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