S515 HW#3 SOLUTION
(due on Monday 10/21/13)
Professor H.K.Hsieh
4.49 Let Y be the time of an arriving call. Then Y has a uniform distribution on the
interval (0, 1), then P(Y in non-busy time period) = 1 – 15/60 =1-1/4 = ¾= 0.75.
4.53
Let Y = time when the defective circuit board was produced. Then, Y has an
approximate uniform distribution on the interval (0, 8).
a. P(0 < Y < 1) = 1/8.
b. P(7 < Y < 8) = 1/8
c. P(4 < Y < 5 | Y > 4) = P(4 < Y < 5)/P(Y > 4) = (1/8)/(1/2) = 1/4.
4.58
a. P(0 ≤ Z ≤ 1.2) = .5 – .1151 = .3849
b. P(–.9 ≤ Z ≤ 0) = .5 – .1841 – .3159.
c. P(.3 ≤ Z ≤ 1.56) = .3821 – .0594 = .3227.
d. P(–.2 ≤ Z ≤ .2) = 1 – 2(.4207) = .1586.
e. P(–1.56 ≤ Z ≤ –.2) = .4207 – .0594 = .3613
f. (omitted)
4.59
a. z0 = 0.
b. z0 = 1.10
c. z0 = 1.645
d. z0 = 2.576
4.62
a. P(Z2 < 1) = P(–1 < Z < 1) = .6826.
b. P(Z2 < 3.84146) = P(–1.96 < Z < 1.96) = .95.
4.71
Let Y = the measured resistance of a randomly selected wire.
.13
.13
 Z  .14.005
) = P(–2 ≤ Z ≤ 2) = .9544.
a. P(.12 ≤ Y ≤ .14) = P( .12.005
b. Let X = # of wires that do not meet specifications. Then, X is binomial with n
= 4 and p = .9544. Thus, P(X = 4) = (.9544)4 = .8297.
4.75 Let Y = the measured resistance of a randomly selected wire.
P(.12 ≤ Y ≤ .14) = Let Y = volume filled, so that Y is normal with mean μ and σ =
.3 oz. They require that P(Y > 8) = .01. For the standard normal, P(Z > z0) = .01
when z0 = 2.33. Therefore, it must hold that 2.33 = (8 – μ)/.3, so μ = 7.301.
4.77 a. Let Y = SAT math score. Then, P(Y < 550) = P(Z < .7) = 0.758.
b. If we choose the same percentile, 18 + 6(.7) = 22.2 would be comparable on the
ACT math test.
4.82 From above we have Γ(1) = 1, so that Γ(2) = 1Γ(1) = 1, Γ(3) = 2Γ(2) = 2(1), and
generally Γ(n) = (n–1)Γ(n–1) = (n–1)! Γ(4) = 3! = 6 and Γ(7) = 6! = 720.
4.88
Let Y have an exponential distribution with β = 2.4.

a. P(Y > 3) =

1  y / 2.4
2.4
e
dy  e 3 / 2.4 = .2865.
3
3
b. P(2  Y  3)   21.4 e  y / 2.4 dy = .1481.
2
4.90
Let Y = magnitude of the earthquake which is exponential with β = 2.4. Let X = #
of earthquakes that exceed 5.0 on the Richter scale. Therefore, X is binomial with

n = 10 and p = P(Y > 5) =

1  y / 2.4
2.4
e
dy  e 5 / 2.4 = .1245. Finally, the probability
5
of interest is
P(X ≥ 1) = 1 – P(X = 0) = 1 – (.8755)10 = 1 – .2646 = .7354.
4.93
Let Y = time between fatal airplane accidents. So, Y is exponential with β = 44
days.
31
a. P(Y ≤ 31) =

1  y / 44
44
e
dy  1  e 31/ 44 = .5057.
0

E (Y )   y
a
a
1
 (  ) 
y
1  y / 
e

dy 
1
 (  ) 
0
y
a 1  y / 
e
dy 
 ( a   ) a  
1
1
 (  ) 
  ((a) )
a
0
2
b. V(Y) = 44 = 1936.
4.96
a. The density function f (y) is in the form of a gamma density with α = 4
and β = 2.
1
1
Thus, k =
.

4
96
(4)2
b. Y has a χ2 distribution with ν = 2(4) = 8 degrees of freedom.
c. E(Y) = 4(2) = 8, V(Y) = 4(22) = 16.
4.103
σ = 16 = 4.
Let R denote the radius of a crater. Therefore, R is exponential w/ β = 10 and the
area is A = πR2. Thus,
E(A) = E(πR2) = πE(R2) = π(100 + 100) = 200π.
V(A) = E(A2) – [E(A)]2 = π2[E(R4) – 2002] = π2[240,000 – 2002] =

200,000π , where E(R ) =
2
4

0
1
10
r 4 e r / 10 dr = 104Γ(5) = 240.000.

4.109
E (Y )   y
a
a
1
 (  ) 
y
1  y / 
e

dy 
1
 (  ) 
0
y
a 1  y / 
e
dy 
 ( a   ) a  
1
1
 (  ) 
  ((a) )
a
0
(See Problem 4.111 (a))
E(L)= E(30Y+2Y**2)=30E(Y)+2E(Y**2)= finish the rest
V(L) =E(L**2) - [E(L)]**2
= E(900Y**2 + 120Y**3+4Y**4) - [E(L)]**2
= Use the general formula in the first line of this problem about
high order moments and complete the rest
y
4.126
a. F ( y )   (6t  6t 2 )dt  3 y 2  2 y 3 , 0 ≤ y ≤ 1.
0
0.0
0.5
1.0
1.5
F(y) = 0 for y < 0 and
F(y) = 1
for y > 1.
0.0
0.2
0.4
0.6
0.8
1.0
y
b. Solid line: f(y); dashed line: F(y)
c. P( .5 ≤ Y ≤ .8 ) = F(.8) – F(.5) = [ 3(.8)2 -2(.8)3 ] - [ 3(.5)2 -2(.5)3 ]
= 1.92 – 1.092 - .75 + .25 = .396.
4.139
Using Ex. 4.137 with a = –3 and b = 4, it is trivial to see that the mgf for X is
mX (t )  e4t m(3t )  e( 43 )t 9 t
2 2
/2
.
By the uniqueness of mgfs, X is normal with mean 4 – 3μ and variance 9σ2.
4.186 Using the substitution u = ym/α in the integrals below, we find:

E (Y )   y e
m  ym / 
m


dy  
1/ m
0
u
1 / m u
e du  1 / m (1  1 / m)
0

E (Y )   y
2
m

m1  y m / 
e

dy  
0
2/ m
u
2 / m u
e du  1 / m (1  2 / m) .
0
Thus,
V (Y )   2 / m [ (1  2 / m)   2 (1  1 / m)] .
4.190 a. For the exponential distribution, f (t )  1 e t /  and F (t )  1  e t /  . Thus,
r(t) = 1/β.
m 1
b. For the Weibull, f ( y )  my e  y /  and F ( y )  1  e  y
which is an increasing function of t when m > 1.
m
m
/
. Thus, r(t ) 
my m 1

,