Transcript Chp 16

Chapter 16
Random Variables
Ranya Kaluarachchi
George Weng
Period 3
Random Variables
A random variable assumes any of several different values as a result of some
random event.
Two Types:
Discrete random variables can take one of a FINITE number of DISTINCT
OUTCOMES.
Ex. Number of credit hours (2 hours, 3 hours, 4 hours)
Continuous random variables can take ANY NUMERIC VALUE within a
RANGE of values.
Ex. Cost of books this term (80.00$ - 120.00$)
What Can Go Wrong?
Probability models are still models.
They may not show what will actually happen in reality.
Don’t assume everything follows a Normal model.
Independent variables
You can add expected values for any two random variables.
You can only add variances of independent random variables.
Variances of independent random variables
Variances of independent random variables add; standard deviations
don’t.
The variance of independent random variable’s sum or difference is always
the sum of their variances.
Expected Value
E(X) = μ
that’s the symbol
they use, mu
Example from textbook: life insurance
An insurance company will pay you 10,000$ if you die that year, 5000$ if
you’re disabled, or 0$ if neither occurs. One out of every 1000 people die, and
two out of every 1000 are disabled.
What are the
outcomes?
Payout x
10000
5000
0
What are the
chances?
Probability P(X = x)
1/1000
2/1000
997/1000
Expected Value = E(X) = 10,000(1/1000) + 5000(2/1000) +
0(997/1000)
E(X) = 20$
in other words, each policy must cost at least 20$ for the company to break even
“just multiply each possible value by the probability that it occurs, and find the sum”
Standard Deviation
So, continuing the example… to find standard deviation, first find deviation.
μ (expected value) = $20
Payout
“x”
10000
5000
0
Probability “P(X=x)”
1/1000
2/1000
997/1000
Deviation “(x - μ)”
10000 - 20 = 9980
5000 - 20 = 4980
0 - 20 = -20
Next, we square each deviation. The variance is the expected value of those
squared deviations.
Var(X) = 9980²(1/1000) + 4980²(2/1000) + (-20)²(997/1000) = 149,600
Then take the square root of the variance to get standard deviation.
SD(X) = sqrt(149,600) ~~ 386.78$
Rules for adding and subtracting
E(X ± c) = E(X) ± c
Var(X ± c) = Var(X)
E(a*X) = a*E(X)
Var(a*X) = a²Var(X)
Homework Problem #13
Kids. A couple plans to have children until they get a girl, but they agree that
they will not have more than three children even if all are boys. (Assume boys
and girls are equally likely.)
13. Find the standard deviation of the number of children the couple may have.
# of Children they’ll have
1
2
3
P(Children)
0.5
0.25
0.25
Deviation “(x - μ)”
1 - 1.75 = -0.75
2 - 1.75 = 0.25
3 - 1.75 =
1.25
E(X) = 1(0.5) + 2(0.25) + 3(0.25) = 1.75
Deviation = (1-1.75)
Var(X) = (-0.75)²(0.5) + (0.25)²(0.25) + (1.25)²(0.25) = 0.6875
SD(X) = √0.6875 =
0.829
Homework Problem #17
Defects. A consumer organization inspecting new cars found that many had appearance
defects(dents, scratches, paint chips, etc.). While none had more than three of these defects, 7% had
three, 11% two, and 21% one defect. Find the expected number of appearance defects in a new car
and the standard deviation.
# of defects
3
2
1
0
P(defects)
7/100
11/100
21/100
61/100
Deviation
3 - 0.64 = 2.36
2 - 0.64 = 1.36
1 - 0.64 = 0.36
0 - 0.64 = -0.64
0.64
E(X) = 3(7/100) + 2(11/100) + 1(21/100) + 0(61/100) =
Var(X) = 2.36²(7/100) + 1.36²(11/100) + 0.36²(21/100) + -0.64²(61/100) = 0.8704
SD(X) = √0.8704 =
0.933