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TUTORIAL 5Chapter 4: Continuous R.V. (Part I) & Assignment
STAT1306 Introductory Statistics(First Semester, 2011-2012)
Jun FU
Department of Statistics and Actuarial ScienceThe University of Hong Kong
October 10, 2011
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 1 / 37
Review of Key Concepts
Game Plan
1 Review of Key Concepts
Basic Definitions
Comparison Between Discrete and Continuous Random Variables
2 Problems
Continuous Random Variables
Some Questions from Assignments
3 Ending
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 2 / 37
Review of Key Concepts Basic Definitions
Game Plan
1 Review of Key Concepts
Basic Definitions
Comparison Between Discrete and Continuous Random Variables
2 Problems
Continuous Random Variables
Some Questions from Assignments
3 Ending
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 3 / 37
Review of Key Concepts Basic Definitions
Definitions
Continuous Random Variables: It can take any value in an interval.
Cumulative Distribution Function (c.d.f.):
FY (y) = P(Y ≤ y) = P(Y < y).
Probability Density Function (p.d.f.):
fY (y) =dFY (y)
dy= F
′Y (y), FY (y) =
ˆ y
−∞fY (s)ds.
It has the following properties:
(a) fY (y) ≥ 0 for any value of y .(b)´∞−∞ fY (y)dy = FY (∞) = 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 4 / 37
Review of Key Concepts Basic Definitions
Definitions
Continuous Random Variables: It can take any value in an interval.
Cumulative Distribution Function (c.d.f.):
FY (y) = P(Y ≤ y) = P(Y < y).
Probability Density Function (p.d.f.):
fY (y) =dFY (y)
dy= F
′Y (y), FY (y) =
ˆ y
−∞fY (s)ds.
It has the following properties:
(a) fY (y) ≥ 0 for any value of y .(b)´∞−∞ fY (y)dy = FY (∞) = 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 4 / 37
Review of Key Concepts Basic Definitions
Definitions
Continuous Random Variables: It can take any value in an interval.
Cumulative Distribution Function (c.d.f.):
FY (y) = P(Y ≤ y) = P(Y < y).
Probability Density Function (p.d.f.):
fY (y) =dFY (y)
dy= F
′Y (y), FY (y) =
ˆ y
−∞fY (s)ds.
It has the following properties:
(a) fY (y) ≥ 0 for any value of y .(b)´∞−∞ fY (y)dy = FY (∞) = 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 4 / 37
Review of Key Concepts Basic Definitions
Definitions
Continuous Random Variables: It can take any value in an interval.
Cumulative Distribution Function (c.d.f.):
FY (y) = P(Y ≤ y) = P(Y < y).
Probability Density Function (p.d.f.):
fY (y) =dFY (y)
dy= F
′Y (y), FY (y) =
ˆ y
−∞fY (s)ds.
It has the following properties:
(a) fY (y) ≥ 0 for any value of y .(b)´∞−∞ fY (y)dy = FY (∞) = 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 4 / 37
Review of Key Concepts Basic Definitions
Definitions
Continuous Random Variables: It can take any value in an interval.
Cumulative Distribution Function (c.d.f.):
FY (y) = P(Y ≤ y) = P(Y < y).
Probability Density Function (p.d.f.):
fY (y) =dFY (y)
dy= F
′Y (y), FY (y) =
ˆ y
−∞fY (s)ds.
It has the following properties:
(a) fY (y) ≥ 0 for any value of y .(b)´∞−∞ fY (y)dy = FY (∞) = 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 4 / 37
Review of Key Concepts Comparison Between Discrete and Continuous Random Variables
Game Plan
1 Review of Key Concepts
Basic Definitions
Comparison Between Discrete and Continuous Random Variables
2 Problems
Continuous Random Variables
Some Questions from Assignments
3 Ending
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 5 / 37
Review of Key Concepts Comparison Between Discrete and Continuous Random Variables
Discrete R.V. vs Continuous R.V.
Basic Definition Discrete ContinuousP(Y ∈ A) =
∑y∈A p(y)
´A fY (y)dy
FY (y) = P(Y ≤ y) P(Y ≤ y) = P(Y < y)
E (Y ) =∑
y y · p(y)´∞−∞ y · fY (y)dy
E [g(Y )] =∑
y g(y) · p(y)´∞−∞ g(y) · fY (y)dy
Var(Y ) = E[(Y − µ)2
]=
∑y (y − µ)2 · p(y)
´∞−∞(y − µ)2 · fY (y)dy
MY (t) = E[etY]
=∑
y ety · p(y)
´∞−∞ ety · fY (y)dy
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 6 / 37
Problems
Game Plan
1 Review of Key Concepts
Basic Definitions
Comparison Between Discrete and Continuous Random Variables
2 Problems
Continuous Random Variables
Some Questions from Assignments
3 Ending
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 7 / 37
Problems Continuous Random Variables
Game Plan
1 Review of Key Concepts
Basic Definitions
Comparison Between Discrete and Continuous Random Variables
2 Problems
Continuous Random Variables
Some Questions from Assignments
3 Ending
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 8 / 37
Problems Continuous Random Variables
For a function
f (x) =
{(ρ− 1)x−ρ x ≥ 1
0 x < 1,
where ρ > 1.
(a) Is f (x) a density function? Why?
(b) Find F (x).
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 9 / 37
Problems Continuous Random Variables
For a function
f (x) =
{(ρ− 1)x−ρ x ≥ 1
0 x < 1,
where ρ > 1.
(a) Is f (x) a density function? Why?
(b) Find F (x).
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 9 / 37
Problems Continuous Random Variables
For a function
f (x) =
{(ρ− 1)x−ρ x ≥ 1
0 x < 1,
where ρ > 1.
(a) Is f (x) a density function? Why?
(b) Find F (x).
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 9 / 37
Problems Continuous Random Variables
Question (a): Is f (x) a density function?
To determine whether a function f (x) is a p.d.f., we just need to check{fX (x) ≥ 0 ∀x ,´∞−∞ fX (x)dx = 1.
First, since ρ > 1, we have
f (x) =
{(ρ− 1)x−ρ ≥ 0 x ≥ 1
0 x < 1.
Second, we haveˆ ∞−∞
f (x)dx =
ˆ ∞1
(ρ− 1)x−ρdx
= −x−ρ+1∣∣∞1
= 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 10 / 37
Problems Continuous Random Variables
Question (a): Is f (x) a density function?
To determine whether a function f (x) is a p.d.f., we just need to check{fX (x) ≥ 0 ∀x ,´∞−∞ fX (x)dx = 1.
First, since ρ > 1, we have
f (x) =
{(ρ− 1)x−ρ ≥ 0 x ≥ 1
0 x < 1.
Second, we haveˆ ∞−∞
f (x)dx =
ˆ ∞1
(ρ− 1)x−ρdx
= −x−ρ+1∣∣∞1
= 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 10 / 37
Problems Continuous Random Variables
Question (a): Is f (x) a density function?
To determine whether a function f (x) is a p.d.f., we just need to check{fX (x) ≥ 0 ∀x ,´∞−∞ fX (x)dx = 1.
First, since ρ > 1, we have
f (x) =
{(ρ− 1)x−ρ ≥ 0 x ≥ 1
0 x < 1.
Second, we haveˆ ∞−∞
f (x)dx =
ˆ ∞1
(ρ− 1)x−ρdx
= −x−ρ+1∣∣∞1
= 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 10 / 37
Problems Continuous Random Variables
Question (b): Find F (x)
By definition, we have for any x ≥ 1
F (x) =
ˆ x
−∞f (y)dy
=
ˆ x
1(ρ− 1)y−ρdy
= −y−ρ+1∣∣x1
= 1− x1−ρ.
Therefore
F (x) =
{1− x1−ρ x ≥ 1
0 x < 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 11 / 37
Problems Continuous Random Variables
Question (b): Find F (x)
By definition, we have for any x ≥ 1
F (x) =
ˆ x
−∞f (y)dy
=
ˆ x
1(ρ− 1)y−ρdy
= −y−ρ+1∣∣x1
= 1− x1−ρ.
Therefore
F (x) =
{1− x1−ρ x ≥ 1
0 x < 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 11 / 37
Problems Continuous Random Variables
A college professor never finishes his lecture before the bell rings butalways finishes his lecture within 2 min after the bell rings. Let X =timethat elapses between the ring and the end of the lecture. Suppose theprobability density function of X is
f (x) =
{kx2 0 ≤ x ≤ 2
0 otherwise.
(a) Find the value of k.
(b) What is the probability that the lecture ends within 1 min of the bellringing?
(c) What is the probability that the lecture continues for at least 90 secbeyond the ring?
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 12 / 37
Problems Continuous Random Variables
A college professor never finishes his lecture before the bell rings butalways finishes his lecture within 2 min after the bell rings. Let X =timethat elapses between the ring and the end of the lecture. Suppose theprobability density function of X is
f (x) =
{kx2 0 ≤ x ≤ 2
0 otherwise.
(a) Find the value of k.
(b) What is the probability that the lecture ends within 1 min of the bellringing?
(c) What is the probability that the lecture continues for at least 90 secbeyond the ring?
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 12 / 37
Problems Continuous Random Variables
A college professor never finishes his lecture before the bell rings butalways finishes his lecture within 2 min after the bell rings. Let X =timethat elapses between the ring and the end of the lecture. Suppose theprobability density function of X is
f (x) =
{kx2 0 ≤ x ≤ 2
0 otherwise.
(a) Find the value of k.
(b) What is the probability that the lecture ends within 1 min of the bellringing?
(c) What is the probability that the lecture continues for at least 90 secbeyond the ring?
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 12 / 37
Problems Continuous Random Variables
A college professor never finishes his lecture before the bell rings butalways finishes his lecture within 2 min after the bell rings. Let X =timethat elapses between the ring and the end of the lecture. Suppose theprobability density function of X is
f (x) =
{kx2 0 ≤ x ≤ 2
0 otherwise.
(a) Find the value of k.
(b) What is the probability that the lecture ends within 1 min of the bellringing?
(c) What is the probability that the lecture continues for at least 90 secbeyond the ring?
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 12 / 37
Problems Continuous Random Variables
Question (a): Find the value of k
By using the property of the p.d.f. we have
1 =
ˆ ∞−∞
f (x)dx
=
ˆ 2
0kx2dx
=x3
3k
∣∣∣∣20
=8
3k.
It implies that
k =3
8.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 13 / 37
Problems Continuous Random Variables
Question (a): Find the value of k
By using the property of the p.d.f. we have
1 =
ˆ ∞−∞
f (x)dx
=
ˆ 2
0kx2dx
=x3
3k
∣∣∣∣20
=8
3k.
It implies that
k =3
8.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 13 / 37
Problems Continuous Random Variables
Question (b&c): End within 1 min? Continue for 90 sec?
For Q (b), the probability can be calculated as
P(X ≤ 1) =
ˆ 1
0
3
8x2dx =
x3
3· 3
8
∣∣∣∣10
=1
8.
For Q (c), the probability can be calculated as
P(X ≥ 1.5) =
ˆ 2
1.5
3
8x2dx
=x3
3· 3
8
∣∣∣∣21.5
=3
8
(8
3− 1.53
3
)= 0.578.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 14 / 37
Problems Continuous Random Variables
Question (b&c): End within 1 min? Continue for 90 sec?
For Q (b), the probability can be calculated as
P(X ≤ 1) =
ˆ 1
0
3
8x2dx =
x3
3· 3
8
∣∣∣∣10
=1
8.
For Q (c), the probability can be calculated as
P(X ≥ 1.5) =
ˆ 2
1.5
3
8x2dx
=x3
3· 3
8
∣∣∣∣21.5
=3
8
(8
3− 1.53
3
)= 0.578.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 14 / 37
Problems Continuous Random Variables
Independent observations are made on a random variable X whoseprobability density function is
f (x) =
{0.125x 0 ≤ x ≤ 4
0 otherwise.
(a) Find the mean and variance of X .
(b) Find P(X > 3).
(c) Find the probability that at least three observations are required toobtain a result of x > 3.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 15 / 37
Problems Continuous Random Variables
Independent observations are made on a random variable X whoseprobability density function is
f (x) =
{0.125x 0 ≤ x ≤ 4
0 otherwise.
(a) Find the mean and variance of X .
(b) Find P(X > 3).
(c) Find the probability that at least three observations are required toobtain a result of x > 3.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 15 / 37
Problems Continuous Random Variables
Independent observations are made on a random variable X whoseprobability density function is
f (x) =
{0.125x 0 ≤ x ≤ 4
0 otherwise.
(a) Find the mean and variance of X .
(b) Find P(X > 3).
(c) Find the probability that at least three observations are required toobtain a result of x > 3.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 15 / 37
Problems Continuous Random Variables
Independent observations are made on a random variable X whoseprobability density function is
f (x) =
{0.125x 0 ≤ x ≤ 4
0 otherwise.
(a) Find the mean and variance of X .
(b) Find P(X > 3).
(c) Find the probability that at least three observations are required toobtain a result of x > 3.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 15 / 37
Problems Continuous Random Variables
Question (a): Find the mean and variance of X
E (X ) =
ˆ ∞−∞
x · f (x)dx =
ˆ 4
00.125x2dx = 0.125 · x
3
3
∣∣∣∣40
=8
3.
Var(X ) = E[(X − EX )2
]= E (X 2)− (EX )2
=
ˆ ∞−∞
x2 · f (x)dx − (8
3)2
=
ˆ 4
00.125x3dx − (
8
3)2
= 0.125 · x4
4
∣∣∣∣40
− 64
9
=8
9.
.Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 16 / 37
Problems Continuous Random Variables
Question (a): Find the mean and variance of X
E (X ) =
ˆ ∞−∞
x · f (x)dx =
ˆ 4
00.125x2dx = 0.125 · x
3
3
∣∣∣∣40
=8
3.
Var(X ) = E[(X − EX )2
]= E (X 2)− (EX )2
=
ˆ ∞−∞
x2 · f (x)dx − (8
3)2
=
ˆ 4
00.125x3dx − (
8
3)2
= 0.125 · x4
4
∣∣∣∣40
− 64
9
=8
9.
.Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 16 / 37
Problems Continuous Random Variables
Question(b): Find P(X > 3)
The P(X > 3) can be calculated as follows
P(X > 3) =
ˆ 4
3f (x)dx
=
ˆ 4
30.125 · xdx
= 0.125 · x2
2
∣∣∣∣43
=7
16.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 17 / 37
Problems Continuous Random Variables
Question(c): At least three observations are required
Let Y =the number of observations required to obtain a result of x > 3,then
Y ∼ Geometric(7
16).
We just need to calculate
P(Y ≥ 3) = (1− 7
16)2 =
81
256,
since that “at least three trials” are required is equivalent with that “thefirst two trials” are failure.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 18 / 37
Problems Continuous Random Variables
Question(c): At least three observations are required
Let Y =the number of observations required to obtain a result of x > 3,then
Y ∼ Geometric(7
16).
We just need to calculate
P(Y ≥ 3) = (1− 7
16)2 =
81
256,
since that “at least three trials” are required is equivalent with that “thefirst two trials” are failure.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 18 / 37
Problems Some Questions from Assignments
Game Plan
1 Review of Key Concepts
Basic Definitions
Comparison Between Discrete and Continuous Random Variables
2 Problems
Continuous Random Variables
Some Questions from Assignments
3 Ending
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 19 / 37
Problems Some Questions from Assignments
Assume that every time you buy an item of the Hong Kong Disney series,you receive one of the four types of cards, each with a cartoon characterMickey, Minnie, Donald and Daisy with an equal probability. Over a periodof time, you buy 6 items of the series. What is the probability that you willget all four cards?
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 20 / 37
Problems Some Questions from Assignments
Method (1): “Three of a kind” (e.g. A× 3,B ,C ,D)
The probability that the six items show A,A,A,B,C ,D in that order is:
(1
4)6
The no. of ways of arranging A,A,A,B,C ,D in a sequence is
6!
3! · 1! · 1! · 1!
The no. of ways of determining which letter appears three times is
4C1
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 21 / 37
Problems Some Questions from Assignments
Method (1): “Three of a kind” (e.g. A× 3,B ,C ,D)
The probability that the six items show A,A,A,B,C ,D in that order is:
(1
4)6
The no. of ways of arranging A,A,A,B,C ,D in a sequence is
6!
3! · 1! · 1! · 1!
The no. of ways of determining which letter appears three times is
4C1
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 21 / 37
Problems Some Questions from Assignments
Method (1): “Three of a kind” (e.g. A× 3,B ,C ,D)
The probability that the six items show A,A,A,B,C ,D in that order is:
(1
4)6
The no. of ways of arranging A,A,A,B,C ,D in a sequence is
6!
3! · 1! · 1! · 1!
The no. of ways of determining which letter appears three times is
4C1
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 21 / 37
Problems Some Questions from Assignments
Method (1): “Two pairs” (e.g. A× 2,B × 2,C ,D)
The probability that the six items show A,A,B,B,C ,D in that order is:
(1
4)6
The no. of ways of arranging A,A,B,B,C ,D in a sequence is
6!
2! · 2! · 1! · 1!
The no. of ways of determining which letters appear twice is
4C2
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 22 / 37
Problems Some Questions from Assignments
Method (1): “Two pairs” (e.g. A× 2,B × 2,C ,D)
The probability that the six items show A,A,B,B,C ,D in that order is:
(1
4)6
The no. of ways of arranging A,A,B,B,C ,D in a sequence is
6!
2! · 2! · 1! · 1!
The no. of ways of determining which letters appear twice is
4C2
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 22 / 37
Problems Some Questions from Assignments
Method (1): “Two pairs” (e.g. A× 2,B × 2,C ,D)
The probability that the six items show A,A,B,B,C ,D in that order is:
(1
4)6
The no. of ways of arranging A,A,B,B,C ,D in a sequence is
6!
2! · 2! · 1! · 1!
The no. of ways of determining which letters appear twice is
4C2
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 22 / 37
Problems Some Questions from Assignments
Method (1): Final result
In sum, the probability that we will collect all of the four letters is
(1
4)6 × 6!
3! · 1! · 1! · 1!× 4C1 + (
1
4)6 × 6!
2! · 2! · 1! · 1!× 4C2.
The final answer is1560
46= 0.3809.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 23 / 37
Problems Some Questions from Assignments
Method (1): Final result
In sum, the probability that we will collect all of the four letters is
(1
4)6 × 6!
3! · 1! · 1! · 1!× 4C1 + (
1
4)6 × 6!
2! · 2! · 1! · 1!× 4C2.
The final answer is1560
46= 0.3809.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 23 / 37
Problems Some Questions from Assignments
Method (2): Consider whether the letter is a different one
Let N denote that the letter is a new one;Let O denote that the letter is an old one.For example, the probability for the result of
NNONN
can be calculated
1× 3
4× 2
4× 2
4× 1
4.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 24 / 37
Problems Some Questions from Assignments
Method (2): Consider whether the letter is a different one
Let N denote that the letter is a new one;Let O denote that the letter is an old one.For example, the probability for the result of
NNONN
can be calculated
1× 3
4× 2
4× 2
4× 1
4.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 24 / 37
Problems Some Questions from Assignments
Method (2) (Cont.)
Number Outcome Probability4 NNNN 1× 3
4 ×24 ×
14
NONNN 1× 14 ×
34 ×
24 ×
14
5 NNONN 1× 34 ×
24 ×
24 ×
14
NNNON 1× 34 ×
24 ×
34 ×
14
NNNOON 1× 34 ×
24 ×
34 ×
34 ×
14
NOONNN 1× 14 ×
14 ×
34 ×
24 ×
14
6 NNOONN 1× 34 ×
24 ×
24 ×
24 ×
14
NONNON 1× 14 ×
34 ×
24 ×
34 ×
14
NONONN 1× 14 ×
34 ×
24 ×
24 ×
14
NNONON 1× 34 ×
24 ×
24 ×
34 ×
14
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 25 / 37
Problems Some Questions from Assignments
If n persons, including A and B, are randomly arranged in a straight line,what is the probability that there are exactly r persons in the line betweenA and B.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 26 / 37
Problems Some Questions from Assignments
Method(1): Select two positions from n positions
It is essentially to select two positions from the n positions in a line.First, the no. of ways to select two positions from the n positions suchthat there are exactly r positions between them is
n − r − 1.
Second, the no. of ways to select two positions from the n positions is
nC2.
Therefore, the probability required to calculate is
n − r − 1
nC2.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 27 / 37
Problems Some Questions from Assignments
Method(1): Select two positions from n positions
It is essentially to select two positions from the n positions in a line.First, the no. of ways to select two positions from the n positions suchthat there are exactly r positions between them is
n − r − 1.
Second, the no. of ways to select two positions from the n positions is
nC2.
Therefore, the probability required to calculate is
n − r − 1
nC2.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 27 / 37
Problems Some Questions from Assignments
Method(1): Select two positions from n positions
It is essentially to select two positions from the n positions in a line.First, the no. of ways to select two positions from the n positions suchthat there are exactly r positions between them is
n − r − 1.
Second, the no. of ways to select two positions from the n positions is
nC2.
Therefore, the probability required to calculate is
n − r − 1
nC2.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 27 / 37
Problems Some Questions from Assignments
Method(1): Select two positions from n positions
It is essentially to select two positions from the n positions in a line.First, the no. of ways to select two positions from the n positions suchthat there are exactly r positions between them is
n − r − 1.
Second, the no. of ways to select two positions from the n positions is
nC2.
Therefore, the probability required to calculate is
n − r − 1
nC2.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 27 / 37
Problems Some Questions from Assignments
Method(2): No. of solutions to an equation
Let L =no. of persons on the left of the two persons;Let M =no. of persons between the two persons;Let R =no. of persons on the right of the two persons. Then we have
L + M + R = n − 2.
First, if we require M = r , then the equation can be deduced to
L + R = n − 2− r .
The no. of solutions to this equation is
n−r−1C1 = n − r − 1.
Second, the no. of solutions to the original equation is
nC2.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 28 / 37
Problems Some Questions from Assignments
Method(2): No. of solutions to an equation
Let L =no. of persons on the left of the two persons;Let M =no. of persons between the two persons;Let R =no. of persons on the right of the two persons. Then we have
L + M + R = n − 2.
First, if we require M = r , then the equation can be deduced to
L + R = n − 2− r .
The no. of solutions to this equation is
n−r−1C1 = n − r − 1.
Second, the no. of solutions to the original equation is
nC2.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 28 / 37
Problems Some Questions from Assignments
Method(2): No. of solutions to an equation
Let L =no. of persons on the left of the two persons;Let M =no. of persons between the two persons;Let R =no. of persons on the right of the two persons. Then we have
L + M + R = n − 2.
First, if we require M = r , then the equation can be deduced to
L + R = n − 2− r .
The no. of solutions to this equation is
n−r−1C1 = n − r − 1.
Second, the no. of solutions to the original equation is
nC2.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 28 / 37
Problems Some Questions from Assignments
Method(2): No. of solutions to an equation
Let L =no. of persons on the left of the two persons;Let M =no. of persons between the two persons;Let R =no. of persons on the right of the two persons. Then we have
L + M + R = n − 2.
First, if we require M = r , then the equation can be deduced to
L + R = n − 2− r .
The no. of solutions to this equation is
n−r−1C1 = n − r − 1.
Second, the no. of solutions to the original equation is
nC2.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 28 / 37
Problems Some Questions from Assignments
A communication system consists of n components each of which will,independently, function with probability p. The whole system will be ableto operate effectively if at least one half-of its components function. Ingeneral, when is a (2k + 1) component system better than a (2k − 1)component system?
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 29 / 37
Problems Some Questions from Assignments
The (2k + 1) component system
We divide it into two parts:
(2k − 1) component sub-system + (2) component sub-system.
Let X1 =No. of components which can function in the (2k − 1) - system
X1 ∼ Bin(2k − 1, p).
Let Y =No. of components which can function in the (2) - system
Y ∼ Bin(2, p).
To make this (2k + 1) component system operate, we need to have
X1 + Y ≥ k + 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 30 / 37
Problems Some Questions from Assignments
The (2k + 1) component system
We divide it into two parts:
(2k − 1) component sub-system + (2) component sub-system.
Let X1 =No. of components which can function in the (2k − 1) - system
X1 ∼ Bin(2k − 1, p).
Let Y =No. of components which can function in the (2) - system
Y ∼ Bin(2, p).
To make this (2k + 1) component system operate, we need to have
X1 + Y ≥ k + 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 30 / 37
Problems Some Questions from Assignments
The (2k + 1) component system
We divide it into two parts:
(2k − 1) component sub-system + (2) component sub-system.
Let X1 =No. of components which can function in the (2k − 1) - system
X1 ∼ Bin(2k − 1, p).
Let Y =No. of components which can function in the (2) - system
Y ∼ Bin(2, p).
To make this (2k + 1) component system operate, we need to have
X1 + Y ≥ k + 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 30 / 37
Problems Some Questions from Assignments
The (2k + 1) component system
We divide it into two parts:
(2k − 1) component sub-system + (2) component sub-system.
Let X1 =No. of components which can function in the (2k − 1) - system
X1 ∼ Bin(2k − 1, p).
Let Y =No. of components which can function in the (2) - system
Y ∼ Bin(2, p).
To make this (2k + 1) component system operate, we need to have
X1 + Y ≥ k + 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 30 / 37
Problems Some Questions from Assignments
The (2k − 1) component system
Let X2 =No. of components which can function in this system, and
X2 ∼ Bin(2k − 1, p).
To make this (2k − 1) component system operate, we need to have
X2 ≥ k.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 31 / 37
Problems Some Questions from Assignments
The (2k − 1) component system
Let X2 =No. of components which can function in this system, and
X2 ∼ Bin(2k − 1, p).
To make this (2k − 1) component system operate, we need to have
X2 ≥ k.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 31 / 37
Problems Some Questions from Assignments
(2k + 1) - system (X1 + Y ) v.s. (2k − 1) - system (X2)
For the (2k + 1) component system, we calculate
P(X1 + Y ≥ k + 1)
= P(X1 ≥ k − 1 ∩ Y = 2) + P(X1 ≥ k ∩ Y = 1) + P(X1 ≥ k + 1 ∩ Y = 0).
Hence,
P(X1 + Y ≥ k + 1)
= P(X1 ≥ k ∩ Y = 2) + P(X1 = k − 1 ∩ Y = 2)
+ P(X1 ≥ k ∩ Y = 1)
+ P(X1 ≥ k ∩ Y = 0)− P(X1 = k ∩ Y = 0).
Therefore, the probability that it can operate is
P(X1 ≥ k) + [P(X1 = k − 1) · P(Y = 2)− P(X1 = k)P(Y = 0)] .
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 32 / 37
Problems Some Questions from Assignments
(2k + 1) - system (X1 + Y ) v.s. (2k − 1) - system (X2)
For the (2k + 1) component system, we calculate
P(X1 + Y ≥ k + 1)
= P(X1 ≥ k − 1 ∩ Y = 2) + P(X1 ≥ k ∩ Y = 1) + P(X1 ≥ k + 1 ∩ Y = 0).
Hence,
P(X1 + Y ≥ k + 1)
= P(X1 ≥ k ∩ Y = 2) + P(X1 = k − 1 ∩ Y = 2)
+ P(X1 ≥ k ∩ Y = 1)
+ P(X1 ≥ k ∩ Y = 0)− P(X1 = k ∩ Y = 0).
Therefore, the probability that it can operate is
P(X1 ≥ k) + [P(X1 = k − 1) · P(Y = 2)− P(X1 = k)P(Y = 0)] .
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 32 / 37
Problems Some Questions from Assignments
(2k + 1) - system (X1 + Y ) v.s. (2k − 1) - system (X2)
For the (2k + 1) component system, we calculate
P(X1 + Y ≥ k + 1)
= P(X1 ≥ k − 1 ∩ Y = 2) + P(X1 ≥ k ∩ Y = 1) + P(X1 ≥ k + 1 ∩ Y = 0).
Hence,
P(X1 + Y ≥ k + 1)
= P(X1 ≥ k ∩ Y = 2) + P(X1 = k − 1 ∩ Y = 2)
+ P(X1 ≥ k ∩ Y = 1)
+ P(X1 ≥ k ∩ Y = 0)− P(X1 = k ∩ Y = 0).
Therefore, the probability that it can operate is
P(X1 ≥ k) + [P(X1 = k − 1) · P(Y = 2)− P(X1 = k)P(Y = 0)] .
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 32 / 37
Problems Some Questions from Assignments
Final result
To make the (2k + 1) - system better than the (2k − 1) - system, we justneed to make
P(X1 = k − 1) · P(Y = 2) > P(X1 = k)P(Y = 0).
This is equivalent to
2k−1Ck−1 · pk−1(1− p)k · p2 ≥ 2k−1Ck · pk(1− p)k−1 · (1− p)2.
Hence,
p ≥ 1− p ⇐⇒ 1
2≤ p ≤ 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 33 / 37
Problems Some Questions from Assignments
Final result
To make the (2k + 1) - system better than the (2k − 1) - system, we justneed to make
P(X1 = k − 1) · P(Y = 2) > P(X1 = k)P(Y = 0).
This is equivalent to
2k−1Ck−1 · pk−1(1− p)k · p2 ≥ 2k−1Ck · pk(1− p)k−1 · (1− p)2.
Hence,
p ≥ 1− p ⇐⇒ 1
2≤ p ≤ 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 33 / 37
Problems Some Questions from Assignments
Final result
To make the (2k + 1) - system better than the (2k − 1) - system, we justneed to make
P(X1 = k − 1) · P(Y = 2) > P(X1 = k)P(Y = 0).
This is equivalent to
2k−1Ck−1 · pk−1(1− p)k · p2 ≥ 2k−1Ck · pk(1− p)k−1 · (1− p)2.
Hence,
p ≥ 1− p ⇐⇒ 1
2≤ p ≤ 1.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 33 / 37
Problems Some Questions from Assignments
Exactly one of six similar keys opens a certain door. If you try the keys oneafter another, what is the expected number of keys that you will have totry before success?
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 34 / 37
Problems Some Questions from Assignments
Solution
Note: The number of keys required to succeed does not follow a geometricdistribution, since the first trial is not independent with the second trial.To try the keys one after another is essentially to arrange six keys in asequence.The correct key is equally likely to be one of the six ones and the numberof trials required follows a uniform distribution with possible values:
1, 2, 3, 4, 5, 6.
Its expectation can be given by
1 + 2 + 3 + 4 + 5 + 6
6= 3.5.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 35 / 37
Problems Some Questions from Assignments
Solution
Note: The number of keys required to succeed does not follow a geometricdistribution, since the first trial is not independent with the second trial.To try the keys one after another is essentially to arrange six keys in asequence.The correct key is equally likely to be one of the six ones and the numberof trials required follows a uniform distribution with possible values:
1, 2, 3, 4, 5, 6.
Its expectation can be given by
1 + 2 + 3 + 4 + 5 + 6
6= 3.5.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 35 / 37
Problems Some Questions from Assignments
Solution
Note: The number of keys required to succeed does not follow a geometricdistribution, since the first trial is not independent with the second trial.To try the keys one after another is essentially to arrange six keys in asequence.The correct key is equally likely to be one of the six ones and the numberof trials required follows a uniform distribution with possible values:
1, 2, 3, 4, 5, 6.
Its expectation can be given by
1 + 2 + 3 + 4 + 5 + 6
6= 3.5.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 35 / 37
Problems Some Questions from Assignments
Solution
Note: The number of keys required to succeed does not follow a geometricdistribution, since the first trial is not independent with the second trial.To try the keys one after another is essentially to arrange six keys in asequence.The correct key is equally likely to be one of the six ones and the numberof trials required follows a uniform distribution with possible values:
1, 2, 3, 4, 5, 6.
Its expectation can be given by
1 + 2 + 3 + 4 + 5 + 6
6= 3.5.
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 35 / 37
Ending
Game Plan
1 Review of Key Concepts
Basic Definitions
Comparison Between Discrete and Continuous Random Variables
2 Problems
Continuous Random Variables
Some Questions from Assignments
3 Ending
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 36 / 37
Ending
Goodbye and See You
See You After Reading Week (Oct 24,26)
Same Time
Same Place
Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 37 / 37