Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

34
Problems Problems 3.75, 3.80, 3.87

Transcript of Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Page 1: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Problems

Problems 3.75, 3.80, 3.87

Page 2: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

Page 3: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

Insurance companies have to take risks.

When you buy insurance you are buying it in case something goes wrong. The insurance company is securing you and making money by betting that you are going to live a long life or that you are not going to crash your car.

Page 4: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

Insurance companies have to take risks.

When you buy insurance you are buying it in case something goes wrong. The insurance company is securing you and making money by betting that you are going to live a long life or that you are not going to crash your car.

It’s important that the insurance company offers it’s insurance at a fair price. How do they calculate it?

Page 5: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

It’s important that the insurance company offers it’s insurance at a fair price. How do they calculate it?

Here is a simple model. An insurance company offers a death and disability policy which pay $10,000 when you die or $5000 if you are disabled. The company charges $50/year for this benefit. Should the company expect a profit?

Page 6: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

Here is a simple model. An insurance company offers a death and disability policy which pay $10,000 when you die or $5000 if you are disabled. The company charges $50/year for this benefit. Should the company expect a profit?

We need to define some terms first.

Page 7: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

A random variable is a way of recording a numerical result of a random experiment. Each sample point is given one numerical value.

Page 8: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

A random variable is a way of recording a numerical result of a random experiment. Each sample point is given one numerical value.

In this case the random variable is the payout of the insurance company.

Page 9: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

A random variable is a way of recording a numerical result of a random experiment. Each sample point is given one numerical value.

In this case the random variable is the payout of the insurance company.

We usually use capital X to represent our random variable.

Page 10: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random VariablesIn this case the random variable is the payout of the insurance company.

We usually use capital X to represent our random variable.

What Happens:

X

the company pays out

Death $10,000

Disability $5000

Healthy $0

Page 11: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

Now we need to know what happens.

What Happens:

X

the company pays out

Death $10,000

Disability $5000

Healthy $0

Page 12: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

Now we need to know what happens. The insurance company uses actuaries to determine the probability of certain events occurring.

What Happens:

X

the company pays out

Death $10,000

Disability $5000

Healthy $0

Page 13: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

The probability the insurance company will have to pay $10,000 is 1 in a thousand is represented with:

What Happens:

X

the company pays out

Death $10,000

Disability $5000

Healthy $0

1000

1)000,10( XP

Page 14: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variables

The probability the insurance company will have to pay $5,000 is 2 in a thousand is represented with:

What Happens:

X

the company pays out

Death $10,000

Disability $5000

Healthy $0

1000

2)000,5( XP

Page 15: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random VariablesIn this case the random variable is the payout of the insurance company.

We usually use capital X to represent our random variable.

What Happens:

X

the company pays out

Probability

P(X=x)

Death $10,000

Disability $5000

Healthy $0

Page 16: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random VariablesIn this case the random variable is the payout of the insurance company.

We usually use capital X to represent our random variable.

What Happens:

X

the company pays out

Probability

P(X=x)

Death $10,000 1/1000

Disability $5000 2/1000

Healthy $0 997/1000

Page 17: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variable

This is called a probability distribution table.

What Happens:

X

the company pays out

Probability

P(X=x)

Death $10,000 1/1000

Disability $5000 2/1000

Healthy $0 997/1000

Page 18: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variable

This is called a probability distribution table.

We may draw the distribution (on a histogram)

What Happens:

X

the company pays out

Probability

P(X=x)

Death $10,000 1/1000

Disability $5000 2/1000

Healthy $0 997/1000

Page 19: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Discrete vs ContinuousRandom Variables

The random variable in the previous example is called a discrete random variable, since X takes an one of a specific number of values

A continuous random variable is one that can take on a range of values inside of an interval. (Example: X represents the height of a randomly selected individual).

Page 20: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

4. Random Variable

Should the company earn be selling these policies?

What Happens:

X

the company pays out

Probability

P(X=x)

Death $10,000 1/1000

Disability $5000 2/1000

Healthy $0 997/1000

Page 21: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Expected value

Mean or expected value of a discrete random variable is:

µ = E(x) = ∑ x P (x)

Page 22: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Expected value

Mean or expected value of a discrete random variable is:

µ = E(x) = ∑ x P (x)

The standard deviation of a discrete random variable is given by

222 )( xpx

: where2

Page 23: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Example: Concert Planning• In planning a huge outdoor concert for June 16, the producer

estimates the attendance will depend on the weather according to the following table. She also finds out from the local weather office what the weather has been like, for June days in the past 10 years.– Weather Attendance Relative Frequency– wet, cold 5,000 .20– wet, warm 20,000 .20– dry, cold 30,000 .10– dry, warm 50,000 .50

– What is the expected (mean) attendance?– The tickets will sell for $9 each. The costs will be $2 per person for

the cleaning and crowd-control, plus $150,000 for the band, plus $60,000 for administration (including the facilities). Would you advise the producer to go ahead with the concert, or not? Why?

Page 24: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Properties of Probability, P( X = xi )

1)(0 (1) ixXP

1)( (2)1

n

iixXP

Page 25: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Example

The random variable x has the following discrete probability distribution:

Find

P (x≤17) P (x =19)

P (x≥17) P(x≤19)

P (x <16 or x >17)

x= 15 16 17 18 19

P(X=x) .2 .3 .2 .1 .2

Page 26: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Example

The random variable x has the following discrete probability distribution:

Find

P (X≤17)= .7 P (X =19)= .2

P (X≥17) =.3 P(X≤19)= 1

P (X <16 or X >17)= .5

x= 15 16 17 18 19

P(X=x) 0.2 0.3 0.2 0.1 0.2

Page 27: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Example

The random variable x has the following discrete probability distribution:

Find:

The expect value and standard deviation of this random variable.

x= 15 16 17 18 19

P(X=x) 0.2 0.3 0.2 0.1 0.2

Page 28: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Example

The random variable x has the following discrete probability distribution:

Find:

The expect value and standard deviation of this random variable.

µ = E(X)=16.8

x= 15 16 17 18 19

P(X=x) 0.2 0.3 0.2 0.1 0.2

Page 29: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Example

The random variable x has the following discrete probability distribution:

Find:

The expect value and standard deviation of this random variable.

µ = E(X)=16.8 and

x= 15 16 17 18 19

P(X=x) 0.2 0.3 0.2 0.1 0.2

222 )( xpx

Page 30: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Example

The random variable x has the following discrete probability distribution:

Find:

The expect value and standard deviation of this random variable.

µ = E(X)=16.8 and

x= 15 16 17 18 19

P(X=x) 0.2 0.3 0.2 0.1 0.2

4.196.1

Page 31: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Empirical Rule and Chebyshev’s Rule

Chebyshev’s Rule and the Empirical Rule for Random Variables. That is

1) The number of points that fall within k standard deviation of the mean is at least:

1-1/k2.

2) If the distribution of the Random Variable is a normal bell shaped curve, 68% of data points are in , 95% are in and 99.7% are in

2.3

Page 32: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Descriptive Phrases

Descriptive Phrases require special care!

– At most– At least– No more than– No less than

Page 33: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

Problems

Problems 4.12, 4.17, 4.36, 4.40, 4.43

Page 34: Problems Problems 3.75, 3.80, 3.87. 4. Random Variables.

34

Homework

• Review Chapter 3, 4.1-4.3

• Read Chapter 4.4, 5.1-5.3

• Have a great Thanksgiving