Oxford Fajar, Mathematics T STPM, Full Worked Answer Chapter 15, Probability Distributions
19
© Oxford Fajar Sdn. Bhd. (008974-T) 2014 Focus on Exam 15 1 (a) P( x ) = 1 1 1 2 3 4 5 6 1 21 1 21 k k k ( ) + + + + + = = = (b) E( ) ( ) ( X x X x x k = + + P 1 1 21 3 1 2 3 2 2 2 + + + = = 4 5 6 91 21 4 2 2 2 ) ∑ ∑ Var P 21 91 21 21 ( ) ( ) ( X X x =∑ - =∑ - = + + + x x 2 2 2 2 3 3 3 3 1 1 2 3 4 m + + - = 5 91 2 9 3 3 2 6 21 2 ) 2 (a) Total number of different pairs = 4! = 24 (b) (i) P( ) X = = 2 6 24 (ii) P( ) X = = 1 8 24 CHAPTER 15
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Transcript of Oxford Fajar, Mathematics T STPM, Full Worked Answer Chapter 15, Probability Distributions
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
Focus on Exam 15
1 (a) P( x) = 1
11 2 3 4 5 6 1
211
21
k
kk
( )+ + + + + =
=
=
(b) E ( ) ( )
(
X x X
xx
k
= + +
P
1
1
21
3
1 2 32 2 2 ++ + +
=
=
4 5 6
91
21
4
2 2 2 )
∑
∑
Var P
21
91
21
21
( ) ( )
(
X X
x
= ∑ −
= ∑
−
= + + +
x
x
2 2
2
2
3 3 3 311 2 3 4
m
++ + −
=
591
2
9
3 3
2
621
2
)
2 (a) Total number of different pairs = 4!
= 24
(b) (i) P( )X = =26
24
(ii) P( )X = =18
24
CHAPTER 15
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEAD Mathematics (T) Third Term2
(c) X 0 1 2 3 4
P(X = x)9
24
8
24
6
240
1
24
(d) E P
( ) ( )
[ ( ) ( ) ( ) ( ) ( )]
X x X= ∑
= + + + +1
240 9 1 8 2 6 3 0 4 1
= 1
3 (a) x = 1 2 3 4 5 6 7 15 30 100, , , , , , , , ,
(b) Y X= −15
(c) E E
( ) ( )
(
Y X= −
= + + + + + + + + +
15
1
101 2 3 4 5 6 7 15 30 1000 15
17 3 15
2 3
)
.
.
−
= −=
The expected gain of a player is RM2.30 (profit)
4 (a) P(X = 1) = P(X = 2)
= 2p
P( X = 3) = P( X = 4)
= P( X = 5)
= p
P(x) =1∑
2(2 p) + 3p = 1
p = 1
7E(X ) = x P(X )
= 1
7+ 4
1
7+ 3
1
7+ 4
1
7+ 5
1
7
= 17
7
(b)
2(2 p) + 3p = 1
p = 1
7E(X ) = x P(X )
= 1
7+ 4
1
7+ 3
1
7+ 4
1
7+ 5
1
7
= 17
7
Var(X) = x2P(X)17
7
2
= 1
7+ 16
1
7+ 9
1
7+16
1
7+ 25
1
7
17
7
2
= 67
7
289
49
= 3.673
∑
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 3
5 (a) X 0 1 2 3
P(X) 2k k 0 k
∑ ===
P
( )
.
X
k
k
1
4 1
0 25
(b) E P
( ) ( )
( )( . ) ( )( . ) ( ) ( . )
X x X= ∑= + + +0 2 0 25 1 0 25 2 0 3 0 25
= 1
Var( ) ( )
( . ) ( . ) ( ) ( .
X x X= ∑ −= + + +
2 2
2 2 2 2
1
0 0 25 1 0 25 2 0 3 0
P
225 1
2 5
2)
.
−=
6 (a) X 0 1 2 3 4 5 6
P(X = x)1
16
1
8
3
16
1
4
3
16
1
8
1
16
(b) Y 0 1 2 3 4 6 9
P(Y = y)7
16
1
16
2
16
2
16
1
16
2
16
1
16
(c) E P
E P
( ) ( )
( )
( ) ( )
(
X x X
Y y Y
= ∑
= + + + + + +
== ∑
=
1
160 2 6 12 12 10 6
3
1
1600 1 4 6 4 2 9
1 625
+ + + + + +
=
)
.
(d) Var Var( ) ( )Y X>
7 (a) X m
X r X r
~ ( )
( ) ( )
Po
P P = = = +1
e m mr
r!= e m mr+1
(r +1)!
1= m
r +1 r +1= m
r = m 1
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEAD Mathematics (T) Third Term4
(b) 2P P
2e e
( ) ( )
( )! !
X r X r
m
r
m
rm
rm
r
= + = =
+
=
−+
−
1
1
1
2 1
2 1
m r
r m
r
= += −
∴ must be an oodd integer.
8 (a) X
X
~
( )
B
P
10 0001
10 000
010 000
0
1
10 000
99990
= =
110 000
0 3679
10 000
= .
(b) P P P( ) ( ) ( )
. ( . )( . )
X X X�1 0 1
0 36791000
10 0001 0 9999 9
= = + =
= +
9999
0 3679 0 03679
0 4047
= +=
. .
.
9 (a) (i) X
X
~ ( )
( )! !
.
Po
P e
3
3 1 33
2
3
3
0 6472
32 3
� = + + +
=
−
(ii) P P( ) ( )
.
.
X X� �4 1 3
1 0 6472
0 3528
= −= −=
(b) P e
P e
P e
( )
( ) ( )
( )
X
X
X
= == =
= =
−
−
−
0
1 3
29
2
3
3
3
= =
= = − + + +
−
−
P e
P e
( )
( )
X
X
39
2
4 1 1 39
2
9
3
3
22
= 1 − e−3(13)
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 5
Y = Number of rented out customers
Y 0 1 2 3 4
P(Y ) e−3 3e−39
23e− 9
23e−
1–13e−3
E e e
e
( )
.
.
.
Y = + + +
+ −
= −= −=
− −
−
3 3
3
0 3 927
24 52
4 26 5
4 1 319
2 681
10 (a) Xt
X X
X X
~
~ ~ ( . )
( ) ( )
Po
Po Po
P P
e
80
20
800 25
3 1 2
1
⇒
= −
= − −
� �
00 252
1 0 250 25
2
0 00216
. ..
!
.
+ +
=
(b) Xk
X
~
( ) .
Po
P
80
0 0 9
= =
ek
80 = 0.9
k = 80 ln 0.9
k = 8.429
11 Area under f(x) for −1 x 1 B not equal to 1 g(x) cannot be negative for any value of x.
12 (a)
01
1
2
f(v )
v
(b)
01
1
2
f(w)
w
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEAD Mathematics (T) Third Term6
(c) f
f d
d
( ),
,
( ) ( )
uu u
u u u u
u u
=−
=
= −( )∫
2 2 0 1
0
2 2
2
0
1
2
� �
otherwise
Var
uu
u u
0
1
3 4
0
12
3 2
2
3
1
21
6
∫
= −
= −
=
13 (a) ( )6 18 10
− =∫ x xk
d
[6x 9x2]
0k = 1
6k 9k 2 = 1
9k 2 6k +1= 0
(3k 1)2 = 0
k = 13
(b) Mean f d
d
=
= −
=
∫
∫
x x x
x x x
k
k
( )
( )
0
0
6 18
33 6
3
9
6
271
9
2 3
0
1
3x x−
= −
=
Variance f d
d
= −
= − −
∫
∫
x x x
x x x
k
k
2
0
2
2
0
6 181
9
( )
( )
µ
= −
−
= −
2
3 4
0
1
3
29
2
1
81
2
27
x x
99
2
1
81
1
81
1
162
−
=
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 7
(c) When x < 0, F(X ) = 0
When 0 x <�13
F( X ) = f (t) dt0
x
= (6 18t)0
xdt
= 6t 9t20
x
= 6x 9x2
When x �1
3, F( x) = 1
F( )
,
,
,
x
x
x x x
x
=
<
− <
0 0
6 9 01
3
11
3
2 �
�
(d)
13
x
F(x)
o
1
14 (a) kx dx = 11
5
kx
k k
k
2
1
5
21
25
2
1
21
1
12
=
− =
=
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEAD Mathematics (T) Third Term8
(b)
1 5
112
512
x
f(x)
(c) P( )x > +41
2
4
12
5
12
9
243
8
=
=
=
(1)
(d) P( )2 31
2
2
12
3
12
5
24
< < = +
=
x (1)
15 a x x x( )2 12
0
2
− =∫ d
a x2 1
3x3
0
2
= 1
a (483
) 0 = 1
a = 34
When x < 0, F(X ) = 0
When 0 x < 2, F(X ) = f d( )t tx
0∫
= −
= −
= −
= −
∫3
42
3
4
1
3
3
4
1
3
3
4
1
4
0
2
2 3
0
2 3
2
x
x
t t t
t t
x x
x x
( ) d
33
When x 2, F(X ) = 1
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 9
F( )
,
,
,
x
x
x x x
x
=
<
−
>
0 0
3
4
1
40 2
1 2
2 3 � �
16 (a) k x x
k x x
k
( )
( ) ( )
− =
−
=
− − −[ ] =
∫ 2 1
1
22 1
8 8 2 4 1
2
4
2
2
4
d
k = 1
2
(b) When x < 2, F(X ) = 0
When 2 < x < 4, F(X ) = f d( )t tx
2∫
= −
= −
= −
− −
∫1
22
1
2
1
22
1
2
1
22 2 4
2
2
2
2
( )
( )
t t
t t
x x
x
x
d
= − +1
412x x
F when
F
( ) .
( )
,
,
,
X x
X
x
x x x
x
= >
=
<
− +
>
1 4
0 2
1
41 2 4
1 4
2 � �
(c) P F( ) ( )X < =3 3
= 94
3+1
= 14
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEAD Mathematics (T) Third Term10
(d)
E f d
d
( ) ( )
( )
X x x x
x x x
x x
=
= −
= − +
∫
∫2
4
2
4
2 3
1
22
1
2
1
3
= − +
− − +
=
2
4
1
216
64
34
8
3
10
3
Var( X ) = x2f (x) dx103
2
2
4
= 12
x2(x 2) dx100
92
4
= 12
14
x4 23
x3
2
4100
9
= 12
64128
34
163
1009
= 29
17 (a) f
otherwise
( )
,
,
,
x
x
x=
−
2
31 0
1
60 2
0
� �
� �
16
23
x
f(x)
−1 20
(b) For
P P P
0 2
0 0
2
3
1
6
<= + < <
= +
x
X x X X x
x
�
� �( ) ( ) ( )
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 11
(c) For
d
−
=
=
−
−
∫
1 0
2
3
2
3
1
1
� �
�
x
X x
t
x
x
,
( )P
t
= +2
3
2
3x
(d) E d d
( )X x x x x
x x
= +
=
+
−
−
∫ ∫2
3
1
6
2
3 2
1
6 2
1
0
0
2
2
1
0 2
= −
+ −
= − +
=
0
2
2
30
1
2
1
62 0
1
3
1
30
( )
(e) P
B
P C
( )
~ ,
( )
X
Y
Y
< = +
=
= =
12
3
1
65
6
105
6
8 105
68
=
8 21
6
0 2907 .
18 (a) F
F
( ),
( ),
,
( )
( )
Xx
b x x x
x
b
=<
−>
=× − =
0 0
6 0 4
1 4
4 1
6 16 64 1
2 3 � �
b = 1
32
(b) fd
dF
otherwise
( ) ( )
( ),
,
xx
X
x x x
=
= −
1
3212 3 0 4
0
2 � �
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEAD Mathematics (T) Third Term12
(c) E d( ) ( )X x x x x
xx
= −
= −
=
∫1
3212 3
1
324
3
4
2
2
0
4
34
0
4
(d) Var d( ) ( )X x x x x
xx
= − −
= −
−
=
∫ 2 2
0
4
45
0
4
1
3212 3 4
1
323
3
54
4
5
19 (a) P
, < 3
3 4 ( ) ,
,
X
x
ax ax b x
x
> = − +
x
1
8
0 4
2 � �
�
P P
F
P
( ) ( )
( )
,
( ),
,
(
X x X x
X
x
ax ax b x
x
X
�
�
� �
�
= − >
= − − +
1
0 3
1 8 3 4
1 4
2
�� 3 1 9 24
0
1 15 0
) ( )= − − +=
+ − =
a a b
a b
(b) P( ) ( )X a b
a b
a
a
b a
� 4 1 1 32
1
16 0
1 0
1
16
16
= − − +=
− =− =
===
6a
(c) P
1 0.25
( . ) ( . ( . ) )X � 3 5 1 3 5 8 3 5 162= − − +−=
= 3
4
P( ) ( )X a b
a b
a
a
b a
� 4 1 1 32
1
16 0
1 0
1
16
16
= − − +=
− =− =
===
6a
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 13
(d) fd
dF
otherwise
othe
( ) ( )
,
,
,
,
xx
ax a x
x x
=
=− +
=− +
X
2 8 3 4
0
2 8 3 4
0
� �
� �
rrwise
(e) E f d
d
( ) ( )
( )
x x x x
x x x
xx
=
= − +
= − +
= − +
∫∫
3
4
3
4
32
3
4
2 8
2
34
128
364
− − +
=
54
336
10
3
(f) Y
Y
Y
=
= =
Number of values less than 3.5
P
~ ,
( )
B 33
4
2 33
4
12
44
39
16
1
4
27
64
=
=
20 (a) f
When F
When
( )
,
,
,
, ( )
x
x
x x
xx
x X
x
=
<
−
>
< =
0 0
5
40 1
1
41
0 0
0
2
� �
� ��1
5
4
5
4
1
2
5
4
1
2
0
0
2
0
,
( ) ( )F f d
d
X t t
t t
t t
x
x
x
x
=
= −
= −
= −
∫
∫
xx
x
X t tt
t
t t
x
2
0
1
21
2
0
1
5
4
1
4
5
4
1
2
When
F d d
>
= −
+
= −
∫ ∫
,
( )
11
1
1
4
1
5
4
1
2
1
4
11
11
4
0 0
−
= −
− −
= −
∴ =
<
t
x
x
X
x
x
F ( )
,
55
4
1
20 1
11
41
2x x x
xx
−
− >
,
,
� �
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEAD Mathematics (T) Third Term14
f
When F
When
( )
,
,
,
, ( )
x
x
x x
xx
x X
x
=
<
−
>
< =
0 0
5
40 1
1
41
0 0
0
2
� �
� ��1
5
4
5
4
1
2
5
4
1
2
0
0
2
0
,
( ) ( )F f d
d
X t t
t t
t t
x
x
x
x
=
= −
= −
= −
∫
∫
xx
x
X t tt
t
t t
x
2
0
1
21
2
0
1
5
4
1
4
5
4
1
2
When
F d d
>
= −
+
= −
∫ ∫
,
( )
11
1
1
4
1
5
4
1
2
1
4
11
11
4
0 0
−
= −
− −
= −
∴ =
<
t
x
x
X
x
x
F ( )
,
55
4
1
20 1
11
41
2x x x
xx
−
− >
,
,
� �
(b) Probability at least two independent values of X is greater than 3
= −= −
= −
= −
=
1 3 3
1 3 3
111
12
1121
14423
144
2
P P
F F
( ). ( )
( ). ( )
X X� �
21
(b)
f
otherwise
E f d
( ),
,
( ) ( )
x b aa x b
X x x x
b a
x
a
b
= −< <
=
=−
∫
1
0
21
2
2
=−
−
= +
+ =
< =
−−
=
−
a
b
b a
b a
a b
a b
X
a
b a
1
2
1
24
35
83 5
824 8
2 2
( )
( )P
aa b a
b a
b b
b
b
= −= += + −= −=
5 5
24 5 3
5 3 4
8 12
36 8
( )
b
a
=
=
= −
= −
36
89
2
49
21
2
(a)
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 15
f
otherwise
E f d
( ),
,
( ) ( )
x b aa x b
X x x x
b a
x
a
b
= −< <
=
=−
∫
1
0
21
2
2
=−
−
= +
+ =
< =
−−
=
−
a
b
b a
b a
a b
a b
X
a
b a
1
2
1
24
35
83 5
824 8
2 2
( )
( )P
aa b a
b a
b b
b
b
= −= += + −= −=
5 5
24 5 3
5 3 4
8 12
36 8
( )
b
a
=
=
= −
= −
36
89
2
49
21
2
22 (a) k
xx
kx
k
k
210
15
10
15
1
11
1
15
1
101
10 15
150
d∫ =
−
=
− −
=
− −
=
=
1
30k
(b) E d
ln
ln ln
ln
( )
( . )
. (
X xx
x
x
=
= [ ]==
∫30
30
30 15 10
30 1 5
210
15
10
15
SShown)
(c) 30
0 5
301
0 5
301 1
100 5
1 1
10
210
10
xx
x
m
m
m
m
d =
−
=
− −
=
− =
∫ .
.
.
−−
= −
=
=
1
601 1
10
1
605
6012
m
m
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEAD Mathematics (T) Third Term16
23
1150
x
f(x)
0
(a) E(X ) = 75
(b) Var( )X x x
x
=
−
=
−
=
∫ 2 2
0
150
3
0
150
2
2
1
15075
1
150 375
150
3
d
−−
=
=
75
1875
1875
2
Standard deviation
= 43 3.
24
P
P
P
( ) .
.
( . ) .
X
z
z
< =
< −
=
< − =
13 0 1151
130 1151
1 2 0 1151
ms
∴ − = −> =
< −
=
13 1 2
36 0 1357
36
m s
ms
.
( ) .P
P
X
z 00 8643
1 1 0 8643
36 1 1
2
.
( . ) .
.
(
P z < =− =m s
)) ( ) : .− === −
1 23 2 3
10
36 11
ssm
= 25
25 X ~ N(303, 42)
P(295 < X < 305) P
P
= − < < −
= − < <=
295 303
4
305 303
3
2 0 5
0
Z
Z( . )
.66687
.... 1
.... 2
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 17
26 X ~ N( m, 1002)
P
P
P
( ) .
.
X
Z
Z
>
> −
< −
1169 0 117
1169
1000 117
1169
100
�
�
�
m
m00 883
1 19 0 883
1169
1001 19
.
( . ) .
.
P
Z < =
∴ − >m
1169 119
1050
− ><
mm
P( X > 879) 0.877
P Z > 879100
0.877
�
�
P
( . ) .
.
Z < =
∴ − < −
1 16 0 877
879
1001 16
m
879 116
995
995 105
− < −>
∴ < <
mmm 00
27 (a) (i) X
X Z
Z
~ ( , )
( . )
.
N
P P
P
P
475 20
500500 475
20
1 25
0 10565
2
>( ) = > −
= >=
rroportion 10.6%=
(ii)
X
X
Z
~ ,
.
.
.
N
P
P
m
m
m
20
500 0 001
500
200 001
500
203 09
2( )>( ) =
> −
=
− = 00
500 20 3 090
438 2
m = − ( )=
.
.
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEAD Mathematics (T) Third Term18
(b) P
Probabilityof nooverflow for 2cups
( ) .
.
X < = −=
=
500 1 0 10565
0 89435
00 89435
0 79986
2.
.
( )=
28 (a) X
X Z
Z
~ , .
.
.
N
P P
P
2 0 4
11 2
0 4
2 5
2( )<( ) = < −
= < −( ) dp
P
= ( )= ( )
>( )
0 00621 5
0 99379
1
2
.
.
X =2
0 98762.
(b) Y
Y Y
~ ( , . )
( ) ( )
. ( . ) ( . ) (
B
P P
10 0 5
4 1 3
1 0 5 10 0 5 45 0 5 2010 10 10
� �= −= − + + + 00 5
1 0 0742
0 9258
10. )
.
.
= −=
29 X
X
Z
Z
~ ( . , )
( . ) .
. ..
( .
N
P
P
P
1 00
1 05 0 01
1 05 1 000 01
1 28
2s
s
> =
> −
=
> 22 0 01
0 051 282
0 039 3
200 0 01
) .
..
. ( )
~ ( , . )
=
=
=s
s d.p
BY
© Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 19
30 (a) X
X
Z
Z
~ ,
.
.
. .
N
P
P
P
5
3 0 123
3 50 123
1 16 0 123
2
2s
s
( )<( ) =
< −
=
< −( ) =−ss
s
= −
== ( )
1 16
1 724
1 72 2
.
.
. d.p
(b) P P
P
d.p
X Z
Z
>( ) = > −
= >( )== ( )
99 5
1 72
2 32
0 01017
0 0102 4
.
.
.
.
(c) Y
Y Y
~ , .
. . .
.
B
P P
5 0 123
2 1 1
1 0 877 5 0 877 0 123
0
5 4
( )( ) = − ( )
= − ( ) − ( ) ( )=
� �
11174
(d) Y
Y
Y Y
~ , .
~ .
B
Po approximation
P P
e
100 0 0102
1 02
3 1 2
1
( )( )
( ) = − ( )
= − −
� �
11 022
1 1 021 02
2
0 0840
. ..
.
+ +
=
