# CS代写 F71SM STATISTICAL METHODS – cscodehelp代写

F71SM STATISTICAL METHODS
Tutorial on Section 4 SPECIAL DISTRIBUTIONS (discrete distributions): SOLU- TIONS
1. X ∼ b(20, 0.6), Y ∼ b(20, 0.4)
P(X ≥ 12) = P(Y ≤ 8) = 0.5956 (NCST p23)
P(X≤12)=P(Y ≥8)=1−P(Y ≤7)=1−0.4159=0.5841(NCSTp23)
−λ λx+1 λ −λ λx λ
2. P(X=x+1)=e (x+1)!=x+1e x! =x+1P(X=x),x=0,1,2,…
f(x)≥f(x+1)⇔f(x)≥ λ f(x)⇔x≥λ−1 x+1
f(x−1)≤f(x)⇔f(x−1)≤ λxf(x−1)⇔x≤λ
So mode satisfies λ − 1 ≤ mode ≤ λ
For instance, Poi(6.4) has mode x = 6; Poi(8) has modes at x = 7 and x = 8.
􏰜n􏰛 􏰆1􏰇r 􏰆3􏰇n−r 􏰜n􏰛 􏰆1􏰇n
3. pA = r 4 4 , pB = r 2 , r = 0, 1, . . . , n.
P (same) = P ({00, 11, 22, 33})
= 0.4219 × 0.1250 + 0.4219 × 0.3750 + 0.1406 × 0.3750 + 0.0156 × 0.1250 = 0.266(= 17/64)
4. (a) P (he needs more than 3 attempts) = P (first 3 attempts are failures) = 0.43 = 0.064
(b) E[value of present] = 􏰊∞k=1 0.9k × 0.6 × 0.4k−1 = 0.54 􏰊∞k=1 0.36k−1 = 0.84375 That is, £843.75.
(c) Let X be number of attempts required to pass. P(X > k) = P(first k are failures) = 0.4k
0.54
1 − 0.36
= 27 = 32
P(X > 6 and X > 2) P(X > 6) 0.46
P(X >6|X >2)= P(X >2) = P(X >2) = 0.42 =0.44 =0.0256
5. Let X be number of passengers with reservations who fail to turn up. X ∼ bi(310,0.04) ∼ Poi(12.4) approx.
P (airline has to ‘bump’ at least one passenger off) = P (X ≤ 9) = 0.209 (from Poisson tables). tX√
6. MX(t)=exp(λ(e −1))andZ=√λ− λ,so
MZ(t) = ⇒lnMZ(t) =
= ⇒limlnMZ(t)=
−t√λ 􏰄√􏰅 􏰄√􏰅 􏰄􏰄t/√λ 􏰅􏰅 􏰄√ 􏰄t/√λ 􏰅􏰅 e MX t/ λ =exp −t λ exp λ e −1 =exp −t λ+λ e −1
√􏰄t/√λ􏰅√􏰜tt2t3 􏰛 −t λ+λ e −1 =−t λ+λ √λ+2λ+3!λ3/2 +···
t2􏰜t3 􏰛 2+ 3!√λ+···
2
X ∼ N (λ, λ) asymptotically for large λ.
λ→∞
t2 2
⇒MZ(t)→et /2 asλ→∞ Limiting distribution of Z is N(0,1).
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