Download Chapter 7 Random Variable- variable whose value is a numerical

Chapter 7
Random Variable- variable whose value is a numerical outcome of a
random phenomenon
Ex. Flip coin 4 times
X = 0,1,2,3,4
X is random variable that represents the number of heads for the 4 flips
****Make sure you always define x *****
Discrete random variable has a countable number of possibilities.
Probability distribution list the values of x and their probabilities
Value of x
x1
x2
x3
x4
……
xk
Probability
p1
p2
p3
p4
…….
Pk
Every probability is a number between 0 and 1
P1+p2+p3+…….pk = 1
Example
Grade
0
1
2
3
4
Probability
.10
.15
.30
.30
.15
.10+.15+.30+.30+.15 = 1
P(grade is 3 or 4) = .30 + .15 = .45
P(x < 2)
.15 + .10 = .25
Look at probability histograms on page 393
X = number of heads when coin is flipped 4 times
TTHH
HTTH
HTTT
HTHT
HHTH
THTT
HHTT
HHHT
TTHT
THHT
HTHH
TTTT
TTTH
THTH
THHH
HHHH
X=0
X=1
X=2
X=3
X=4
X
0
1
2
3
4
P
1/16
4/16
6/16
4/16
1/16
P(X = 3) 4/16 = .25
P(X 2)
6/16 + 4/16 + 1/16 = 11/16
P(X > 2)
4/16 + 1/16 = 5/16
P(X
1)
1 – P(X = 0)
1 – 1/16 = 15/16
Homework read pages 391 – 395 do problems 1a, 2 – 5
Continuous random variables
All numbers between 0 and 1 we cannot assign probabilities because
there are infinite amount of numbers between 0 and 1.
We will use areas under the curve to assign probabilities.
Uniform distribution
P (0.3
_______________________
0
.3
.7
1
x
0.7) = 0.4
_______________________
0
.5
.8
P(x
.5 or x
.8)
1
.5 + .2 = .7
Area = P(A)
______________________________________
Event A
Continuous random variable x takes all values in an interval of numbers.
The probability distribution of x is described by a density curve. The
probability of an event in this area under the density curve and above the
values of x that make up the event.
P(x = .3) = 0
P(x
.3 ) = .3
P(x
.3 ) = .3
Normal distributions as probability distributions
_______________________________
Z=
now z =
Find z then use table to find area
Ex. N(.3, 0.0118) remember N(p, σ)
P(p(hat) < .28) or p(p(hat) > .32)
Z=
z=
Z=
z=
Z = -1.69
Need to find two z-scores
z = 1.69
.0455
.0455
______________________________
-1.69
1.69
P( p(hat) < .28 or p(hat) > .32) = .0455 + .0455 = .0910
Homework read pages 397 – 403 and do problems 6 – 12 and 14 – 16
7.2 means and variances of random variables
Ex. Lottery 3 numbers
1000 possibilities
Payoff x
$0
$500
Probability
999/1000
1/1000
The average payoff is 0(999/1000) + 500(1/1000)
0 + .5
.5 (50 cents)
Mean of random variable x is .50
Tickets cost $1.00 so in the long run the state keeps half of the money
you wager.
μx = mean of probability distribution.
******Expected value if x is μx **********
Mean of discrete random variable
Value of x
x1
x2
x3
x4 ……. xk
Probability
p1
p2
p3
p4 …….. pk
To find the mean of x, multiple each possible value by its probability,
then add all the products.
μx = x1p1 + x2p2 + x3p3 + x4p4 + …. + xkpk
μx = Σ xipi
Variance of a discrete random variable
Value of x
x1
x2
x3
xk
Probability
p1
p2
p3
pk
σ2 = (x1 – μx)2 p1 + (x2 – μx)2 p2 + (x3 – μx)2 p3 + …+ (xk – μx)2 pk
σ2 = Σ(xi – μx)2 pi
the standard deviation σ2 of x is the square root of variance
ex. Selling aircraft parts
units sold
1000
3000
5000
10000
probability
.1
.3
.4
.2
xi
pi
xipi
(xi – μx)2pi
1000
.1
100
(1000-5000)2(.1) = 1,600,000
3000
.3
900
(3000-5000)2(.3) = 1,200,000
5000
.4
2000
(5000-5000)2(.4) = 0
10,000
.2
2000
(10000-5000)2(.2)= 5,000,000
μx= 5000
σ2 = 7,800,000 (variance)
standard deviation =
= 2792.8
homework read pages 411 – 413 do problems 22-26, 29
Statistical estimation and the law of large numbers
Law of large numbers
Draw independent observations at random from any population
with finite mean μ. Decide how accurately you would like to estimate μ.
As the number of observations drawn increases, the mean x of the
observed values eventually approaches the mean μ of the population as
closely as you specified and then stays that close.
Rules of means
Rule 1 if x is a random variable and a and b are fixed numbers,
then μa + bx = a + bμx
rule 2 if x and y are random variables, then μx+ y = μx + μy
Ex 7.10 page 419
Rules for variance
Rule 1 if x is a random variable and a and b are fixed numbers, then
σ2a + bx = b2σ2x
Rule 2 if x and y are independent random variables, then
σ²x + y = σ2x +σ2y
σ²x - y = σ2x +σ2y
Rule 3 if x and y have correlations p, then
σ²x + y = σ2x +σ2y + 2pσxσy
σ²x + y = σ2x +σ2y - 2pσxσy
general addition rule for variances of random variables
ex. 7.12 SAT scores 422
Ex. 7.14 page 424
Homework read pages 425 – 426 do problems 34-39, 41