F71SM STATISTICAL METHODS
Tutorial on Section 4 SPECIAL DISTRIBUTIONS (continuous distributions): SO- LUTIONS
1. (a)
MX(t) = eμt+σ2t2/2, and Z ∼ N(0,1) so
2. (a)
P(life of one tube exceeds y) = P(X > y) = exp(−0.01y) (cdf of exponential distribu- tion)
P (Y > y) = P (life of each of 5 tubes exceeds y) = (exp(−0.01y))5 = exp(−0.05y) ⇒ Y is exponentially distributed with mean 20 hours.
t2/2 MZ (t) = e
t2 1 t22 1 t23
= 1 + 2 + 2! 2 + 3! 2 + · · ·
t2 t4 t6
= 1+2+8+48+···
t2 t4 t6
= 1+2!+3 4! +15 6! +···
E[Zk] is coefficient of tk/k!, so that E[Z3] = 0 ⇒ E[(X − μ)3] = 0 E[Z4] = 3 ⇒ E[(X − μ)4] = 3σ4 E[Z5] = 0 ⇒ E[(X − μ)5] = 0
(b) E[Z6]=15⇒E[(X−μ)6]=15σ6
(b) Foronedevice,P(Y <40)=1−exp(−0.05×40)=1−exp(−2)=0.8647.
Number of devices that fail within 40 hours is distributed as Bin(3, 0.8647).
So P (2 or 3 devices fail) = 3 × 0.86472 × 0.1353 + 0.86473 = 0.950 2
(c) (i)P(X
1
⇒ 5.06 − μ − 4.94 − μ = 3.42 ⇒ 0.12 = 3.42 ⇒ σ = 0.12 = 0.035 σσ σ 3.42
⇒ 5.06−μ =1.25⇒μ=5.016 0.035
So μ = 5.016mm, σ = 0.035mm
(b) E[sale value of bearing] = 8 × 0.88 + 0.5 × 0.12 = 7.1p
⇒ E[profit] = 100 000 × (7.1 − 5) = 210 000p = £2100
4.99 − 5.016 5.01 − 5.016
(c)P(4.99 < X < 5.01) = P 0.035