miles. What is the probability that City A will experience at least one tornado during the next 5 years? Solution for Example 11
Let X = number of tornados that City A experiences PER YEAR X ~ Poisson (λ = μ= n p= 40 x 0.000444 = 0.01776) Let Y = number of tornados that City A experiences PER 5 YRS Y = X1 + X 2 + X 3 + X 4 + X 5 Y ~ Poisson (λ = 40 x 0.000444 x 5 = 0.0888) λ y e − λ 0.0888 y e −0.0888 P[Y = y ] = = for y=0,1,2,………….. y! y! P (at least 1 tornado) = 1 – P ( No tornadoes ) = 1 0.915028561 = 0.08497
Note: Both spellings – tornados and tornadoes – are correct for the plural of tornado. ...
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This note was uploaded on 02/22/2010 for the course FBE STAT0302 taught by Professor Unknown during the Spring '10 term at HKU.
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