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Method 1 If X is binomially distributed, f (x) Then the moment generating function is given by M(t) E(etx) a etxf (x) n n a etx a x b pxqn n n a a x b( pet)xqn
n a bpxqn x
(q Method 2 For a sequence of n Bernoulli trials, define Xj e 0 1 if failure in jth trial if success in jth trial X1 et0 q X2 et1 p c q pet)n
1, 2, . . . , n) Then the Xj are independent and X Mj (t) Then by Theorem 39, page 80, Xn. For the moment generating function of Xj , we have pet (j 1, 2, . . . , n) M(t) M1(t)M2(t) cMn(t) pet)n
4.8. Prove that the mean and variance of a binomially distributed random variable are, respectively, 2 npq. Proceeding as in Method 2 of Problem 4.7, we have for j E(Xj) Var (Xj) E[(Xj p2q Then s2 m E(X ) 0q p)2] q2p 1, 2, . . . , n, 1p (0 pq( p E( X2) Var ( X2) p p)2q q) c c (1 pq E(Xn) np npq p)2p np and
E( X1) Var ( X1) Var (X ) Var ( Xn) where we have used Theorem 37 for 2. The above results can also be obtained (but with more difficulty) by differentiating the moment generating function (see Problem 3.38) or directly from the probability function. 4.9. If the probability of a defective bolt is 0.1, find (a) the mean, (b) the standard deviation, for the number of defective bolts in a total of 400 bolts. (a) Mean (b) Variance np
(400) (0.1) npq
40, i.e., we can expect 40 bolts to be defective. 36. Hence, the standard deviation s !36 6. (400)(0.1)(0.9) CHAPTER 4 Special Probability Distributions
The law of large numbers for Bernoulli trials 4.10. Prove Theorem 41, the (weak) law of large numbers for Bernoulli trials. By Chebyshev s inequality, page 83, if X is any random variable with finite mean (1) P( u X mu ks) 1 k2 np, s 1 k2 1 k2 !npq and (1) becomes and variance then
In particular, if X is binomially or Bernoulli distributed, then m (2) or (3) pq , (3) becomes An X Pa 2 n p2 k pq b An P( u X np u k!npq ) If we let P
and taking the limit as n S ` we have, as required, The result also follows directly from Theorem 319, page 83, with Sn
4.11. Give an interpretation of the (weak) law of large numbers for the appearances of a 3 in successive tosses of a fair die. The law of large numbers states in this case that the probability of the proportion of 3s in n tosses differing from 1> 6 by more than any value P 0 approaches zero as n S ` . X lim P a 2 n
X Pa 2 n
pq nP2
0 X, m np, s !npq.
The normal distribution 4.12. Find the area under the standard normal curve shown in Fig. 43 (a) between z 0 and z (b) between z 0.68 and z 0, (c) between z 0.46 and z 2.21, (d) between z 0.81 and z (e) to the right of z 1.28. 1.2, 1.94, (a) Using the table in Appendix C, proceed down the column marked z until entry 1.2 is reached. Then proceed right to column marked 0. The result, 0.3849, is the required area and represents the probability that Z is between 0 and 1.2 (Fig. 43). Therefore, P(0 Z 1.2) 1 22p 30 e

