>
)Q1.
a) Let X be the monthly income of HK resident.
X~N (19000, 41502)
P
(X>20000) = P (
>
)
=P (Z>0.24)
=0.5 - 0.0948
=0.4052
b) P (15000<X<25000) = P (
<
<
)
=P (-0.96<Z<1.45)
=0.3315 + 0.4265
=0.758
c) P (X<24000) = P (
<
)
= P (Z<1.20)
=0.5+0.3849
=0.8849
d) P (X<31000) = P (
<
)
= P (Z<2.89)
=0.5 + 0.4981
=0.9981
e) P (9000<X<18000) = P (
<
<
)
= P (-2.41<Z<-0.24)
=0.4920 - 0.0948
=0.3972
f) P (X<12000) = P (
<
)
= P (Z<-1.69)
=0.5 - 0.4545
=0.0455
g) P (X>10000) = P (
>
)
= P (Z>-2.17)
=0.5 + 0.4850
=0.9850
h) P (21000<X<30000) = P (
<
<
)
= P (0.48<Z<2.65)
=0.4960 - 0.1844
=0.3116
i) P (X>16500) = P (
>
)
= P (Z>-0.60)
=0.5 + 0.2257
=0.7257
j) P (X>19000) = P (
>
)
= P (Z>0)
=0.5
Q2.
a) P (punctual) = P (does not oversleeps and no traffic jam)
= P (does not oversleeps) × P (no traffic jam)
=0.9 × 0.8
=0.72
Let X be the number of days he is punctual.
X~ Bin (5, 0.72)
f(X) = 5CX (0.72)X (1-0.72)5-X for X = 0,1,2,…5
P (X=4) = f (4) = 5C4 (0.72) 4 (1-0.72) 5-4 =0.3762
b) P (X≧3) = f (3) + f (4) + f (5)
=
5C3 (0.72) 3 (1-0.72) 5-3 + 5C4 (0.72) 4 (1-0.72)
5-4 + (0.72) 5
=0.2926 + 0.3762 + 0.1935
=0.8623
c) Let Y be the number of week in which he is punctual in at least 3 days in the week.
Y~ Bin (14, 0.8623)
f(Y) = 14CY (0.8623) Y (1-0.8623)14- Y for Y = 0,1,2,…14
P (Y=11) = f
(11) = 14C11 (0.8623) 11 (1-0.8623) 3
=0.1863
d) E (Y) = np
=14×0.8623
=12.0722
Therefore, Peter is expected to be punctual for 12 weeks.
e) P (punctual in the 1st semester and punctual in the 2nd semester)
= P (punctual in the 1st semester) × P (punctual in the 2nd semester)
=0.1863×0.1863
=0.0347
Q3
a) Let X be the number of accidents per day.
X~ Poisson (λ=1.5)
f(X) =
for X= 0,1,2,…….
P(X≧2) = 1 - f(0) - f(1)
= 1-
- 
(1.5)
= 1- 0.2231- 0.3347
= 0.4422
Let Y be the number of day that at least 2 accidents in a period.
Y~ Bin (10, 0.4422)
f(Y) = 10CY (0.4422)Y (1-0.4422)10-Y for Y = 0,1,2,3,…10
P (Y>6)= f(7) + f(8) + f(9) + f(10)
=10C7 (0.4422)7 (1-0.4422)10-7 + 10C8 (0.4422)8 (1-0.4422)10-8 +
10C9
(0.4422)9 (1-0.4422)10-9 +(0.4422)10
=0.0688 +
0.0205 + 3.6045×10-3 + 2.857×10-4
=0.0932
b) Let U be the number of weeks in which the road is labeled as ‘traffic black point’.
U~ Bin (25, 0.0932
E (Y) = np
=25 (0.0932)
=2.3298
c) P (labeled as a “traffic black point” in 3 consecutive periods)
= 0.09323
=8.0933×10-4
Q4
a) The proportion of students who get score ’A’,
Let X be the total marks that a student can get.
X~ N (73.2, 102)
P (X>85) =
P(
>
)
=P (Z> 1.18)
=0.5-0.3810
=0.1190
Let Y be the number of students who get score “A”.
Y~ Bin (25, 0.1190)
E(Y)= 25 (0.1190)
=5.95
Therefore, the expected number of students who get ‘A’ is 6.
b) X~N (73.2, 102)
P (X>x) =1-0.9
P
(
>
)=0.10000000000000
P
(0≦Z≦
)=0.5-0.1000000
∴
=1.2800
000∴x=86
c) Let Q be the mark of the student.
X~N (73.2, 102)
P (X> Q) =1-0.75
P(
>
)=0.250000000000000
P
(0≦Z≦
)=0.5-0.25000000
∴
=0.6700
000∴Q=79.9
Therefore the mark of the student is 79.9.
d) P (X>60)
= P (
>
)
=P (Z>-1.32)
=0.5 + 0.4066
=0.9066
e) Let W be the number of student who can pass the course.
W ~ Bin (20, 0.9066)
f(W)= 20CW (0.9066)W (1-0.9066)20-W for Y = 0,1,2,3,…20
f(18) = 20C18
(0.9066)18 (1-0.9066)20-18
=0.2837
f) Let U be the number of student who can pass the course.
U ~ Bin (200, 0.9066)
f(U)= 200CU (0.9066)U (1-0.9066)200-U for U = 0,1,2,3,…200
f(180) = 200C180
(0.9066)180 (1-0.9066)200-180
=0.0890
Q5. Let X be the number of enquiries per week.
X~ Poisson (λ=1.2×7=8.4)
f (X) =
for X= 0,1,2,…….
f (5)=
= 0.0784
Let Y be the week that there are exactly 5 enquiries in the week.
Y~ Bin (52, 0.0784)
E (Y)= 52 (0.0784)
=4.0751