Question

co id Assume that X1,..., X. V(X) = 1. Poisson(X). The following facts about the Poisson distribution may be useful: E(X) =

Answer 1

finswer Summarized the letters, alphabets and symbols provided in the problem, @ Possible passwords using (16,5 length) 26 different letters, ten digits and 14 symbols; total symbols = 26+14 +10 combinations a 3 (letter length) +4+5 +15014 + - 50: (50)3 (50)5 - 125000 + 6250000 + 312500000 Ans. 318,875,000 ul

(2 Since, for the above Poisson probleme X,,---, Xn are lid and E(X) = Var(83) = Loo &k=1(1) - accooding to CLT, Sn-n Elxx) & NCoisas In var (XK) where x = sny un (Fox) JA Jolie .Xx -> asy GN (0,1) as justified ) + Now we define what the pivotal quantity's A sample &= (x,--, Xn) from a population p, governed by unknown parameter o is choosen. À function h(X,,---,n;o) is called pivotal if the distribution ofh (xo doesn't depepend on o Based on this definition, we can suggest h(x, ...- Xn; 4) = sn (x-x) Ja pivot for confidence interval of x. because the asymptotite dist of h(x,---, Xn;!) is (0,1) and Independent of x, (from *) as a

3.8) Nowo, che have already guessed h(x10-80% ) and its asymptotic disth. Now for 95% confidence interval if we use the above prvot, it will be problamatic to seperate ž ex- ~ To resolve the problem; we will are variance stabilizing transformation, to find another pivot using square-root transformation. Xin Poisson (2) Hi=16n. n is large (x;-) BN (0,x) Let gc.) be a func: 7 g 6 exists at x, then by delta method, g(xi)-g (x) (6, & (6 (2) 2) take, ( gt (x))? = e, >9! (X) = e = g() = Retk Put e=1/k=0 3 Varg (xi)) becomes constant ... g(^)= X => 9 (xi) = 5xi ti = (5xi_57) D > N(0,7) -74 (JX; -JA)* nxn as n => 0 as no

here a is unknown and estimated by; Jх, (â)- Sxn Yu Yu ĥ å sri + + n i=1 14 and E(F) = 0 So under null, T=4 (588- São Xn- so, 100 (1-x)% confidence interval fort is obtainedo Ñ by n (san) —> (* =) i sa Cafe 1-2 Yun K=005,1-0=0.95 | the new pivotis ht (x,x)= Ę Take=71095 î -5T upper 0.025 point Yun of N(o,)) = Santi Sai –v)3N (0,1) T0025 ༢ Jཏོ -0.95 1. C.Iis 70025 with 250 confidence coeffecient 95% (ans) 0.02 20 ñ = ) [(17- a) Cão 025 + 25n

3.9) Interpretation: the above given interval will contain the true value of a with confidence coefficient 95%. as n is large *
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