ssrs 2008 r2 barcode font Special Probability Distributions in Software

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CHAPTER 4 Special Probability Distributions
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Then 0.5 in standard units Required probability 0.5 20 4.08 (area between z (area between z 0.5 0.1443 5.02. 18.5 in standard units 5.02 and z 0.37 0.37)
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(area under normal curve between z 0 and z 0 and z 0.3557 5.02) 0.37)
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(b) We proceed as in part (a), replacing 18 by 14. Then 0.5 in standard units Required probability 5.02 14.5 in standard units 14.5 20 4.08 5.02 and z 1.35 1.35)
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(area under normal curve between z (area between z (area between z 0.5 0.4115 0 and z 0 and z 0.0885 5.02) 1.35)
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It follows that if we were to take repeated samples of 120 tosses of a die, a 4 should turn up 14 times or less in about one-tenth of these samples.
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The Poisson distribution 4.20. Establish the validity of the Poisson approximation to the binomial distribution.
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If X is binomially distributed, then (1) where E(X) np. Let P(X np so that p x) P(X x) n a b p xq n x
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>n. Then (1) becomes l n nb
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n l x a b a n b a1 x n(n a1 1)(n 1 n b a1
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2) c (n x!nx 2 c a1 nb x!
lx a1 1 b
l n nb
lx a1
l n nb
Now as n S ` ,
while a1 l n nb
1 n b a1
2 c a1 nb
l n n b a1 l nb
1 S b 1
S (e l)(1)
using the well-known result from calculus that
lim a1
u n nb
It follows that when n S ` but (2) which is the Poisson distribution.
stays fixed (i.e., p S 0), P(X lxe x) S x!
CHAPTER 4 Special Probability Distributions
Another method The moment generating function for the binomial distribution is (3) If (4) As n S ` this approaches (5) el(et
(q np so that p
pet)n
pet)n
p(et
1)]n
>n, this becomes c1 l(et n 1) d
which is the moment generating function of the Poisson distribution. The required result then follows on using Theorem 3-10, page 77.
4.21. Verify that the limiting function (2) of Problem 4.20 is actually a probability function.
First, we see that P(X x)
` x 0
0 for x x)
0, 1, . . . , given that lxe a x! x 0
0. Second, we have e
a P(X and the verification is complete.
lx a x!
el
4.22. Ten percent of the tools produced in a certain manufacturing process turn out to be defective. Find the probability that in a sample of 10 tools chosen at random, exactly 2 will be defective, by using (a) the binomial distribution, (b) the Poisson approximation to the binomial distribution.
(a) The probability of a defective tool is p 0.1. Let X denote the number of defective tools out of 10 chosen. Then, according to the binomial distribution, P(X (b) We have np P(X (10)(0.1) x) 2) a 10 b(0.1)2(0.9)8 2 0.1937 or 0.19
1. Then, according to the Poisson distribution, P(X 2) (1)2e 1 2! np 5. 0.1839 or 0.18
lxe l or x! In general, the approximation is good if p
0.1 and
4.23. If the probability that an individual will suffer a bad reaction from injection of a given serum is 0.001, determine the probability that out of 2000 individuals, (a) exactly 3, (b) more than 2, individuals will suffer a bad reaction.
Let X denote the number of individuals suffering a bad reaction. X is Bernoulli distributed, but since bad reactions are assumed to be rare events, we can suppose that X is Poisson distributed, i.e., P(X (a) (b) P(X 2) x) lxe x!
where 3) [P(X c 20e 0!
(2000)(0.001) 0.180 1) 22e 2 d 2! P(X 2)]
P(X 1 1 1
23e 3! 0)
P(X 21e 1!
An exact evaluation of the probabilities using the binomial distribution would require much more labor.
The central limit theorem 4.24. Verify the central limit theorem for a random variable X that is binomially distributed, and thereby establish the validity of the normal approximation to the binomial distribution.
CHAPTER 4 Special Probability Distributions
np)> !npq, and the moment generating function for X* is
np)>1npq)
The standardized variable for X is X* E(etX*)
E(et(X e e e e [e (qe
tnp>1npq
E(etX>1npq )
n n a etx>1npq a bpx qn x x 0 n n a a b( pet>1npq)x qn x 0 x
tnp>1npq
tnp>1npq
tnp>1npq(q tp>1npq(q tp>1npq
pet>1npq)n pet>1npq)]n petq>1npq)n
Using the expansion eu we find qe
tp>1npq
u2 2! tp
u3 3!
petq>1npq
qa1 pa1 q 1 p t2 2n a1
!npq tq
t2p2 2npq
cb cb
!npq pq(p q)t2 2npq c t2 2n cb
t2q2 2npq c
Therefore,
E(etX*)
But as n S ` , the right-hand side approaches et2/2, which is the moment generating function for the standard normal distribution. Therefore, the required result follows by Theorem 3-10, page 77.
4.25. Prove the central limit theorem (Theorem 4-2, page 112).
For n Thus 1, 2, . . . , we have Sn X1 X2 E(X1) c Xn. Now X1, X2, . . . , Xn each have mean and variance s2. c E(Xn) nm
E(Sn) and, because the Xk are independent, Var (Sn)
E(X2)
Var (X1)
Var (X2)
Var (Xn)
It follows that the standardized random variable corresponding to Sn is S* n The moment generating function for
* E(etSn)
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