STAT1306_11-12_T05_Slides

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

Review of Key Concepts Basic Definitions

Game Plan

Basic Definitions

2 Problems

3 Ending

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)

′Y (y), FY (y) =

−∞fY (s)ds.

It has the following properties:

(a) fY (y) ≥ 0 for any value of y .(b)´∞−∞ fY (y)dy = FY (∞) = 1.

Definitions

FY (y) = P(Y ≤ y) = P(Y < y).

fY (y) =dFY (y)

′Y (y), FY (y) =

−∞fY (s)ds.

Definitions

FY (y) = P(Y ≤ y) = P(Y < y).

fY (y) =dFY (y)

′Y (y), FY (y) =

−∞fY (s)ds.

Definitions

FY (y) = P(Y ≤ y) = P(Y < y).

fY (y) =dFY (y)

′Y (y), FY (y) =

−∞fY (s)ds.

Definitions

FY (y) = P(Y ≤ y) = P(Y < y).

fY (y) =dFY (y)

′Y (y), FY (y) =

−∞fY (s)ds.

Review of Key Concepts Comparison Between Discrete and Continuous Random Variables

Game Plan

Basic Definitions

2 Problems

3 Ending

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

Problems

Game Plan

Basic Definitions

2 Problems

3 Ending

Problems Continuous Random Variables

Game Plan

Basic Definitions

2 Problems

3 Ending

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).

For a function

f (x) =

{(ρ− 1)x−ρ x ≥ 1

0 x < 1,

where ρ > 1.

(b) Find F (x).

For a function

f (x) =

{(ρ− 1)x−ρ x ≥ 1

0 x < 1,

where ρ > 1.

(b) Find F (x).

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

f (x) =

{(ρ− 1)x−ρ ≥ 0 x ≥ 1

0 x < 1.

f (x)dx =

ˆ ∞1

(ρ− 1)x−ρdx

= −x−ρ+1∣∣∞1

f (x) =

{(ρ− 1)x−ρ ≥ 0 x ≥ 1

0 x < 1.

f (x)dx =

ˆ ∞1

(ρ− 1)x−ρdx

= −x−ρ+1∣∣∞1

Question (b): Find F (x)

By definition, we have for any x ≥ 1

F (x) =

−∞f (y)dy

1(ρ− 1)y−ρdy

= −y−ρ+1∣∣x1

= 1− x1−ρ.

Therefore

F (x) =

{1− x1−ρ x ≥ 1

0 x < 1.

Question (b): Find F (x)

By definition, we have for any x ≥ 1

F (x) =

−∞f (y)dy

1(ρ− 1)y−ρdy

= −y−ρ+1∣∣x1

= 1− x1−ρ.

Therefore

F (x) =

{1− x1−ρ x ≥ 1

0 x < 1.

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?

f (x) =

{kx2 0 ≤ x ≤ 2

0 otherwise.

f (x) =

{kx2 0 ≤ x ≤ 2

0 otherwise.

f (x) =

{kx2 0 ≤ x ≤ 2

0 otherwise.

Question (a): Find the value of k

By using the property of the p.d.f. we have

ˆ ∞−∞

f (x)dx

0kx2dx

∣∣∣∣20

It implies that

Question (a): Find the value of k

By using the property of the p.d.f. we have

ˆ ∞−∞

f (x)dx

0kx2dx

∣∣∣∣20

It implies that

Question (b&c): End within 1 min? Continue for 90 sec?

For Q (b), the probability can be calculated as

P(X ≤ 1) =

8x2dx =

∣∣∣∣10

For Q (c), the probability can be calculated as

P(X ≥ 1.5) =

∣∣∣∣21.5

3− 1.53

)= 0.578.

Question (b&c): End within 1 min? Continue for 90 sec?

For Q (b), the probability can be calculated as

P(X ≤ 1) =

8x2dx =

∣∣∣∣10

For Q (c), the probability can be calculated as

P(X ≥ 1.5) =

∣∣∣∣21.5

3− 1.53

)= 0.578.

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.

f (x) =

{0.125x 0 ≤ x ≤ 4

0 otherwise.

(b) Find P(X > 3).

f (x) =

{0.125x 0 ≤ x ≤ 4

0 otherwise.

(b) Find P(X > 3).

f (x) =

{0.125x 0 ≤ x ≤ 4

0 otherwise.

(b) Find P(X > 3).

Question (a): Find the mean and variance of X

E (X ) =

ˆ ∞−∞

x · f (x)dx =

00.125x2dx = 0.125 · x

∣∣∣∣40

Var(X ) = E[(X − EX )2

]= E (X 2)− (EX )2

ˆ ∞−∞

x2 · f (x)dx − (8

00.125x3dx − (

= 0.125 · x4

∣∣∣∣40

− 64

.Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 16 / 37

Question (a): Find the mean and variance of X

E (X ) =

ˆ ∞−∞

x · f (x)dx =

00.125x2dx = 0.125 · x

∣∣∣∣40

Var(X ) = E[(X − EX )2

]= E (X 2)− (EX )2

ˆ ∞−∞

x2 · f (x)dx − (8

00.125x3dx − (

= 0.125 · x4

∣∣∣∣40

− 64

.Jun FU (S&AS, HKU) STAT1306 Introductory Statistics. October 10, 2011 16 / 37

Question(b): Find P(X > 3)

The P(X > 3) can be calculated as follows

P(X > 3) =

3f (x)dx

30.125 · xdx

= 0.125 · x2

∣∣∣∣43

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

We just need to calculate

P(Y ≥ 3) = (1− 7

16)2 =

since that “at least three trials” are required is equivalent with that “thefirst two trials” are failure.

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

We just need to calculate

P(Y ≥ 3) = (1− 7

16)2 =

since that “at least three trials” are required is equivalent with that “thefirst two trials” are failure.

Problems Some Questions from Assignments

Game Plan

Basic Definitions

2 Problems

3 Ending

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?

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:

The no. of ways of arranging A,A,A,B,C ,D in a sequence is

3! · 1! · 1! · 1!

The no. of ways of determining which letter appears three times is

3! · 1! · 1! · 1!

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:

The no. of ways of arranging A,A,B,B,C ,D in a sequence is

2! · 2! · 1! · 1!

The no. of ways of determining which letters appear twice is

2! · 2! · 1! · 1!

Method (1): Final result

In sum, the probability that we will collect all of the four letters is

4)6 × 6!

3! · 1! · 1! · 1!× 4C1 + (

4)6 × 6!

2! · 2! · 1! · 1!× 4C2.

The final answer is1560

46= 0.3809.

Method (1): Final result

In sum, the probability that we will collect all of the four letters is

4)6 × 6!

3! · 1! · 1! · 1!× 4C1 + (

4)6 × 6!

2! · 2! · 1! · 1!× 4C2.

The final answer is1560

46= 0.3809.

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

can be calculated

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

can be calculated

Method (2) (Cont.)

Number Outcome Probability4 NNNN 1× 3

4 ×24 ×

NONNN 1× 14 ×

5 NNONN 1× 34 ×

NNNON 1× 34 ×

NNNOON 1× 34 ×

NOONNN 1× 14 ×

6 NNOONN 1× 34 ×

NONNON 1× 14 ×

NONONN 1× 14 ×

NNONON 1× 34 ×

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.

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

Therefore, the probability required to calculate is

n − r − 1

n − r − 1.

n − r − 1

n − r − 1.

n − r − 1

n − r − 1.

n − r − 1

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

L + M + R = n − 2.

L + R = n − 2− r .

n−r−1C1 = n − r − 1.

L + M + R = n − 2.

L + R = n − 2− r .

n−r−1C1 = n − r − 1.

L + M + R = n − 2.

L + R = n − 2− r .

n−r−1C1 = n − r − 1.

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?

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.

X1 ∼ Bin(2k − 1, p).

Y ∼ Bin(2, p).

X1 + Y ≥ k + 1.

X1 ∼ Bin(2k − 1, p).

Y ∼ Bin(2, p).

X1 + Y ≥ k + 1.

X1 ∼ Bin(2k − 1, p).

Y ∼ Bin(2, p).

X1 + Y ≥ k + 1.

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.

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.

(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)] .

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).

P(X1 ≥ k) + [P(X1 = k − 1) · P(Y = 2)− P(X1 = k)P(Y = 0)] .

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).

P(X1 ≥ k) + [P(X1 = k − 1) · P(Y = 2)− P(X1 = k)P(Y = 0)] .

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.

Final result

P(X1 = k − 1) · P(Y = 2) > P(X1 = k)P(Y = 0).

Hence,

p ≥ 1− p ⇐⇒ 1

2≤ p ≤ 1.

Final result

P(X1 = k − 1) · P(Y = 2) > P(X1 = k)P(Y = 0).

Hence,

p ≥ 1− p ⇐⇒ 1

2≤ p ≤ 1.

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?

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.

Solution

1, 2, 3, 4, 5, 6.

1 + 2 + 3 + 4 + 5 + 6

6= 3.5.

Solution

1, 2, 3, 4, 5, 6.

1 + 2 + 3 + 4 + 5 + 6

6= 3.5.

Solution

1, 2, 3, 4, 5, 6.

1 + 2 + 3 + 4 + 5 + 6

6= 3.5.

Ending

Game Plan

Basic Definitions

2 Problems

3 Ending

Ending

Goodbye and See You

See You After Reading Week (Oct 24,26)

Same Time

Same Place

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