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Chapter 16

STAB22H3 Chapter Notes - Chapter 16: Random Variable, Squared Deviations From The Mean, Normal Distribution


Department
Statistics
Course Code
STAB22H3
Professor
Ken Butler
Chapter
16

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CHAPTER 16 - RANDOM VARIABLES
WHERE ARE WE GOING?
- random variable - used to model probability of outcomes
- helps us discuss about, & predict random behaviour
MAIN TEXT
[1]
(ex)
- insurance company offers "death and disability" policy
- pays $10,000 when you die
- pays $5,000 when you are permanently disabled
- charges $50/yr for having this benefit
Q: is company likely to make profit selling this plan?
- to ans., need to know probability that clients will die/be disabled in any yr
- using thta actuarial info, company can calc. expected val. that it gains from this
policy
WHAT IS AN ACTUARY?
- ppl who estimate likelihood & costs of rare events in order for them to be insured
- req. financial, statistical and business skills
p423

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EXPECTED VALUE: CENTRE
[1]
RANDOM VARIABLE - var. that has its numerical val. based on outcome of random
event
- denoted with X
(ex) amt the company pays out on an individual policy
---
NOTATION ALERT
- X, Y, Z are most common letters used for random var's
- CAPITAL LETTERS
---
- denote particular val has by using "x"
- ex. for insuracnce company, x = 10,000 (if you die this yr), $5000 (if disabled),
or $0 (if neither occurs)
DISCRETE RANDOM VARIABLE
CONTINUOUS RANDOM VARIABLE
- random var. s.t. we can list all of its
outcomes
(ex) x = 10,000 or 5000, or 0 are its
outcomes for insurance company ex.
- random var s.t. you cannot list all of its
outcomes (has too many)
PROBABILITY MODEL (for a random var.) = collection of all possible val's and
probabilities that they occur at
[2]

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probability model for insurance company example
Policyholder
Outcome
Payout
(x)
Probability
P(X = x)
Death
10,000
1/1000
Disability
5,000
2/1000
Neither
0
997/1000
- suppose that company ensures 1000 ppl
- 1 policyholder dies
- 2 are disabled
- remaining 997 live the year unharmed
=> has to pay $10k to one client (die), and $5,000 to each of the 2 clients
=> 20k in total, or an average:
- $20,000 / 1000 policies = $20/policy
- expected value E(X)
- recall: company charges $50 for this policy
- then, in a year it loses $20/policy, it has made profit of $30 per policy
[3]
- cannot predict what WILL happen in any given yr, but can say what we EXPECT to
happen (E(X))
- req. probability model
- parameter of probability model is EXPECTED VALUE
- E(X) or μ
- which is mean
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