Solução_Calculo_Stewart_6e

F.
17. P (μ − 2σ ≤ X ≤ μ +2σ) =
2
−2
μ+2σ
μ−2σ
1
σ √ 2π e− t2 /2 (σdt)= 1 √
2π
2
−2
TX.10
SECTION 8.5 PROBABILITY ¤ 373
1
σ √ 2π exp (x − μ)2
− dx. Substituting t = x − μ and dt = 1 dx gives us
2σ 2
σ
σ
e − t2 /2 dt ≈ 0.9545.
19. (a) First p(r) = 4 a 3 0
r 2 e −2r/a 0
≥ 0 for r ≥ 0. Next,
∞
−∞
∞
p(r) dr =
0
4
a 3 0
r 2 e −2r/a 0
dr = 4 a 3 0
lim
t→∞
By using parts, tables, or a CAS , we find that x 2 e bx dx =(e bx /b 3 )(b 2 x 2 − 2bx +2).
Next, we use ()(withb = −2/a 0 ) and l’Hospital’s Rule to get
a function to be a probability density function.
(b) Using l’Hospital’s Rule,
4
a 3 0
lim
r→∞
r 2
e 2r/a 0 = 4 a 3 0
To find the maximum of p, we differentiate:
p 0 (r) = 4 a 3 0
lim
r→∞
2r
4
a 3 0
(2/a 0 )e 2r/a 0 = 2 a 2 0
t
0
r 2 e −2r/a 0
dr
()
a
3
0
−8 (−2)
=1.Thissatisfies the second condition for
lim
r→∞
2
(2/a 0 )e 2r/a 0 =0.
r 2 e −2r/a 0
− 2
+ e −2r/a 0
(2r) = 4
e −2r/a 0
(2r) − r
+1
a 0 a 3 0 a 0
p 0 (r) =0 ⇔ r =0or 1= r a 0
⇔ r = a 0 [a 0 ≈ 5.59 × 10 −11 m].
p 0 (r) changes from positive to negative at r = a 0 ,sop(r) has its maximum value at r = a 0 .
(c) It is fairly difficult to find a viewing rectangle, but knowing the maximum
value from part (b) helps.
p(a 0 )= 4 a 2
a
0e −2a 0/a 0
= 4 e −2 ≈ 9,684,098,979
3
0
a 0
With a maximum of nearly 10 billion and a total area under the curve of 1,
we know that the “hump” in the graph must be extremely narrow.
r
(d) P (r) =
0
4
a 3 0
P (4a 0 )= 4 a 3 0
∞
4a0
s 2 e −2s/a 0
ds ⇒ P (4a 0)=
0
e
−2s/a 0
−8/a 3 0
=1− 41e −8 ≈ 0.986
4
s 2 e −2s/a 0
ds. Using() from part (a) [with b = −2/a 0],
a 3 0
4
s 2 + 4 4a0
s +2 = 4 a
3
0
[e −8 (64 + 16 + 2) − 1(2)] = − 1
a 2 0 a 0 0
a 3 0 −8
2 (82e−8 − 2)
t
lim
t→∞
0
(e) μ = rp(r) dr = 4 r 3 e −2r/a 0
dr. Integrating by parts three times or using a CAS, we find that
−∞ a 3 0
x 3 e bx dx = ebx
b 3 x 3 − 3b 2 x 2 +6bx − 6 .Sowithb = − 2 , we use l’Hospital’s Rule, and get
b 4 a 0
μ = 4
− a4 0
a 3 0 16 (−6) = 3 a 2 0.