SM025 Self-Directed Learning

NSO/KKW/CCW

a) (1 < < 5)
b) (2 ≤ ≤ 4)
c) (3 < ≤ 6)
d) ( ≤ 3)

Solution 2:
a) (1 < < 5) = ( = 2) + ( = 3) + ( = 4)

= 0.2 + 0.3 + 0.25
= 0.75

b) (2 ≤ ≤ 4) = ( = 2) + ( = 3) + ( = 4)
= 0.2 + 0.3 + 0.25
= 0.75

c) (3 < ≤ 6) = ( = 4) + ( = 5) + ( = 6)
= 0.25 + 0.1 + 0.05
= 0.4

d) ( ≤ 3) = ( = 1) + ( = 2) + ( = 3)
= 0.1 + 0.2 + 0.3
= 0.6

Cumulative Distribution Function

( ) is found by adding together all the probabilities for those outcomes that are equal to
or less than .
( ) = ( ≤ )

Example 1:

The discrete random variable has the probability function

0.1 = −2, −1
( = ) = { = 0,1
= 2
0.3

a) Find the value of .

b) Find the cumulative distribution function ( ).

c) Find (0.3).

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Solution 1:
a) Since this is a discrete random variable

∑ ( = ) = 1

0.1 + 0.1 + + + 0.3 = 1 Substitute the value of
0.5 + 2 = 1
2 = 0.5
= 0.25

0.1, = −2, −1
b) ( = ) = { , = 0,1
= 2
0.3,

For < −2, ( ) = 0 ( ) = ( ≤ )

For −2 ≤ < −1, ( ) = 0.1

For −1 ≤ < 0, ( ) = 0.1 + 0.1 = 0.2

For 0 ≤ < 1, ( ) = 0.1 + 0.1 + 0.25 = 0.45

For 1 ≤ < 2, ( ) = 0.1 + 0.1 + 0.25 + 0.25 = 0.7

For ≥ 2, ( ) = 0.1 + 0.1 + 0.25 + 0.25 + 0.3 = 1.0

0, < −2
, −2 ≤ < −1
0.1
− 1 ≤ < 0
∴ ( ) = 0.2, 0 ≤ < 1
0.45, 1 ≤ < 2
≥ 2
0.7,

{ 1,

c) (0.3) = (0) = 0.45 (0.3) means ( ≤ 0.3) but does not take any values
between 0 and 1 so ≤ 0.3 is the same as ≤ 0 and
thus (0.3) = (0).

Example 2:

The discrete random variable has a cumulative distribution function ( ) defined by

0, = 0
= 1,2,3,4,5
( ) = {1 + ,
6 > 5

1,

a) Find the value of ( ≤ 4).

b) Show that ( = 4) is 16.
c) Find the probability distribution table for .

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Solution 2: ( ) = ( ≤ )

a) ( ≤ 4) = (4)
1+4

=6
5

=6

b) ( = 4) = (4) − (3) ( = ) = ( ≤ ) − ( ≤ − 1)
= ( ) − ( − 1)
1+4 1+3
=6−6

54
=6−6

1
=6

c) ( = 1) = (1) − (0) = 1+1 − 0 = 2 = 1
6 6 3

( = 2) = (2) − (1) = 1 + 2 − 1 + 1 = 3 − 2 = 1
6 6 6 6 6

1+3 1+2 4 3 1
( = 3) = (3) − (2) = 6 − 6 = 6 − 6 = 6

( = 4) = 1
6

( = 5) = (5) − (4) = 1 + 5 − 1 + 4 = 6 − 5 = 1
6 6 6 6 6

1 2 3 4 5

( = ) 1 1 1 1 1

36666

Expectation and Variance

( ) = ∑ ( = )
( 2) = ∑ 2 ( = )
( ) = ( 2) − [ ( )]2

Example 1:

A biased die has the probability distribution

1 2 3 4 5 6

( = ) 0.1 0.1 0.1 0.2 0.4 0.1

Find ( ) and ( )

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Solution 1:
( ) = 1(0.1) + 2(0.1) + 3(0.1) + 4(0.2) + 5(0.4) + 6(0.1) = 4
( 2) = 12(0.1) + 22(0.1) + 32(0.1) + 42(0.2) + 52(0.4) + 62(0.1) = 18.2
( ) = ( 2) − [ ( )]2 = 18.2 − 42 = 2.2

Example 2:

The random variable has the following probability distribution.

1 2 3 4 5
0.1
( = ) 0.1 0.2

Given ( ) = 2.9, find the value of and . Hence, calculate ( ).

Solution 2:

∑ ( = ) = 1

0.1 + + + 0.2 + 0.1 = 1
+ = 0.6
= 0.6 −

( ) = 2.9 ( ) = ∑ ( = )
1(0.1) + 2 + 3 + 4(0.2) + 5(0.1) = 2.9

2 + 3 = 1.5
2(0.6 − ) + 3 = 1.5

= 0.3

= 0.6 − 0.3
= 0.3

1 2 3 4 5

( = ) 0.1 0.3 0.3 0.2 0.1

( 2) = 12(0.1) + 22(0.3) + 32(0.3) + 42(0.2) + 52(0.1) ( 2) = ∑ 2 ( = )
= 9.7

( ) = ( 2) − [ ( )]2
= 9.7 − 2.92
= 1.29

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Example 3: = 1,2,3,4,5,6
ℎ
The random variable has probability function
1

( = ) = {6 ,
0,

a) Find ( ) and show that ( ) = 1325.
b) Find (2 − 1).

c) Find (3 − 2 ).

Solution 3:

a) ( ) = 1 (1) + 2 (1) + 3 (1) + 4 (1) + 5 (1) + 6 (1)

666666 ( ) = ∑ ( = )

7
=2

( 2) = 12 1 + 22 1 + 32 1 + 42 1 + 52 1 + 62 1 ( 2) = ∑ 2 ( = )
(6) (6) (6) (6) (6) (6)

91
=6

( ) = ( 2) − [ ( )]2
91 7 2

= 6 − (2)
35

= 12

b) (2 − 1) = 2 ( ) − 1 ( + ) = ( ) +
7

= 2 (2) − 1
=6

c) (3 − 2 ) = (−2)2 ( ) ( + ) = 2 ( )
35

= 4 (12)
35

=3

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Exercise

1. The random variable has probability function

= 1,2,3,4,5,6
( = ) = {21 , ℎ

0,

a) Construct the probability distribution table of .

b) (2 < ≤ 5)

c) ( )

d) ( )

e) (3 − 2 )

2. The discrete random variable has the probability distribution as given below.

-2 -1 0 1 2 3
0.1
( = ) 0.1 0.2 0.3 0.1

Find

a)

b) (−1 ≤ < 2)

c) (0.6)

d) (2 + 3)

e) (2 + 3)

3. The discrete random variable has probability distribution function

(1 − ), = 0,1
( = ) = { ( − 1), = 2,3 where is a constant.

0, ℎ

a) Show that = 1
4

b) Find ( ) and show that ( 2) = 5.5

c) Find (2 − 2).

4. The random variable has a probability distribution table as given below.

1 2 3 4 5

( = ) 0.1 0.3 0.1

Given that ( ) = 3.1. Find

a) The value of and

b) ( )

c) (2 − 3)

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Solution
1.
a)

1 234 56

( = ) 1 2 1 4 5 2
21 21 7 21 21 7

b) (2 < ≤ 5) = ( = 3) + ( = 4) + ( = 5)
14 5

= 7 + 21 + 21
4

=7

c) ( ) = 1 ( 1 ) + 2 ( 2 ) + 3 (1) + 4 ( 4 ) + 5 ( 5 ) + 6 (2) ( ) = ∑ ( = )

21 21 7 21 21 7
13
=3

d) ( 2) = 12 ( 1 ) + 22 ( 2 ) + 32 (1) + 42 ( 4 ) + 52 ( 5 ) + 62 (2) ( 2) = ∑ 2 ( = )

21 21 7 21 21 7

= 21

( ) = ( 2) − [ ( )]2
13 2

= 21 − ( 3 )
20

=9

e) (3 − 2 ) = (−2)2 ( ) ( + ) = 2 ( )
20

= 4( 9 )
80

=9

2.
a) ∑ ( = ) = 1
0.1 + 0.2 + 0.3 + + 0.1 + 0.1 = 1
0.8 + = 1
= 0.2

-2 -1 0 1 2 3
( 0.1 0.2 0.3 0.2 0.1 0.1
= )

b) (−1 ≤ < 2) = ( = −1) + ( = 0) + ( = 1)
= 0.2 + 0.3 + 0.2
= 0.7

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c) (0.6) = ( ≤ 0.6)
= ( = −2) + ( = −1) + ( = 0)
= 0.1 + 0.2 + 0.3
= 0.6

d) ( ) = −2(0.1) − 1(0.2) + 0(0.3) + 1(0.2) + 2(0.1) + 3(0.1)
= 0.3

(2 + 3) = 2 ( ) + 3 ( + ) = ( ) +
= 2(0.3) + 3
= 3.6

e) ( 2) = (−2)2(0.1) + (−1)2(0.2) + 02(0.3) + 12(0.2) + 22(0.1) + 32(0.1)
= 2.1

( ) = ( 2) − [ ( )]2
= 2.1 − 0.32

= 2.01

(2 + 3) = 22 ( ) ( + ) = 2 ( )
= 4(2.01)
= 8.04

3.
a) ∑ ( = ) = 1
(1) + (0) + (1) + (2) = 1

4 = 1
1

= 4

0 1 2 3
( = ) 0
1 1 1
4 4 2

b) ( ) = 0 (1) + 1(0) + 2 (1) + 3 (1)

4 42

=2

( 2) = 02 1 + 12(0) + 22 1 + 32 1 = 5.5
(4) (4) (2)
c) ( ) = ( 2) − [ ( )]2

= 5.5 − 22

= 1.5

(2 − 2) = 22 ( ) ( + ) = 2 ( )
= 4(1.5)
=6

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4.
a) ∑ ( = ) = 1
0.1 + + + 0.3 + 0.1 = 1
+ = 0.5
= 0.5 −

( ) = 3.1
1(0.1) + 2 + 3 + 4(0.3) + 5(0.1) = 3.1

1.8 + 2 + 3 = 3.1
2 + 3 = 1.3

2(0.5 − ) + 3 = 1.3
= 0.3

= 0.5 − 0.3
= 0.2

1 2 3 4 5

( = ) 0.1 0.2 0.3 0.3 0.1

b) ( ) = 3.1
( 2) = 12(0.1) + 22(0.2) + 32(0.3) + 42(0.3) + 52(0.1) = 10.9

( ) = ( 2) − [ ( )]2
= 10.9 − 3.12

= 1.29

c) (2 − 3) = 22 ( ) ( + ) = 2 ( )
= 4(1.29)
= 5.16

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9.2 Continuous Random Variable

∫−∞∞ ( ) = 1
∫ ( ) = ( ≤ < )

= ( < ≤ )
= ( ≤ ≤ )
= ( < < )
Mode: highest point of the probability density function, ( )

Example 1:

Let be a continuous random variable with the probability density function given by:

( ) = { 0 (,5 − 2), −2 ≤ ≤ 2
ℎ

a) Find the value of .

b) Calculate

i) ( > 1.5)

ii) (| | ≥ 1)

Solution 1:
a) ∫−∞∞ ( ) = 1

−2 2 ∞

∫ 0 + ∫ (5 − 2) + ∫ 0 = 1

−∞ −2 2

3 2
[5 − ] = 1
3
−2

23 (−2)3
[5(2) − 3 − 5(−2) + 3 ] = 1

44
[ 3 ] = 1

= 3

44

b)

i) ( > 1.5) = ∫12.5 3 (5 − 2)
44

= 3 3 2
44 [5 − ]
3
1.5

3 23 1.53 23
= 44 [5(2) − 3 − 5(1.5) + 3 ] = 352

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ii) (| | ≥ 1) = ( ≤ −1) + ( ≥ 1) NSO/KKW/CCW

−1 2 3 Use the basic definition:
3 44 | | > ⇔ < − >
= ∫ 44 (5 − 2) + ∫ (5 − 2)

−2 1

3 3 −1 3 3 2
= 44 [5 − ] + [5 − ]
3 44 3
−2 1

3 (−1)3 (−2)3
= 44 [5(−1) − 3 − 5(−2) + 3 ]

3 (2)3 (1)3
+ 44 [5(2) − 3 − 5(1) + 3 ]

22
= 11 + 11

4
= 11

Example 2:

The continuous random variable has probability density function
, 0 ≤ ≤ 2

( ) = {2 (3 − ), 2 ≤ ≤ 3
0, ℎ

a) Find the value of .

b) Calculate

i) ( ≤ 1)
ii) ( ≥ 0.5)
iii) (| − 2| ≤ 0.5)

Solution 2:
a) ∫−∞∞ ( ) = 1

02 3 ∞

∫ 0 + ∫ + ∫ 2 (3 − ) + ∫ 0 = 1

−∞ 0 2 3

2 2 − 2 3
[ ] + 2 [3 ] = 1
2 2
0 2

22 32 22
[ 2 − 0] + 2 [3(3) − 2 − 3(2) + 2 ] = 1

2 + = 1

1
= 3

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b) ( ≤ 1) = ∫01 1
i) 3

1 2 1
= [ ]
3 2
0

11
= 3 [2 − 0]

1
=6

ii) ( ≥ 0.5) = ∫02.5 1 + ∫23 2 (3 − )
3 3

1 2 2 2 2 3
= 3[2 ] + [3 − ]
3 2
0.5 2

1 22 0.52 2 32 22
= 3 [ 2 − 2 ] + 3 [3(3) − 2 − 3(2) + 2 ]

51
=8+3

23
= 24

iii) (| − 2| ≤ 0.5) = (−0.5 ≤ − 2 ≤ 0.5) Use the basic definition:
= (1.5 ≤ ≤ 2.5) | | < ⇔ − < <

2 2.5

12
= ∫ 3 + ∫ 3 (3 − )

1.5 2

1 2 2 2 − 2 2.5
= 3[ 2 ] + 3 [3 ]
2
1.5 2

1 22 1.52 2 2.52 22
= 3 [ 2 − 2 ] + 3 [3(2.5) − 2 − 3(2) + 2 ]

71
= 24 + 4

13
= 24

Example 3:

The continuous random variable has probability density function

( ) = 1 , 0 ≤ ≤ 4

{8
0, ℎ

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a) Sketch the probability density function.
b) Find the mode of .
c) Find ( > 3)

Solution 3: ( )
a) 1/2

04

b) Mode = 4 The highest point is at = 4

c) ( > 3) = ∫34 1
8

1 2 4
= [ ]
8 2
3

1 42 32
= 8[2 − 2]

7
= 16

Cumulative Distribution Function

( ) = ( ≤ ) = ∫− ∞ ( ) ( ) has the same distribution as ( )
( ) = 0 ( ) = 1 and we use rather than to avoid
( ≤ ≤ ) = ( ) − ( )
( ) = ( ) confusion with the limit of integration



Median: ( ) = 0.5

Example 1:

The continuous random variable has probability density function

( ) = { 1 , 1 ≤ ≤ 3
4

0, ℎ

a) Find the cumulative distribution function of .

b) Find ( ≤ 1.5)

c) Find (1.2 ≤ ≤ 2.5)

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Solution 1:

a) For < 1, ( ) = ∫− ∞ ( ) ( ) has the same distribution as ( )
and we use rather than to avoid

confusion with the limit of integration
= ∫ 0

−∞

=0

For 1 ≤ ≤ 3, ( ) = ∫− ∞ ( ) Calculate for every interval of

1

1
= ∫ 0 + ∫ 4

−∞ 1

[ 2
= ]
8
1

2 − 1
=8

For > 3, ( ) = ∫− ∞ ( )

13 1

= ∫ 0 + ∫ 4 + ∫ 0
−∞ 1 3

= (3)

32 − 1
=8

=1

0, < 1

∴ ( ) = { 2 − 1 , 1 ≤ ≤ 3
8 > 3
1,

b) ( ≤ 1.5) = (1.5) ( ≤ ) = ( )
1.52 − 1

=8
5

= 32

c) (1.2 ≤ ≤ 2.5) = (2.5) − (1.2) ( ≤ ≤ ) = ( ) − ( )

2.52 − 1 1.22 − 1
=8−8

481
= 800

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Example 2:

The continuous random variable has probability density function

2 0 ≤ < 3
27 ,
( ) = 1 3 ≤ ≤ 5
3, ℎ
{ 0,

a) Find the cumulative distribution function of .

b) Find ( > 2.5)

Solution 2:
a) For < 0, ( ) = ∫− ∞ ( )



= ∫ 0

−∞

=0

For 0 ≤ < 3, ( ) = ∫− ∞ ( )
0 2

= ∫ 0 + ∫ 27

−∞ 0

3
= [81]0

3
= 81

For 3 ≤ ≤ 5, , ( ) = ∫− ∞ ( )

0 3 2

1
= ∫ 0 + ∫ 27 + ∫ 3
−∞ 0 3


= (3) + [3]3

33 − 3
= 81 + ( 3 )

− 2
=3

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For > 5, ( ) = ∫− ∞ ( )

0 3 2 5

1
= ∫ 0 + ∫ 27 + ∫ 3 + ∫ 0
−∞ 0 35

= (5)

5−2
=3

= 1 ( ) = 1

0, < 0
0 ≤ < 3
∴ ( ) = 3
81 , 3 ≤ ≤ 5
− 2 > 5

3,
{ 1,

b) ( > 2.5) = 1 − (2.5) ( > ) = 1 − ( ≤ )
2.53

= 1 − 81
523

= 648

Example 3:

The continuous random variable has cumulative distribution function given by

0, < 0
( ) = { 2 2 − 4,
0 ≤ ≤ 1
1, > 1

a) Find the probability density function of .

b) Find the median of .

Solution 3:

a) For < 0, ( ) = (0)
( ) = ( )

=0

For 0 ≤ ≤ 1, ( ) = (2 2 − 4)


= 4 − 4 3

= 4 (1 − 2)

For > 1, ( ) = (1)


=0

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∴ ( ) = {04, (1 − 2), 0 ≤ ≤ 1
ℎ

b) ( ) = 0.5

2 2 − 4 = 1
2

4 − 2 2 + 1 = 0
2

2 4 − 4 2 + 1 = 0

2 = 4 ± √(−4)2 − 4(2)(1) Use formula:
2(2) − ± √ 2 − 4
= 2

= 4 ± √8
4

2 = 1 ± √2
2

= √1 ± √2
2

= 1.31 or = 0.541

Since 0 ≤ ≤ 1, hence = 0.541

Expectation and Variance

( ) = ∫−∞∞ ( )
( 2) = ∫−∞∞ 2 ( )
( ) = ( 2) − [ ( )]2

Example 1:

A random variable has probability density function

1 ≤ ≤ 3
( ) = {4 , ℎ

0,

Find

a) ( )

b) ( )

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Solution 1:

a) ( ) = ∫−∞∞ ( )

1 3∞

= ∫ . 0 + ∫ . 4 + ∫ . 0
−∞ 1 3

3 2 Simplify first before integrate
= ∫ 4

1

1 3 3
= [ ]
4 3
1

1 33 13
= 4[3 − 3]

13
=6

b) ( 2) = ∫−∞∞ 2 ( )

1 3∞

= ∫ 2. 0 + ∫ 2. 4 + ∫ 2. 0

−∞ 1 3

3 3 Simplify first before integrate
= ∫ 4

1

1 4 3
= [ ]
4 4
1

1 34 14
= 4[4 − 4]

=5

( ) = ( 2) − [ ( )]2
13 2

=5−(6)
11

= 36

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Example 2:

A random variable has probability density function

2 0 ≤ < 3
15 ,
( ) = 1 3 ≤ ≤ 5
5 (5 − ), ℎ

{0,

Find

a) ( )

b) ( )

c) (3 + 2)

d) (2 − 3)

Solution 2:

a) ( ) = ∫−∞∞ ( )

03 5 ∞
2 1
= ∫ . 0 + ∫ . 15 + ∫ . 5 (5 − ) + ∫ . 0
−∞ 0 3 5

= 3 2 2 + 5 1 (5 − 2)
15 5
∫ ∫

03

2 3 3 1 5 2 3 5
= [ ] + [ − ]
15 3 5 2 3
0 3

2 33 03 1 5(52) 53 5(32) 33
= 15 [ 3 − 3 ] + 5 [ 2 − 3 − 2 + 3 ]

6 22
= 5 + 15

8
=3

b) ( 2) = ∫−∞∞ 2 ( )

03 2 5 1 ∞
15 5
= ∫ 2. 0 + ∫ 2. + ∫ 2 . (5 − ) + ∫ 2. 0

−∞ 0 3 5

= 3 2 3 + 5 1 (5 2 − 3)
15 5
∫ ∫

03

2 4 3 1 5 3 4 5
= [ ] + [ − ]
15 4 5 3 4
0 3

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2 34 04 1 5(53) 54 5(33) 34
= 15 [ 4 − 4 ] + 5 [ 3 − 4 − 3 + 4 ]

27 82
= 10 + 15

49
=6

( ) = ( 2) − [ ( )]2
49 8 2

= 6 − (3)
19

= 18

c) (3 + 2) = 3 ( ) + 2 ( + ) = ( ) +
8

= 3 (3) + 2
= 10

d) (2 − 3) = 22 ( ) ( + ) = 2 ( )
19

= 4 (18)
38

=9

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Exercise

1. The random variable has probability density function ( ) given by

0 ≤ ≤ 2
( ) = { (1 + 2) , ℎ

0,

Find

a) The value of .

b) (0.5 < < 1.5)

2. The random variable has probability density function ( ) given by

( ) = {0 , 2, 0 ≤ ≤ 2
ℎ

where is a positive constant.

a) Show that = 3.

8

b) Find the median.

c) Calculate ( ) and (3 + 2).

3. The continuous random variable has cumulative distribution function given by

0, < 1
1 ≤ ≤ 3
( ) = {1 ( 2 − 1),
8 > 3
1,

a) Find the probability density function of .

b) Sketch the probability density function of .

c) Find the mode of .

4. The continuous random variable has probability density function given by

( ) = { 3 (3 − 2), 0 ≤ ≤ 2
10 ℎ

0,

a) Find the cumulative distribution function of .

b) Find the ( < 1)

c) Calculate ( ) and (3 − 2 )

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Solution

1.

a) ∫−∞∞ ( ) = 1
02 ∞

∫ 0 + ∫ (1 + 2) + ∫ 0 = 1
−∞ 0 2
2 2
[ + ] = 1
4
0
22 02
[2 + 4 − 0 − 4 ] = 1

(3) = 1

1
= 3

b) (0.5 < < 1.5) = ∫01.5.5 1 (1 + 2 )
3

1 2 1.5
= 3 [ + ]
4
0.5
1 1.52 0.52
= 3 [1.5 + 4 − 0.5 − 4 ]

= 0.5

2. ∞
a) ∫−∞∞ ( ) = 1

02

∫ 0 + ∫ 2 + ∫ 0 = 1

−∞ 0 2 3 2

[ 3 ] = 1

0
23 03
[ 3 − 3 ] = 1

b) ( ) = 0.5 8
(3) = 1

3
= 8

3
8
∫ 2 = 0.5

0 3

[ 8 ] = 0.5

0
3 03
[ 8 − 8 ] = 0.5

3 = 4

= 3√4

= 1.587

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c) ( ) = ∫−∞∞ ( )
02 ∞
3
= ∫ . 0 + ∫ . 8 2 + ∫ . 0

−∞ 0 2
2
3
= ∫ 8 3

0 2

= 3 4]
[32
0
3
= 32 [24 − 04]

= 1.5

(3 + 2) = 3 ( ) + 2
= 3(1.5) + 2
= 6.5

3.

a) For < 1, ( ) = (0)



=0

For 1 ≤ ≤ 3, ( ) = [81 ( 2 − 1)]


1
= 8 (2 )
=4

For > 3, ( ) = (1)

=0

1 ≤ ≤ 3
∴ ( ) = { 4 , ℎ

0,

b) ( )

3
4

1 3
4

1

c) Mode is 3

4.
a) For < 0, ( ) = ∫− ∞ ( )



= ∫ 0

−∞

=0

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For 0 ≤ ≤ 2, ( ) = ∫− ∞ ( )
0
3
= ∫ 0 + ∫ 10 (3 − 2)

−∞ 0
3 3 2 3
= 10 [ 2 − ]
3
0
3 3 2 3 3(0)2 03
= 10 [ 2 − 3 − 2 + 3 ]

3 3 2 3
= 10 ( 2 − 3 )

= 1 2(9 − 2 )
20

For > 2, ( ) = ∫− ∞ ( )
02
3
= ∫ 0 + ∫ 10 (3 − 2) + ∫ 0

−∞ 0 2

= (2)

= 1 22(9 − 2(2))
20

=1

0, < 0
0 ≤ ≤ 2
∴ ( ) = {1 2(9 − 2 ),
20 > 2
1,

b) ( < 1) = (1)

= 1 12(9 − 2(1))
20
7
= 20

c) ( ) = ∫−∞∞ ( )
02 ∞
3
= ∫ . 0 + ∫ . 10 (3 − 2) + ∫ . 0

−∞ 0 2
2
3
= ∫ 10 (3 2 − 3)

0 4 2

= 3 ( 3 − )]
[10
4 0

= 3 [23 − 24 − 03 + 04
10 4 4]

= 1.2

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∞

( 2) = ∫ 2 ( )

−∞
02 ∞
3
= ∫ 2. 0 + ∫ 2. 10 (3 − 2) + ∫ 2. 0

−∞ 0 2
2
3
= ∫ 10 (3 3 − 4)

0 3 3 4 5 2

= [10 ( 4 − 5 )]

0
3 3(2)4 25 3(0)4 05
= 10 [ 4 − 5 − 4 + 5 ]

= 1.68

( ) = ( 2) − [ ( )]2
= 1.68 − 1.22

= 0.24

(3 − 2 ) = (−2)2 ( )
= 4(0.24)
= 0.96

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Chapter 10: Special Probability Distributions

10.1 Binomial Distribution ~ ( , )

A binomial distribution consists of a number of successive, independent trials of a random
experiment where each trial has only two possible outcomes.
( = ) = −
, ( ) =
, ( ) =

Example 1:
The probability that Alan wins a game is 0.3. Given that he plays only 5 games a week, find the
probability that in any week he wins
a) exactly four games
b) at least two games

Solution 1:
Let be the number of games he wins in a week
~ (5,0.3)

a) ( = 4) = 5 4(0.3)4(0.7) Use the formula
= 0.0284

b) ( ≥ 2) = 0.4718 Use the binomial table

Example 2:
A multiple choice test consists of 30 questions. There are four choices for each question and only
one choice is correct. A student makes a random guess for each question. Find the probability that
the student obtained
a) no correct answer
b) more than 16 correct answers

Solution 2:

Let be the number of correct answers

~ (30,0.25) One out of 4 answer is correct

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a) ( = 0) = 30 0(0.25)0(0.75)30
= 0.0002

b) ( > 16) = ( ≥ 17) =0.0002 To use table, change to ≥

Example 3:
In an orchid farm, 60% of the orchids will produce flowers. Find the probability that if a person
purchases 40 plants, at least 20 plants will produce flowers.

Solution 3:

Let be the number of plants that will produce flowers.

~ (40,0.6) Since the probability is more than 0.5,
̅ ~ (40,0.4) use ̅ as the opposite of
̅
( ≥ 20) = ( ̅ ≤ 20)
= 1 − ( ̅ ≥ 21) 20 20
= 1 − 0.0744
= 0.9256 21 19

22 18

Example 4: ̅

The manager of a carbonated drink company claims that 20% o2f 0the popul2a0tion drinks his product. If

10 people are randomly selected from the population, find the 22m21ean and 11va89riance of the number of

people drinking his product.

Solution 4:
Let be the number of people that drink the carbonated drink.
~ (10,0.2)
mean = = 10(0.2) = 2
Variance = = 10(0.2)(0.8) = 1.6

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10.2 Poisson Distribution ~ 0( )

It is used for independent events which occur at a constant rate within a given interval of
time.
( = ) = −

!

Mean =
Variance =

Example 1:
On average, three traffic accidents occur per month at a certain junction. Find the probability that in
any given month
a) exactly five accidents will occur
b) at least 1 accident will occur

Solution 1:

Let be the number of accidents occur in a month

~ 0(3) Use formula
Use Poisson table
a) ( = 5) = −335 = 0.1008
5!

b) ( ≥ 1) = 0.9502

Example 2:
The mean number of customers who arrive at a counter is two per minute. For any given minute,
find the probability that
a) no customers arrive
b) at least four customers arrive
c) less than three customers arrive

Solution 2:

Let be the number of customers arrive in a minute

~ 0(2)

a) ( = 0) = −220 = 0.1353 Use formula
0! Use Poisson table

b) ( ≥ 4) = 0.1429

c) ( < 3) = 1 − ( ≥ 3)

= 1 − 0.3233 = 0.6767

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Exercise

1. Suppose it is known that the probability of recovery from infection by a type of virus is 0.10.
From a random sample of 15 patients, find
a) the probability that at most 5 will recover
b) the probability that at least 5 will recover
c) the expected number who will recover

2. The probability of a man hitting the target is 0.25.
a) If he shoots 7 times, find the probability that he hits the target at least twice.
b) How many times must he shoot so that the probability of hitting the target at least once is
more than 2/3?

3. 12% of the packages delivered by a courier do not meet the targeted standards of delivery. If
a company sends 20 packages through this courier, find the mean and variance of the
number of packages that do not meet the targeted standards of delivery.

4. There are 6 rainy days during the month of January on average. Find the probability that in a
given month of January, there are
a) less than 4 rainy days
b) 6 to 8 rainy days

5. The mean number of cars owned by each family is two. Find the probability that a family has
a) no car
b) at least one car
c) at most two cars

6. The average number of vehicles breaking down at any hour has a Poisson distribution with a
mean of 2. Find the probability that
a) exactly 2 vehicles break down in a particular hour.
b) at most 5 vehicles break down in a particular hour
c) 9 to 12 vehicles break down in 12 hours.

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Solution
1. Let be the number of patients recover from infection by a type of virus .
~ (15,0.1)

a) ( ≤ 5) = 1 − ( ≥ 6) = 1 − 0.0022 = 0.9978
b) ( ≥ 5) = 0.0127
c) ( ) = = 15(0.1) = 1.5

2. a) Let be the number of times he hits the target .
~ (7,0.25)
( ≥ 2) = 0.5551

b) ~ ( , 0.25)

( ≥ 1) > 2
3

2
1 − ( = 0) > 3

( = 0) < 1
3

0(0.25)0(0.75) < 1 Use formula
3 Do not forget to change the inequality sign

1
log 0.75 < log 3

> log 1
3

log 0.75

> 3.82

= 4

3. Let be the number of packages delivered by a courier that do not meet the targeted
standards of delivery.
~ (20,0.12)

( ) = 20(0.12) = 2.4
( ) = 2.112

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4. Let be the number of rainy days in a month of July
~ 0(6)

a) ( < 4) = 1 − ( ≥ 4) = 1 − 0.8488 = 0.1512
b) (6 ≤ ≤ 8) = ( ≥ 6) − ( ≥ 9) = 0.5543 − 0.1528 = 0.4015

5. Let be the number of cars owned by each family
~ 0(2)

a) ( = 0) = −220 = 0.1353

0!

b) ( ≥ 1) = 0.8647

c) ( ≤ 2) = 1 − ( ≥ 3) = 1 − 0.3233 = 0.6767

6. Let be the number of vehicles which break down in any hour.
~ 0(2)

a) ( = 2) = −222 = 0.2707 2 × 12 = 24 vehicles

2!

b) ( ≤ 5) = 1 − ( ≥ 6) = 1 − 0.0166 = 0.9834

c) Let be the number of vehicles which break down in 12 hour

~ 0(24)
(9 ≤ ≤ 12) = ( ≥ 9) − ( ≥ 13) = 0.0054 − 0.0002 = 0.0052

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10.3 Normal Distribution ~ ( , 2)

A normal distribution is a probability distribution that is symmetric about the mean, showing

that data near the mean are more frequent in occurrence than data far from the mean.

In graph form, normal distribution will appear as a bell curve.

The standard normal random variable = − has mean 0 and variance 1, that is ~ (0,1)


Example 1:
X is normally distributed with a mean of 84 and a variance of 12. By using tables, find
a) ( > 80)
b) ( < 92)
c) (| − 84| < 2.9)

Solution 1:

~ (84,12)

a) ( > 80) = ( > 80−84) −
=
√12

= ( > 1.15)

= 0.1251

b) ( < 92) = ( < 92−84)

√12

= ( < 2.31)

= 1 − ( > 2.31)

= 1 − 0.0104

= 0.9896

c) (| − 84| < 2.9) = (−2.9 < − 84 < 2.9)

= (81.1 < < 86.9)

= ( 81.1 − 84 < < 86.9 − 84 )

√12 √12

= (−0.84 < < 0.84)

= 1 − 2 ( > 0.84)

= 1 − 2(0.2005)

= 0.599

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Example 2:
If ~ (30,16), find if
a) ( > ) = 0.1587
b) ( > ) = 0.8413
c) ( < ) = 0.6915
d) ( < ) = 0.3

Solution 2:

a) ( > ) = 0.1587
− 30

( > 4 ) = 0.1587
From the table ( > 1.00) = 0.1587

− 30
4 = 1.00
= 34

b) ( > ) = 0.8413 Symmetric around the mean
( < ) = 0.1587
− 30

( < 4 ) = 0.1587
From the table, ( > 1.00) = 0.1587

Hence,
− 30

( < 4 ) = ( > 1.00)
− 30

( < 4 ) = ( < −1.00)
− 30
4 = −1.00
= 26

c) ( < ) = 0.6915
( > ) = 0.3085
− 30

( > 4 ) = 0.3085
From the table, ( > 0.50) = 0.3085

− 30
4 = 0.50
= 32

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d) ( < ) = 0.3085 NSO/KKW/CCW
− 30 Symmetric around the mean

( < 4 ) = 0.3085
From the table, ( > 0.50) = 0.3085

Hence,
− 30

( < 4 ) = ( > 0.50)
− 30

( < 4 ) = ( < −0.50)
− 30
4 = −0.5
= 28

Example 3:
The scores of a certain examination are found to be normally distributed with a mean of 55 and
a standard deviation of 10. Calculate the probability that a candidate selected at random has a
score
a) between 45 and 65
b) more than 70
c) not less than 50

Solution 3:
Let be the score of the candidate
~ (55,102)

a) (45 < < 65) = (45−55 < < 65−55)

10 10

= (−1 < < 1)

= 1 − 2 ( > 1)

= 1 − 2(0.1587)

= 0.6826

70 − 55
b) ( > 70) = ( > 10 )

= ( > 1.5)
= 0.0668

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50 − 55
c) ( ≥ 50) = ( ≥ 10 )

= ( ≥ −0.5)
= 1 − ( ≤ −0.5)
= 1 − ( ≥ 0.5)
= 1 − 0.3085
= 0.6915

Example 4:
A machine is used to produce butter whose masses are normally distributed with a mean of g
and a standard deviation of g. It is found that 5.05% of the output of the machine have a mass
greater than 85 and 10% have a mass less than 25 . Find the values of and .

Solution 4:
Let be the mass of the butter.
~ ( , 2)

( > 85) = 0.0505 --------------------1
85 −

( > ) = 0.0505
From the table, ( > 1.64)

85 −
∴ = 1.64

85 − = 1.64

( < 25) = 0.1 Symmetric around the mean
25 −

( < ) = 0.1
From the table, ( > 1.28) = 0.1003

25 −
( < ) = ( > 1.28)

25 −
( < ) = ( < −1.28)

25 −
∴ = −1.28

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25 − = −1.28 -------------------2

1-2 Solve the simultaneous equations
60 = 2.92
= 20.5

85 − = 1.64 (20.5)
= 51.4

10.4 Normal Approximation to the Binomial Distribution

Can be done when ≥ 5 and 0.1 ≤ ≤ 0.9
Use = , 2 =
A continuity correction is needed for the approximation from discrete case (binomial
distribution) to continuous case (normal distribution)

Example 1:
A fair coin is tossed 400 times. Find the probability of getting
a) between 185 and 210 tails (inclusive)
b) exactly 205 tails
c) less than 176 or more than 227 tails

Solution 1: Since n is large, we can use
Let be the number of tails normal approximation
~ (400,0.5)

= = 400(0.5) = 200
2 = = 400(0.5)(0.5) = 100
~ (200,100)

a) (185 ≤ ≤ 210) = (184.5 < < 210.5) Continuity correction

184.5 − 200 < < 210.5 − 200
= ( 10 10 )

= (−1.55 < < 1.05)

= 1 − ( > 1.55) − ( > 1.05)

= 1 − 0.0606 − 0.1469

= 0.7925

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b) ( = 205) = (204.5 < < 205.5) Continuity correction

204.5 − 200 205.5 − 200
= ( 10 < < 10 )

= (0.45 < < 0.55)

= ( > 0.45) − ( > 0.55)

= 0.3264 − 0.2912

= 0.0352

c) ( < 176) + ( > 227) = ( < 175.5) + ( > 227.5) Continuity correction

175.5 − 200 227.5 − 200
= ( < 10 ) + ( > 10 )

= ( < −2.45) + ( > 2.75)

= ( > 2.45) + ( > 2.75)

= 0.0071 + 0.00298

= 0.01008

Example 2:
For a short domestic flight, an airline has three different choices on its snack menu-pretzels,
potato chips, and cookies. Based on past experience, the airline feels that each snack is equally
likely to be chosen. If there are 150 passengers on a particular flight, what is the approximate
probability that
a) at least 60 will choose pretzels for dessert?
b) exactly 60 will choose pretzels for dessert?
c) fewer than 60 will choose pretzels for dessert?

Solution 2:

Let be the number of customers that will choose pretzels for dessert

1
~ (150, 3)

= = 150 (1) = 50 Since n is large, we can use
normal approximation
3

2 = = 150 1 2 = 100
(3) (3) 3

100
~ (50, 3 )

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a) ( ≥ 60) = ( > 59.5) Continuity correction

59.5 − 50
= >

( √1300 )
= ( > 1.65)
= 0.0493

b) ( = 60) = (59.5 < < 60.5) Continuity correction

59.5 − 50 60.5 − 50
= < <
( √1300 √1300 )

= (1.65 < < 1.82)

= ( > 1.65) − ( > 1.82)

= 0.0493 − 0.0344

= 0.0149

c) ( < 60) = ( < 59.5) Continuity correction

59.5 − 50
= <

( √1300 )
= ( < 1.65)
= 1 − ( > 1.65)
= 1 − 0.0493
= 0.9507

Example 3:
It is known that in a sack of mixed grass seed 35% of seeds are ryegrass. In a sample of 400
seeds, find the probability that there are
a) at most 120 ryegrass seeds
b) between 120 and 150 (inclusive) ryegrass seeds
c) more than 160 ryegrass seeds

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Let be the number of ryegrass seeds
~ (400,0.35) Since n is large, we can use
normal approximation
= = 400(0.35) = 140
2 = = 400(0.35)(0.65) = 91
~ (140,91)

a) ( ≤ 120) = ( < 120.5) Continuity correction

= ( < 120.5 − 140 )

√91

= ( < −2.04)

= ( > 2.04)

= 0.0207

b) (120 ≤ ≤ 150) = (119.5 < < 150.5) Continuity correction

= ( 119.5 − 140 < < 150.5 − 140 )

√91 √91

= (−2.15 < < 1.10)

= 1 − ( > 2.15) − ( > 1.10)

= 1 − 0.0158 − 0.1357

= 0.8485

c) ( > 160) = ( > 160.5) Continuity correction

= ( < 160.5 − 140 )

√91

= ( > 2.15)

= 0.0158

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Exercise

1. A customer calling a call center spends an average of 45 minutes on hold during the peak
season, with a standard deviation of 12 minutes. Suppose these times are normally
distributed. Find the probability that the customer will be on hold for
a) more than 54 minutes
b) less than 24 minutes
c) between 24 and 54 minutes
d) more than 39 minutes

2. The time taken to assemble a car in a certain plant is a random variable having a normal
distribution of 20 hours and a standard deviation of 2 hours. What is the probability that
the car can be assembled at this plant in
a) less than 19.5 hours?
b) between 20 and 22 hours?

3. The annual salaries of employees in a large company are normally distributed with a
mean of RM50,000 and a standard deviation of RM20,000. What percent of people earn
a) less than RM40,000?
b) between RM45,000 and RM65,000?
c) more than RM70,000?

4. According to a survey, 70% of adults between 50 and 64 years old use the Internet. You
randomly select 80 adults in that range and ask them if they use the internet. Find the
probability that
a) 70 or more people say they use the Internet
b) from 50 to 70 people say they use the Internet
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Solution

1. Let be the time that customer will be on hold
~ (45,122)

a) ( > 54) = ( > 54−45)

12

= ( > 0.75)

= 0.2266

24 − 45
b) ( < 24) = ( < 12 )

= ( < −1.75)

= ( > 1.75)

= 0.0401

c) (24 < < 54) = (24−45 < < 54−45)

12 12

= (−1.75 < < 0.75)

= 1 − ( > 1.75) − ( > 0.75)

= 1 − 0.0401 − 0.2266

= 0.7333

d) ( > 39) = ( > 39 − 45
12 )

= ( > −0.5)

= 1 − ( > 0.5)

= 1 − 0.3085

= 0.6915

2. Let be the time taken to assemble a car
~ (20,22)

a) ( < 19.5) = ( < 19.52−20)
= ( < −0.25)
= ( > 0.25)
= 0.4013

b) (20 < < 22) = (20−20 < < 22−20)

22

= (0 < < 1)

= 0.5 − ( > 1)

= 0.5 − 0.1587

= 0.3413

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3. Let be the annual salaries of employees in a large company
~ (50000,200002)

a) ( < 40000) = ( < 40000−50000)

20000

= ( < −0.5)

= ( > 0.5) Symmetric around the mean
= 0.3085

Hence, 30.85% earn less than RM40,000

b) (45000 < < 65000) = (45000−50000 < < 65000−50000)
20000 20000

= (−0.25 < < 0.75)

= 1 − ( > 0.25) − ( > 0.75)

= 1 − 0.4013 − 0.2266

= 0.3721

Hence, 37.21% earn between RM45,000 and RM65,000

c) ( > 70000) = ( > 7000200−00500000)
= ( > 1.0)
= 0.1587

Hence, 15.87% earn more than RM70,000

4. Let be the number of people say they use the Internet

~ (80,0.7)

= = 80(0.7) = 56 Since n is large, we can use
2 = = 80(0.7)(0.3) = 16.8 normal approximation

~ (56,16.8)

a) ( ≥ 70) = ( > 69.5) Continuity correction

= ( > 69.5 − 56 )

√16.8

= ( > 3.29)

= 0.0005

b) (50 ≤ ≤ 70) = (49.5 < < 70.5) Continuity correction

= ( 49.5 − 56 < < 70.5 − 56 )

√16.8 √16.8

= (−1.59 < < 3.54)

= 1 − ( > 1.59) − ( > 3.54)

= 1 − 0.0559 − 0.0002

= 0.9439

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