1. The probability that a randomly chosen tire will have a lifetime (a) between 30000 and
40000 miles (b) more than 25000 miles.
R∞ c
∞
−1
(x)|−∞ = cπ = 1 ⇒ c = π1 .
5. (a) −∞ 1+x
2 dx = c tan
1
R1
tan−1 (x) 1
(b) −1 π(1+x
dx
=
= 12 .
2)
π
−1
R∞
R 10
0.1dx = 1. f (x) ≥ 0 for all x.
R 10
2
= 5.
(b) −∞ xf (x)dx = 0 0.1xdx = 10
20
200
∞
R 200 −t/µ
R ∞ −t/µ
(a) 0 e µ dt = −e−t/1000 0 = 1 − e−0.2 . 800 e µ dt = −e−t/1000 800 = e−0.8 .
∞
R ∞ −t/µ
−m
= − ln 2 ⇔ m = 1000 ln 2.
(b) m e µ dt = −e−t/1000 m = e−m/1000 = 21 ⇔ 1000
R∞ 1
2
2
√ e−(x−9.4) /(2·4.2 ) dx ≈ 0.4432 = 44.32%.
10 4.2 2π
R 480
2
2
(a) −∞ 12√1 2π e−(x−500) /(2·12 ) dx ≈ 0.0478 = 4.78%.
R 500
R 500
2
2
2
2
(b) −∞ 12√1 2π e−(x−µ) /(2·12 ) dx ≤ 5%. Note that if µ = 520, then −∞ 12√1 2π e−(x−µ) /(2·12 ) dx =
R 480 1 −(x0 −500)2 /(2·122 ) 0
√ e
dx ≤ 5%. Hence we can choose the target weight 520 g.
−∞ 12 2π
R 100
2
2
(a) −∞ 8√12π e−(x−112) /(2·8 ) dx ≈ 6.68%.
R∞
2
2
(b) 125 8√12π e−(x−112) /(2·8 ) dx ≈ 5.21%.
2
2
2
2
−2(x−µ)
d
√1 e−(x−µ) /(2·σ ) = √1 e−(x−µ) /(2·σ ) ·
.
dx σ 2π
2·σ 2
σ 2π
7. (a)
−∞
f (x)dx =
0
R∞
10.
13.
14.
15.
16.
d2
d2 x
1
2
2
√ e−(x−µ) /(2·σ )
σ 2π
−2
−2(x − µ) −2(x − µ)
1
−(x−µ)2 /(2·σ 2 )
·
+
·
= √ e
2 · σ2
2 · σ2
2 · σ2
σ 2π
2
2
e−(x−µ) /(2·σ )
√
(4(x − µ)2 − 4σ 2 )
=
3
 4σ 2π
 > 0 ,x < µ − σ
< 0 ,µ − σ < x < µ + σ
=

> 0 ,x > µ + σ
18.
Z
∞
(t − µ)
0
1/2 Z ∞
Z ∞
Z ∞ −t/µ 1/2
e
1
2 −t/µ
−t/µ
2
dt
=
te
dt −
2te
dt + µ
dt
µ
µ 0
µ
0
0
1/2
Z
Z ∞
1 2
1 ∞
−t/µ ∞
−t/µ
−t/µ
2
=
t (−µe
)0 −
2t(−µe
)dt −
2te
dt + µ
µ
µ 0
0
1/2
t2
2
= lim − t/µ + µ
=µ
t→∞
e
−t/µ
2e
1
Remark. If f (x) is a normal distribution, one should expect that this definition of
standard deviation coincides the one defined in f (x). And indeed it’s true:
"Z
2
∞
2
e−(x−µ) /(2·σ )
√
dx
(x − µ)2
σ 2π
−∞
19. (a)
R∞
0
4 2 −2r/a0
r e
dr
a30
=
4
a30
h
#1/2
r2
#1/2
−X 2 /(2·σ 2 )
e
√
dX
=
X2
σ 2π
−∞
∞
"
#1/2
Z ∞
−X 2 /(2·σ 2 )
−X 2 /(2·σ 2 ) e
e
√
√
(−σ 2 )
= X · (−σ 2 )
−
dX
σ 2π −∞
σ 2π
−∞
1/2
= 0 − (−σ 2 )
=σ
"Z
∞
−a0 −2r/a0
e
2
− 2r
a20 −2r/a0
e
22
+2
−a30 −2r/a0
e
23
and p(r) ≥ 0 for all r ≥ 0.
(b) lim a43 r2 e−2r/a0 =
r→∞
0
4
2re−2r/a0 + r2 ·
a3
0
2
4
lim r
a30 r→∞ e2r/a0
−2 −2r/a0
e
=
a0
= 0 by l’Hospital’s Rule.
0 ⇔ 2r + r2 ·
−2
a0
d
dr
i∞
= 1
0
4 2 −2r/a0
r e
a30
=
= 0 ⇔ r = 0 or r = a0 . Clearly
r = 0, p(r) is not maximum and notice that r > a0 ,
Hence it’s maximal at r = a0 .
d
p(r)
dr
< 0; r < a0 ,
d
p(r)
dr
> 0.
(c) Near 0, it behaves like r2 . And it behaves like e−2r/a0 as r → ∞.
2
3
i4a0
h
R 4a0 4 2 −2r/a
a0 −2r/a0
−a0 −2r/a0
4
2 −a0 −2r/a0
0
e
− 2r 22 e
+ 2 23 e
dr = a3 r
=
(d) 0 a3 r e
2
0
0
3
3 i
h 0
2
a
−a
−a
4
16a20 −a2 0 e−8 − 8a0 220 e−8 + 2 230 e−8 − 2 230 = 1 − 41e−8 ≈ 98.63%.
a30
h
2
3
4
i∞
R∞
a
−a
a
(e) 0 r· a43 r2 e−2r/a0 dr = a43 r3 −a2 0 e−2r/a0 − 3r2 220 e−2r/a0 + 6r 230 e−2r/a0 − 6 240 e−2r/a0
0
=
0
0
3a0
.
2
2