STA 130A Homework #3 Solutions. Textbook Problems: Exercise 3.3-2. (a) (i) 1 =
∫ c. 0 x3/4 dx = x4/16∣∣ c. 0. = c4/16, so c4 = 16 or c = 2. (ii) F(x) = 0 when x ...
STA 130A Homework #3 Solutions Textbook Problems: Exercise 3.3-2 c Rc (a) (i) 1 = 0 x3 /4 dx = x4 /16 0 = c4 /16, so c4 = 16 or c = 2. Rx (ii) F (x) = 0 when x < 0; F (x) = 0 t3 /4 dt = t4 /16 when 0 ≤ x ≤ 2; F (x) = 1 when x > 2. (iii) See below.
(b)
c (3/16)x2 dx = x3 /16 −c = c3 /8, so c3 = 8 or c = 2. x Rx (ii) F (x) = 0 when x < −2; F (x) = −2 (3/16)t2 dt = t3 /16 −2 = −2 ≤ x ≤ 2; F (x) = 1 when x > 2. (i) 1 =
Rc
−c
x3 16
+
1 2
when
(iii) See below.
(c)
√ √ 1 c/ x dx = 2c x|0 = 2c so c = 1/2. This p.d.f. is unbounded. √ x √ Rx 1 (ii) F (x) = 0 when x < 0; F (x) = 0 2√ dt = t 0 = x when 0 ≤ x ≤ 1; F (x) = 1 t when x > 1. (i) 1 =
R1 0
(iii) See below.
Exercise 3.3-4 R2 (a) (i) µ = 0 x.x3 /4dx = (1/4).(1/5).x5 |20 = 1.6 R2 (ii) σ 2 = E(X 2 ) − µ2 = 0 x2 .x3 /4dx − µ2 = (1/4).(1/6).x6 |20 −µ2 = 8/3 − (8/5)2 = 0.1067 √ (iii) σ = 0.1067 = 0.3266 R2 (b) (i) µ = −2 (3/16).x.x2 dx = (3/16)(1/4)x4 |2−2 = 0 R2 (ii) σ 2 = E(X 2 ) = −2 (3/16).x2 .x2 dx = (3/16)(1/5)x5 |2−2 = 2.4 √ (iii) σ = 2.4 = 1.5492 R1 R1√ √ (b) (i) µ = 0 x.1/(2 x)dx = (1/2) 0 xdx = (1/2).(2/3).x3/2 |10 = 1/3 R1 √ (ii) σ 2 = E(X 2 ) − µ2 = 0 x2 .1/(2 x)dx − 1/9 = (1/2).(2/5).x5/2 |10 −1/9 = 1/5 − 1/9 = 0.0889 (iii) σ = 0.2982 Exercise 3.3-8 R∞ ∞ (a) 1 = 1 c/x2 dx = −c/x|1 = 0 + c = c, so c = 1. R∞ R∞ ∞ (b) E(X) = 1 x(1/x2 ) dx = 1 1/x dx = ln x|1 = limx→∞ ln x = ∞, so E(X) is unbounded (not finite). Exercise 3.3-12 M 0 (0) M 0 (t) , so R0 (0) = = M 0 (0) = µ (since M (0) = 1). (a) R0 (t) = M (t) M (0) (b) R00 (t) = so R00 (0) =
M (t)M 00 (t) − [M 0 (t)]2 [M (t)]2
M (t)M 00 (0) − [M 0 (0)]2 = M 00 (0) − [M 0 (0)]2 = σ 2 . [M (0)]2
Exercise 3.4-2 See below
Integrating f (x) we get, F (x) = R1 2 x f (x)dx = 1/3 −1 Exercise 3.4-5
1 2 (1
+ x)1[0,1] + 1(1,∞) , µ =
R1 −1
x.f (x)dx = 0, σ 2 =
(a) w−a w−a = , G(W ) = P [W ≤ w] = P [a + (b − a)Y ≤ w] = P Y ≤ b−a b−a
a ≤ w ≤ b.
(b) By (a), G(W ) is the distribution function of U (a, b). Therefore, W ∼ U (a, b). Exercise 3.4-6 (a) X follows exponential distribution with mean 20 (hence λ = 1/20). P (10 < X < 30) = P (X < 30) − P (X ≤ 10) = (1 − exp(−30/20)) − (1 − exp(−10/20)) = e−0.5 − e−1.5 = 0.3834 (b) P (X > 30) = exp(−30/20) = 0.2231 (c) P (X > 40|X > 10) = P ((X > 40) ∩ (X > 10))/P (X > 10) = P (X > 40)/P (X > 10) = e0.5−2 = 0.2231 (d) Variance of X = 1/λ2 = 400, MGF of X =
λ λ−t
=
1 1−20t
Exercise 3.4-21a P (X > 10) = e−10/80 = 0.8825 P (X > 90|X > 80) = P (X > 90)/P (X > 80) = e10/80−90/80 = 0.8825 1 Exercise 3.5-2 X has p.d.f. f (x) = 16 xe−x/4 for x > 0. So Z 5 5 1 −x/4 P (X < 5) = dx = − 14 xe−x/4 0 + 16 xe 0
1 4
Z
5
e−x/4 dx
0
5 = − 45 e−5/4 − e−x/4 0 = − 45 e−5/4 − e−5/4 + 1 =
1 − 94 e−5/4 ≈ 0.35536.
Exercise 3.5-5 Identifying this with the m.g.f. of a gamma function, we see that this has to be a Gamma(20,7) distribution. So, mean = 20.7 = 140, Variance = 20.72 = 980. Exercise 3.5-9 (a) 0.02499674 (b) 0.0499629 (c) 0.9400349 (d) 8.67176 (e) 30.19101 Exercise 3.5-10 Using Table IV with r = 12, we have a = 5.226 and b = 21.03. Exercise 3.6-2 Using Table Va and Table Vb: (a) P (0 ≤ Z ≤ 0.87) = P (Z ≤ 0.87) − P (Z < 0) = .8078 − 0.5 = 0.3078 (b) P (−2.64 ≤ Z ≤ 0) = P (Z ≤ 0) − P (Z < −2.64) = 0.5 − 0.0041 = 0.4949 (c) P (−2.13 ≤ Z ≤ −0.56) = P (Z ≤ −0.56) − P (Z < −2.13) = 0.2877 − 0.0166 = 0.2711 (d) P (|Z| > 1.39) = P (Z < −1.39) + P (Z > 1.39) = 2(0.0823) = 0.1646 (e) P (Z < −1.62) = 0.0526 (f) P (|Z| > 1) = P (Z < −1) + P (Z > 1) = 2(0.1587) = 0.3174 (g) P (|Z| > 2) = P (Z < −2) + P (Z > 2) = 2(0.0228) = 0.0456 (h) P (|Z| > 3) = P (Z < −3) + P (Z > 3) = 2(0.0013) = 0.0026 Exercise 3.6-3 (a) 2.326348 (b) -2.575829 (c) 1.669593 (d) -2.17009 Exercise 3.6-14 √ (a) P (X > 22.07) = P (Z > (22.07 − 21.37)/ 0.16) = P (Z > 1.75) = 0.0401; √ (b) P (X < 20.857) = P (Z < (20.857 − 21.37)/ 0.16) = P (Z < −1.28) ≈ 0.10. Thus Y ∼ B(15, 0.10), so using Table II, P (Y ≤ 2) = 0.8159. Exercise 3.7-20 Rx
2
(2t/502 )e−(t/50) dt. R (x/50)2 −y Substituting y = (t/50)2 , we get dy = (2t/(502 ))dt. Hence, F (x) = 0 e dy = 2 1 − e−(x/50) . For x < 0, distribution function F (x) = 0. So let x > 0. Then F (x) =
0
(a) P (40 < X < 60) = F (60) − F (40) = 0.2903647 (b) P (X > 80) = 1 − F (80) = 0.07730474 Additional Problems: Problem 1. R1 (a) We require f (x) ≥ 0 for all x and 0 f (x) dx = 1 so that f (x) is a legitimate density. If k > 0 then the first condition is satisfied. For the second condition, Z 1 Z 1 2 1 = kx (1 − x) dx = k (x2 − x3 ) dx 0
x3 x4 k − 3 4 k 12
= =
0
1
=k
0
1 1 − 3 4
so k = 12. (b) For x < 0 we have F (x) = 0. For 0 ≤ x ≤ 1 we have 3 x Z x t t4 F (x) = 12 t2 (1 − t) dt = 12 − 3 4 0 0 3 4 x x = 12 − = 4x3 − 3x4 = x3 (4 − 3x) 3 4 For x > 1 we have F (x) = 1. (c) P (0.2 < X < 0.3) = F (0.3) − F (0.2) = (0.3)3 (4 − 3(0.3)) − (0.2)3 (4 − 3(0.2)) = 0.0565 (d) The mean of X is Z
1
x3 (1 − x) dx = 12
E(X) = 12 0
x5 x4 − 4 5
1
= 12
0
1 1 − 4 5
=
12 = 0.6 20
Then 2
Z
E(X ) = 12 0
1
x5 x6 x (1 − x) dx = 12 − 5 6 4
1
= 12
0
1 1 − 5 6
=
2 = 0.4 5
so Var(X) = E(X 2 ) − [E(X)]2 = 0.4 − 0.36 = 0.04. Finally, Z 1 Z 1 1 X x E = 12 x2 (1 − x) dx = 12 x3 dx = 3x4 0 = 3 1−X 1−x 0 0 Problem 2.
(a) We have mX (t) = E[etX ]. So mY (t)
= =
(b) If X ∼ Γ(3, 2) then mX (t) =
i h E etY = E et(a+bX) = E eat+btX i h E eat ebtX = eat E e(bt)X = eat mX (bt) 1 (1−2t)3 .
From part (a),
mY (t) = e0t mX (5t) =
1 , (1 − 10t)3
so Y ∼ Γ(3, 10). 2
Problem 3. MGF of Z ∼ N (0, 1) is mZ (t) = et /2 . If X ∼ N (µ, σ 2 ), then X = µ + σZ. 2 2 2 2 Thus, using 2a, mX (t) = eµt mZ (σt) = eµt eσ t /2 = eµt+σ t /2
Problem 4. R∞
R ∞ α+1/2−1 −x/θ 1 1 α−1 −x/θ x1/2 . Γ(α)θ e dx = Γ(α)θ x e dx = αx α 0 R Γ(α+1/2)θ α+1/2 1 1 α+1/2−1 −x/θ α+1/2 ∞ x e dx = .1 = Γ(α)θ α Γ(α+1/2)θ Γ(α)θ α 0 Γ(α+1/2)θ α+1/2
E(X 1/2 ) =
0
Γ(α+1/2)θ 1/2 Γ(α)
= 0.9940, so x − 14 = 2.51, from Table Va. Thus the Problem 5. We have P Z < x−14 1 diameter of the pizza is 16.51 inches.
