Giải bài tập chương 3 xác suất thống kê trong sách bài tập

A
P (A) = 0, 51
X ∼ B(n = 5; p = 0, 51)
P (X = 2) = C
2
5
.0, 51
2
.0, 49
3
= 0, 306
P (X  2) = P (X = 0) + P (X = 1) + P (X = 2) = C
0
5
.0, 49
5
+ C
1
5
.0, 51.0, 49
4
+
0, 306 = 0, 481
X ∼ B(n = 12; p = 1/3)
X ∼ B(n = 12; p = 1/3)
P (X = 4) = C
4
12
(1/3)
4
.(2/3)

8
= 0, 238
P (3  X  6) = P (X = 3) + P (X = 4) + P(X = 5) + P (X = 6) = = 0, 751
X ∼ B(n = k; p = 0, 8)
⇒ E(X) = np = k.0, 8  10 ⇒ k  10/0, 8 = 12, 5 ⇒ k = 13
X ∼ B(n = 5; p = 0, 1)
P (X  2) = P (X = 0)+P (X = 1)+P (X = 2) =

2
i=0
C
i
5
0, 1
i
0, 9
5−i
= 0, 99144
E(X) = np = 5.0, 1 = 0, 5
m
0
np + p − 1  m
0
 np + p ⇔ 5.0, 1 + 0, 1 − 1  m
0
 5.0, 1 + 0, 1 ⇔ m
0
= 0
X ∼ B(n = 10000; p = 0, 85)
E(X) = np = 10000.0, 85 = 8500

V (X) = npq = 10000.0, 85.0, 15 = 1275
2
X ∼ B(n = 855; p = 0, 02)
m
0
np+p−1  m
0
 np+p ⇔ 855.0, 02+0, 02−1  m
0
 855.0, 02+0, 02 ⇔ m
0
= 17
⇒ X ∼ B(n = 5; p = 3/4)
P (X = 3) = C
3
5
.(3/4)
3
.(1/4)
2
= 0, 263
P (X  1) = 1 − P (X = 0) = 1 − C
0
5
.(1/4)
5
= 0, 999902
P (X  2) =

2

i=0
C
i
5
.(3/4)
i
.(1/4)
5−i
= 0, 1035
X ∼ B(n = 2; p)
E(X) = np = 2p = 1, 2 ⇒ p = 0, 6
V (X) = npq = 2.0, 6.0, 4 = 0, 48
⇒ X ∼ B(n = 3; p)
V (X) = npq = 3p(1 − p) = 0, 63 ⇔ p
2
− p + 0, 21 = 0 ⇔ p = 0, 7 p = 0, 3
⇒ X ∼ B(n = 12; p = 0, 3)
m
0
np + p − 1  m
0
 np + p ⇔ 12.0, 3 + 0, 3 − 1  m
0
 12.0, 3 + 0, 3 ⇔ m
0
= 3
P (X = 3) = C
3
12
.0, 3

3
.0, 7
9
= 0, 2397
X ∼ B(n = 3; p = 0, 7)
P () = P (X = 2) + P (X = 3) = C
2
3
.0, 7
2
.0, 3 + C
3
3
.0, 7
3
= 0, 784
→ X ∼ B(n = 15; p = 0, 7)
P (X  10) =
15

i=10
C
i
15
.0, 7
i
.0, 3
15−i
= 0, 72162
P (X = x) =

C
x
3
.C
4−x
7
C
4
10
; x = 0, 1, 2, 3
⇒ X ∼ M(N = 10; M = 3; n = 4)
X ∼ M(N = 20; M = 12; n = 5)
P (X  3) = P (X = 3) + P (X = 4) + P (X = 5) =
C
3
12
C
2
8
C
5
20
+
C
4
12
C
1
8
C

5
20
+
C
5
12
C
0
8
C
5
20
= 0, 703818
⇒ X ∼ M(N = 20; M = 5; n = 3)
E(X) = n
M
N
= 3.
5
20
= 0, 75
Y = 2.50.X = 100X
E(Y ) = 100E(X) = 100.0, 75 = 75
→ X ∼ M(N = 20; M = 3; n = 5)
P (X = x) =
C
x
3
C
5−x

17
C
5
20
; x = 0, 1, 2, 3
E(X) = np = 5 ·
3
20
= 0, 75
V (X) = npq
N − n
N − 1
= 5 ·
3
20
·
17
20
·
15
19
= 0, 5033
→ X ∼ M(N = 100; n = 5)
P (X = 3) =
C
3
40
C
2
60

C
5
100
= 0, 23228
X ∼ B(n = 100; p = 0, 02)
E(X) = np = 100.0, 02 = 2
X ∼ P (λ)
E(X) = λ = 8
P (X > 4) = 1−P (X  4) = 1−e
−8
8
0
0!
−e
−8
8
1
1!
−e
−8
8
2
2!
−e
−8
8
3
3!
−e
−8

8
4
4!
= 0, 90037
→ X ∼ B(n = 150; p = 0, 04)
np = 150.0, 04 = 6 ≈ np(1 − p) X ∼ P (λ = 6)
P (X  2) = P (X = 0)+P (X = 1)+P (X = 2) = e
−6
6
0
0!
+e
−6
6
1
1!
+e
−6
6
0
2
2!
= 0, 06197
⇒ X ∼ P (λ = 2)
P
a
= P (X  6) =
6

i=0

P (X = i) =
6

i=0
e
−2
2
i
i!
= 0, 9947
P
b
= P (X  12) = 1 − P (X < 12) = 1 −
12

i=0
e
−2
2
i
i!
= 0, 0000013646
P (X > 7) > 0, 1
P (X > 7) = 1−P (X  7) = 1−P (X  6)−P (X = 7) = 1−0, 99547−e
−2
2
7
7!
= 0, 0011
→ X ∼ B(n = 1800; p = 1/5000)

np = 1800.1/5000 = 0, 36 ≈ np(1 − p)
X ∼ P (λ = 0, 36)
⇒ P (X  2) = e
−0,36
0, 36
0
0!
+ e
−0,36
0, 36
1
1!
+ e
−0,36
0, 36
2
2!
= 0, 99405
X ∼ U[30; 50]
f(x) =



1
20
; x ∈ [30; 50]
0 ; x /∈ [30; 50]
F (x) =
x


−∞
f(t)dt =









0 ; x < 30
x − 30
20
; x ∈ [30; 50]
1 ; x > 50
E(X) =
30 + 50
2
= 40; V (X) =
(50 − 30)
2
12
= 33, 333
P (X  45) = F (45) =
45 − 30
20
= 0, 75
X ∼ U(a, b)






E(X) =
a + b
2
= 30
σ
X
=

(b − a)
2
12
= 5
⇔



a + b = 60
b − a = 10
√
3
⇔



a = 30 − 5
√

3
b = 30 + 5
√
3
f(x) =





1
10
√
3
x ∈ (30 − 5
√
3; 30 + 5
√
3)
0 ; x /∈ (30 − 5
√
3; 30 + 5
√
3)
P (X  32) =
+∞

32
f(x)dx =
30+5

√
3

32
1
10
√
3
dx =
30 + 5
√
3 − 32
10
√
3
= 0, 6155
f(x) =



0 ; x < 0
1
1500
e
−x/1500
F (x) =



0 ; x < 0;

1 − e
−x/1500
; x  0
P (X < 1500) = F (1500) = 1 − e
−1500/1500
= 1 − e
−1
= 0, 632
f(x) =



5e
−5x
; x  0
0 ; x < 0
⇒ λ = 5
P (0, 4 < X < 1) =
1

0,4
f(x)dx =
1

0,4
5e
−5x
dx = −e
−5x



1
0,4
= 0, 1286
E(X) =
1
λ
=
1
5
= 0, 2
f(x) =



2e
−2x
; x > 0
0 ; x < 0
⇒ λ = 2
E(X) =
1
λ
=
1
2
= 0, 5
→ X ∼ E(λ)
E(X) = 1/λ = 3 ⇒ λ = 1/3
P (X  2) =

+∞

2
f(x)dx =
+∞

2
(1/3)e
−(1/3)x
dx = −e
−(1/3)x


+∞
2
= 0, 51342
P (−2, 33 < U < 2, 33) = P (U > −2, 33) − P (U > 2, 33) = 1 − α − α
= 1 − 2.0, 0099 = 0, 9802
α u
α
= 2, 33 ⇒ α = 0, 0099
P (−2 < U < 1) = P (U > −2)−P(U > 1) = 1−α
1
−α
2
= 1−0, 0228 −0, 1587
= 0, 8185
α
1
u

α
1
= 2 ⇒ α
1
= 0, 0228
α
2
u
α
2
= 1 ⇒ α
2
= 0, 1587
P (−0, 89 < U < 2, 5) = P (U > −0, 89) − P (U > 2, 5) = 1 − α
1
− α
2
= 1 − 0, 1867 − 0, 0062 = 0, 8071
u
α
1
= 0, 89 ⇒ α
1
= 0, 1867; u
α
2
= 2, 5 ⇒ α
2
= 0, 0062
P (U > 3, 02) = 0, 0013 u

α
= 3, 02 ⇒ α = 0, 0013
P (U < 2, 5) = 1 − P (U > 2, 5) = 1 − 0, 0062 = 0, 9938
→ X ∼ N(µ, σ
2
)
E(X) = µ = 100(g); σ = 1(g)
P (98 < X < 102) = P

98 − 100
1
<
X −100
1
<
102 − 100
1

= P (−2 < U < 2)
= P (U > −2) − P (U > 2) = 1 − 2α
α P (U > 2) = α ⇒ u
α
= 2 ⇒ α = 0, 0228
⇒ P (98 < X < 102) = 1 − 2α = 1 − 2.0, 0228 = 0, 9544
P
b
= 1 − P
a
= 1 − 0, 9544 = 0, 0456
⇒ X ∼ N(µ = 15000, σ

2
= 500
2
)
P (X > 16000) = P

X−15000
500
>
16000−15000
500

= P (U > 2) = 0, 0228
P (X < 14500) = P

X−15000
500
<
14500−15000
500

= P(U < −1) = P (U > 1) =
0, 1587
P (14500 < X < 16500) = P (−1 < U < 3) = P (U > −1) − P (U > 3) =
1 − α
1
− α
2
α
1

, α
2
u
α
1
= 1 → α
1
= 0, 1587; u
α
2
= 2 → α
2
= 0, 0013
⇒ P (14500 < X < 16500) = 1 − 0, 1587 − 0, 0013 = 0, 84
→ X ∼ N(µ = 200; σ
2
= 40
2
)
P (X > 250) = P

X −200
40
>
250 − 200
40

= P (U > 1, 25) = 0, 1056
P (X < 180) = P (U <
180 − 200

40
) = P (U < −0, 5) = P (U > 0, 5) = 0, 3085
→ X ∼ N(µ = 160; σ
2
= 6
2
)
P (X < 155) = P (U <
155 − 160
6
) = P (U < −5/6 = −0, 83) = P (U > 0, 83) = 0, 2033
0, 2033
4
P
b
= 1 − 0, 2033
4
= 0, 9983
→ X ∼ N(µ = 50; σ
2
)
P (32  X  68) = 1
⇔ P (
32 − 50
σ
< U <
68 − 50
σ
) = 1
⇔ 1 = P (−

18
σ
< U <
18
σ
) = 1 − 2P (U >
18
σ
)
⇔ P (U >
18
σ
) = 0 ⇔
18
σ
≈ 5 ⇔ σ = 3, 6
P (X > 55) = P (U >
55 − 50
3, 6
) = P (U > 1, 39) = 0, 0823
P (X < 40) = P (U <
40 − 50
3, 6
) = P (U < −2, 78) = P (U > 2, 78) = 0, 0027
X =
1
n

n
i=1

X
i
→ E(
X) =
1
n
n

i=1
E(X
i
) =
1
n
n

i=1
m = m
V (
X) =
1
n
2
n

i=1
V (X
i
) =
σ

2
n
→ σ
X
=
σ
√
n
⇒
X ∼ N(m,
σ
2
n
)
P (|
X −m| < ε) = 2Φ
0
(
ε
σ
X
) = 2Φ
0
(
ε
√
n
σ
)
X ∼

B(n = 1000; p = 0, 75)
X ∼ N(µ = np = 750; σ
2
= np(1 − p) = 187, 5)
f =
X
n
E(f) = p = 0, 75; V (f) =
p(1 − p)
n
=
0, 75.0, 25
1000
= 0, 0001875 = σ
2
P (|f − p| < 0, 02) = 2φ
0

0, 02
√
0, 0001875

= 2Φ
0
(1, 46) = 2.0, 4279 = 0, 8558
X ∼ B(n = 900; p = 0, 9)
E(X) = µ = np = 900.0, 9 = 810
V (X) = σ
2
= npq = 810.0, 1 = 81

(a, b) = (|X −µ| < 2σ) = (µ − 2σ < X < µ + 2σ)
⇒ a = µ − 2σ = 810 − 2.9 = 792; b = µ + 2σ = 810 + 2.9 = 828
f ∼ N(µ = p; σ
2
=
p(1 − p)
n
)
P (|f − p|  0, 02) = 0, 7698
P (|f − p|  0, 02) = 2Φ
0
(
0, 02
σ
) = 0, 7698
⇔ Φ
0
(
0, 02
σ
) = 0, 3849 ⇒
0, 02
σ
= 1, 2 ⇒ σ =
0, 02
12
= 0, 01667
n =
p(1 − p)
σ

2
=
0, 5.0, 5
0, 01667
2
= 89964
→ X ∼ N(µ =
d
1
+ d
2
2
; σ
2
=
(d
2
− d
1
)
2
4
2
)
d
1
< X < d
2
P () = 1 − P (d
1

< X < d
2
) = 1 − P

d
1
−
d
1
+ d
2
2
d
2
− d
1
4
< U <
d
2
−
d
1
+ d
2
2
d
2
− d
1

4

= 1 − P (−2 < U < 2) = 1 − [P (U > −2) − P (U > 2)]
= 1 − [1 − 2P (U > 2)] = 2P (U > 2) = 2.0, 0228 = 0, 0456
P (A) = 0, 01
X ∼ B(n = 365; p = 0, 01)
⇒ E(X) = np = 365.0, 01 = 3, 65
•
E(X).1000 = 3, 65.1000 = 3650
•
E(X)
2
.1000 + 12.120 = 3265
⇒
3650 − 3265 = 385
X ∼ N(µ = 4, 2; σ
2
= 1, 8
2
)
P (X  3) = P (U <
3 − 4, 2
1, 8
) = P (U < −0, 67) = P (U > 0, 67) = 0, 2514
u
α
= 0, 67 → α = 0, 2514
→ Y = 150
Y = 150 − 500 = −350
E(Y ) = −350.0, 2514 + 150.0, 7486 = 24, 3

P (Z = 150) = P (X  a) = 1 − P (X < a); P (Z = −350) = P (X < a)
E(Z) = −350.P (X < a) + 150(1 − P (X < a)) = 150 − 500.P (X < a)
E(Z) = 50 ⇔ 150−500.P (X < a) = 50 ⇔ P(X < a) = 0, 2 ⇔ P (U <
a − 4, 2
1, 8
) = 0, 2
⇔ P (U >
4, 2 − a
1, 8
) = 0, 2 ⇒
4, 2 − a
1, 8
= 0, 84 ⇒ a = 4, 2 −18.0, 84 = 2, 688
X ∼ P (λ = 12)
P (X > 8) = 1 − P (X  8) = 1 −
8

i=0
e
−12
12
i
i!
= 0, 845
P (X > 15) = 1 − P (X  15) = 1 −
15

i=0
e
−12

12
i
i!
= 0, 1556
P (X < 10) =
9

i=0
e
−12
12
i
i!
= 0, 2424
X ∼ N(µ = 11; σ
2
= 2
2
)
P (X  10) = P (U <
10 − 11
2
) = P (U < −0, 5) = P (U > 0, 5) = 0, 3805
0, 1 = P (X  a) = P (U <
a − 11
2
) = P (U >
11 − a
2
) ⇒

11 − a
2
= 1, 28 ⇒ a = 8, 44
⇒ X ∼ N(µ; σ
2
= 9
2
)
0, 8413 = P (X  84) = P (U 
84 − µ
9
) ⇒ P (U >
84 − µ
9
) = 0, 1587 ⇒
84 − µ
9
= 1
⇒ µ = 75
P (X  80) = P (U >
80 − 75
9
) = P (U > 5/9) = 0, 2877 ⇒ P (X < 80) = 0, 7123
P () = 1 − P (X < 80)
3
= 1 − 0, 7123
3
= 0, 6386
→ X ∼ N(µ
1

= 25, 2; σ
2
1
= 3, 9
2
)
→ X ∼ N(µ
1
= 25, 2; σ
2
1
= 1, 9
2
)
P (X  30) = P (U 
30 − 25, 2
3, 9
= 1, 23) = 1−P (U > 1, 23) = 1−0, 1093 = 0, 8907
P (Y  30) = P (U 
30 − 25, 2
1, 9
= 2, 53) = 1−P (U > 2, 53) = 1−0, 0057 = 0, 9953
Z ∼ N(0; 1); X = α + βZ + δZ
2
⇒ E(X) = α + βE(Z) + δE(Z
2
) = α + δ E(Z
2
) = V (Z) + [E(Z)]
2

= 1 + 0 = 1
V (X) = e(X
2
) − [E(X)]
2
= E(α + βZ + δZ
2
)
2
− (α + δ)
2
= α
2
+ β
2
E(Z
2
) + δ
2
E(Z
4
) + 2αβE(Z) + 2αδE(Z
2
) + 2βδE(Z
3
) − (α + δ)
2
= δ
2
E(Z

4
) + 2βδE(Z
3
) + α
2
+ β
2
.1 + 2αβ.0 + 2αδ.1 − α
2
− 2αδ −δ
2
= δ
2
E(Z
4
) + 2βδE(Z
3
) + β
2
− δ
2
E(Z
3
) =
1
√
2π
+∞

−∞

z
3
e
−z
2
/2
dz = 0
E(Z
4
) =
1
√
2π
+∞

−∞
z
4
e
−z
2
/2
dz
2
√
2π
+∞

0
z

4
e
−z
2
/2
dz =
2
√
2π
+∞

0
(2t)
3/2
e
−t
dt
=
4
√
π
Γ(5/2) =
4
√
π
3
4
√
π = 3 (t := z
2

/2)
V (X) = 3δ
2
+ β
2
− δ
2
= β
2
+ 2δ
2
Γ(a) =
+∞

0
x
a−1
e
−x
dx
Γ(a + 1) = aΓ(a); Γ(1/2) =
√
π
⇒ Γ(5/2) = (3/2)Γ(3/2) = (3/2).(1/2)Γ(1/2) = (3/4)
√
π
X ∼ N(µ = 8%; σ
2
= 10
2

)
P (X > 12) = P (U >
12 − 8
10
) = P (U > 0, 4) = 0, 3446
→ X ∼ U[5; 9]
f(x) =



1
4
; x ∈ [5; 9]
0 ; x /∈ [5; 9]
P (X < 8) =
9

8
1
4
dx =
8 − 5
4
= 0, 75
E(X) =
5 + 9
2
= 7
2/5 = 0, 4
→ X ∼ B(n = 20; p = 0, 4)

P (X = 2) = C
2
20
.0, 4
2
.0, 6
18
= 0, 003087
P (X  2) = C
0
20
.0, 4
0
.0, 6
20
+ C
1
20
.0, 4
1
.0, 6
19
+ 0, 003087 = 0, 0036115
p ≈ f = 0, 1
p = 0, 4
→ X ∼ N(µ; σ
2
)
0, 2 = P (X > 20) = P (U >
20 − µ

σ
) ⇒
20 − µ
σ
= 0, 84
0, 1 = P (X < 10) = P (U <
10 − µ
σ
= P (U >
µ − 10
σ
) ⇒
µ − 10
σ
= 1, 28



20−µ
σ
= 0, 84
µ−10
σ
= 1, 28
⇔



σ =
10

2,12
= 4, 717
µ = 16, 038
P (X  14) = P (U 
14 − 16, 038
4, 717
= −0, 43) = 1−P (U > 0, 43) = 1−0, 3336 = 0, 6664
X ∼ N(µ
x
= 8; σ
2
x
= 0, 3
2
); Y ∼ N(µ
y
= 4; σ
2
y
= 0, 2
2
)
P
a
= P (|X −µ
x
|  0, 1; |Y −µ
y
|  0, 1) = P (|X −µ
x

|  0, 1)P (|Y −µ
y
|  0, 1)
= 2Φ
0

0, 1
0, 3
)2Φ
0

0, 1
0, 2
) = 4Φ
0
(0, 33)Φ
0
(0, 5) = 4.0, 1293.0, 1915 = 0, 099044
P
b
= 1 − 0, 900956
3
= 0, 268674
→ X ∼ N(µ = 4300; σ
2
= 250
2
)
10.360 = 3600
P (X  3600) = P (U 

3600 − 4300
250
) = P (U  −2, 8) = P (U > 2, 8) = 0, 0026
2.3600 = 7200
µ
1
Y ∼ N(µ
1
; 250
2
)
0, 0026 = P (Y  7200) = P (U 
7200 − µ
1
250
) = P (U >
µ
1
− 7200
250
)
⇒
µ
1
− 7200
250
= u
0,0026
= 2, 8 ⇒ µ
1

= 7200 + 2, 8.250 = 7900
A
P (A) = 0, 5; P (
A) = 0, 5
→ X ∼ B(n = 6; p = 0, 5)
X  3
P (X  3) = 1−P (X < 3) = 1−
2

k=0
P (X = k) = 1−
2

k=0
C
k
6
0, 5
k
.0, 5
6−k
= 0, 65625
H
1
H
2
⇒ P (B) = P (H
1
)P (B/H
1

) + P (H
2
)P (B/H
2
)
• P (H
1
); P (H
2
)
→ Y ∼ B(n = 2; p = 0, 5)
P (H
2
) = P (Y = 0) = 0, 5
2
= 0, 25
P (H
1
) = 1 − P (H
2
) = 1 − 0, 25 = 0, 75
• P (B/H
1
)
→ Z ∼ B(n = 4; p = 0, 5)
P (B/H
1
) = P (Z  1) = 1 − P (Z = 0) = 1 − C
0
4

.0, 5
0
.0, 5
4
= 0, 9375
• P (B/H
2
)
P (B/H
2
) = 0, 5
4
= 0, 0625
⇒
P (B) = 0, 75.0, 9375 + 0, 25.0, 0625 = 0, 71875
→ X ∼ N(µ
x
= 0; σ
2
x
= 4
2
); Y ∼ N(µ
y
= 0; σ
2
y
= 6
2
)

A
i
(i = 1, 2) A
i
= (|X|  2; |Y |  4)
P (A
i
) = P (|X|  2; |Y |  4) = P (|X|  2)P (|Y |  4) = 2Φ
0
(
2
4
)2Φ
0
(
4
6
)
= 2.0, 1915.2.0, 2486 = 0, 190428
⇒ P (
A
i
) = 1 − P (A
i
) = 1 − 0, 190428 = 0, 809572
→ B = A
1
A
2
+ A

1
A
2
+ A
1
A
2
= U − A
1
A
2
P (B) = 1 − P (
A
1
A
2
) = 1 − P (A
1
)P (A
2
) = 1 − 0, 809572
2
= 0, 344593
⇒ p ≈ f =
120
72.20
= 0, 083
→ X ∼ B(n = 72; p = 0, 083)
P (X = 1) = C
1

72
.0, 083
1
.0, 917
71
= 0, 0127
P (X = 2) = C
2
72
.0, 083
2
.0, 917
70
= 0, 0409
E(X) = np = 72.0, 083 = 5, 976
X
A
, X
B
→ X
A
∼ N(µ
A
= 11; σ
2
A
= 4
2
); X
B

∼ N(µ
B
= 10, 4; σ
2
B
= 2, 6
2
)
P (X
A
 10) = P (U 
10 − 11
4
= −0, 25) = 1−P (U > 0, 25) = 1−0, 4013 = 0, 5987
P (X
B
 10) = P (U 
10 − 10, 4
2, 6
= −0, 15) = 1−P (U > 0, 15) = 1−0, 4404 = 0, 5596
A : B = p : (1 − p); (0  p  1)
Y = pX
A
+ (1 − p)X
B
V (Y ) = p
2
V (X
A
)+(1−p)

2
V (X
B
) = p
2
.4
2
+(1−p)
2
.2, 6
2
= 22, 76p
2
−13, 52p+6, 76
f(p) = 22, 76p
2
− 13, 52p + 6, 76
f
′
(p) = 45, 52p − 13, 52 = 0 ⇔ p =
13, 52
45, 52
≈ 0, 297
f”(p) = 45, 52 > 0