S515 HW#7 Solution
6.72 The pdf is f(y)= 1, and the cdf is F(y) = y for 0 ≤ y ≤ 1,
a. U_1=min(Y_1,Y_2), its pdf is , 0 ≤ u ≤ 1.
b. E(U_1)=Int_0^1 u2(1-u) du= [u^2-(2/3)u^3]_{0, 1}=1-2/3= 1/3.
E(U_1)^2=Int_0^1 u^2 2(1-u) du= [(2/3)u^2-(2/4)u^4]_{0, 1}=2/3-1/2= 1/6.
V(U_1)= (1/6) – (1/3)^2 = 1/6-1/9= 1/18.
6.74 Since the pdf of Y is f (y) = 1/θ for 0 ≤ y ≤ θ, the cdf is F(y) = y/θ for 0 ≤ y ≤ θ,
a. , 0 ≤ y ≤ θ.
b. , 0 ≤ y ≤ θ.
c. It is easily shown that E(Y(n)) = , V(Y(n)) = .
6.76 Following Ex. 6.74 with f (y) = 1/θ for 0 ≤ y ≤ θ,
a. By Theorem 6.5, , 0 ≤ y ≤ θ.
b. E(Y(k)) = . To evaluate this integral, apply the transformation z = and relate the resulting integral to that of a beta density with α = k + 1 and β = n – k + 1. Thus, E(Y(k)) = .
c. Using the same techniques in part b above, it can be shown that so that V(Y(k)) = .
d. E(Y(k) – Y(k–1)) = E(Y(k)) – E(Y(k–1)) = – = . Note that this is constant for all k, so that the expected order statistics are equally spaced.
6.79 The joint density of Y(1) and Y(n) is given by (see Ex. 6.77 with j = 1, k = n),
, 0 ≤ y1 ≤ yn ≤ θ.
Applying the transformation U = Y(1)/Y(n) and V = Y(n), we have that y1 = uv, yn = v and the Jacobian of transformation is v. Thus,
, 0 ≤ u ≤ 1, 0 ≤ v ≤ θ.
Since this joint density factors into separate functions of u and v and the support does not depend each other, thus Y(1)/Y(n) and V = Y(n) are independent.
6.85 If Y_1 and Y_2 are independent uniform (0,1), the joint density of Y(1) and Y(2) is given by
, 0 ≤ y1 ≤ y2 ≤ 1.
Thus, P(2Y(1) < Y(2)) = = .5.
6.98 The joint distribution of Y1 and Y2 is , y1 > 0, y2 > 0. Let U1 = , U2 = Y2. The inverse transformations are y1 = u1u2/(1 – u1) and y2 = u2 so the Jacobian of transformation is
.
Thus, the joint distribution of U1 and U2 is
=, 0 ≤ u1 ≤ 1, u2 > 0.
Therefore, the marginal distribution for U1 is
= 1, 0 ≤ u1 ≤ 1.
Note that the integrand is a gamma density function with α = 1, β = 1 – u1.
6.108 The system will operate provided that C1 and C2 function and {C3 or C4 functions}.
P{ C3 or C4 functions}=1- P{ C3 and C4 does not function}=1-F(y)F(y)
Defining the system as S, then
P(S)=[1-F(y)][1-F(y)][1-F(y)**2]=[1-F(y)]**2 [1-F(y)**2]
6.114 The volume of the sphere is V = , or R = , so that . Thus, , 0 ≤ v ≤ .
7.42 Let denote the sample mean strength of 100 random selected pieces of glass. Thus, the quantity (– 14.5)/.2 has an approximate standard normal distribution.
a. P(> 14) ≈ P(Z > 2.5) = .0062.
b. We have that P(–1.96 < Z < 1.96) = .95. So, denoting the required interval as (a, b) such that P(a < < b) = .95, we have that –1.96 = (a – 14)/.2 and 1.96 = (b – 14)/.2. Thus, a = 13.608, b = 14.392.
7.52 Let denote the average resistance for the 25 resistors. With μ = 200 and σ = 10 ohms,
a. P(199 ≤ ≤ 202) ≈ P(–.5 ≤ Z ≤ 1) = .5328.
b. Let X = total resistance of the 25 resistors. Then,
P(X ≤ 5100) = P(≤ 204) ≈ P(Z ≤ 2) = .9772.
7.81 Let Y = # of non–conforming items in our lot. Thus, with n = 50:
a. With p = .1, P(lot is accepted) = P(Y ≤ 5) = P(Y ≤ 5.5) = = = .5948.
b. With p = .2 and .3, the probabilities are .0559 and .0017 respectively.
7.96 Note that Y has a beta distribution with α = 3 and β = 1. So, μ = 3/4 and σ2 = 3/80. By the Central Limit Theorem, = .9484.