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

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.

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?

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?

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.

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.

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.

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.

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

4. Random Variables

Now we need to know what happens.

What Happens:

X

the company pays out

Death $10,000

Disability $5000

Healthy $0

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

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

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

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

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

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

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

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).

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

Expected value

Mean or expected value of a discrete random variable is:

µ = E(x) = ∑ x P (x)

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

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?

Properties of Probability, P( X = xi )

1)(0 (1) ixXP

1)( (2)1

n

iixXP

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

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

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

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

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

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

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

Descriptive Phrases

Descriptive Phrases require special care!

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

Problems

Problems 4.12, 4.17, 4.36, 4.40, 4.43

34

Homework

• Review Chapter 3, 4.1-4.3

• Read Chapter 4.4, 5.1-5.3

• Have a great Thanksgiving