| Hand in: p. 268, 10.44 Benford One more discrete probability Continuous sample spaces: p. 259, 10.15 Iowa Test Scores p. 269, 10.49 Did you vote? p. 269, 10.51 NAEP scores p. 269, 10.53 Friends READ Personal Probability, pp. 261-2. All our theory will be developed using the "frequentist" point of view (probability = proportion in the long run). But there is another theory based on Personal Probability, sometimes called "Bayesian". p. 262, 10.18 On a separate sheet using http://www.whfreeman.com/bps4e "Probability " applet: : If you do it jointly, one sheet for both people (I'll aggregate the results) p. 270, 10.55 runs of free throws p. 270, 10.56 a. For b, do 20 people 10 times, but do 320 people only twice. Record not only the proportion (.63 or whatever) but the fraction (like 201/320. 208/320 = .65 exactly) - - - - - - - - - Ch. 11- - - p. 272 11.1 caffeine (Param./Stat.) p. 272 11.2 voters(Param./Stat.) Postpone these two: p. 275, 11.4 means in action (LLN) p. 275, 11.5 insurance (LLN) p. 277, 11.6 sampling distribution of exam scores Do a and a modified version of b; Do b this way. Close your eyes and put your finger down somewhere on table B (Don't use row 116!! unless you land there.). Start reading the table where your fingertip lands. Record your sampleof 4, and find xbar for your sample. Now Repeat part b, to get a total of 3 values of xbar. (You can just keep reading the table where you left off, or you can put your finger in a different spot). Record your 3 xbars, Make a dotplot of your 3 xbars and bring the values to class to be compiled with everyone else's.. |
Read, to
discuss
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Optional
More practice: p. 261, 10.17, ACT scores - - - - p. 272,11.3 Bearings (Param./Stat.) |
Probability rules:
A an event in sample space S, P(A)
is "the probability
that A occurs"
These rules are all true for
proportions
in long run (Probabilities), prop.of counts, proportions of areas.
1. 0 <
P(A) < 1
2. P(S) = 1
3. For any event A,
P(A
does not occur) = 1 - P(A)
4. A and B
are
disjoint if they have no outcomes in common (can't happen
simultaneously.)
If
A and B are disjoint, their probabilities add: P(A or B) =
P(A)
+ P(B)
Finite sample
spaces:
Assign a probability to each outcome (>0)
so they add to 1.
Prob. of an event is sum of
prob's of its outcomes.
Two principles for assigning
probabilities:
--Sometimes, a properly chosen sample space will
have equally likely
outcomes. You can use this to find other probabilities.
Phenomenon: Flip coin twice.
S1 = {HH, HT, TH,
TT} S2
= {0, 1, 2} number of heads
S3 = {Y, N} both are heads?
Sample space | HH | HT | TH |
TT
|
EQUALLY LIKELY
Prob's
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.25| .25| .25| .25| P(tail followed by head)=?
Sample space | 2
|
1 | 0 | P(at
least 1 tail)=? P(1 of each) = ?
Prob's
|
.25| .50 | .25| P(at least 1 Head)=
?
P(2 Heads) = ?
Sample space | Y
|
N |
Prob's
|
.25| .75 |
--If you pick an individual
at random from a population, the
probability that one individual will be XYZ is the same as the
proportion of XYZ's in the population.
Pick one person from U.S. Pop. (Age 25
+)
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Random Variable:
(X, Y, Z...) Variable whose value is a numerical outcome of a
random
phenomenon.
Probability distribution of X tells
us what values X can take and how to assign probabilities to them.
If X has a finite number of
possible values (Discrete distributions), nothing new except notation.
P(X < 2) is "Prob.
that X is less than 2."
Flip coin twice. R.V. X
= number of heads:
Distribution given by table.
x| 2 | 1 | 0 |
P(X=x)
|
.25| .50 | .25|
P(X >
1) = ?
Words: Prob that #
heads is >
1
P(X = 2)
=
?
Prob that # heads is
2
Looking ahead (back)
Random variables with intervals
of outcomes
("continuous") Ch.10 (p. 256 on)
If the sample space is an interval of values (or the whole
line),
the way we assign probabilities to events is with a density curve
(Ch. 3, cf. Day
7 on) (remember density curves were
idealizations
of
histograms--of repeating the "experiment" many many times)
P(a < X < b) = the
probability that X is between a and b is the
area
under the density curve, between a and b.
We declare P (X = a) = 0 , so P(a < X < b)
= P(a < X < b)
Notation: Use capital letter for the random variable, the "label"
of the phenomenon. Use small letters for particular values it can
have. But this rule is often broken--Moore uses x-bar where many
(I)
would use X-bar.
Review "Density curves" HW day
7, restating these parts as probability questions, and adding a bit:
(Copies of the HW handout
are in class/ outside my door if you can't find yours.)
Change language from "description of a population of data" to "pick
an individual from the population, call the value X"
A. ("Uniform") X = number the spinner points to.
a) (example) The probability that the spinner points to a number
less than .6 = P( X < .6) = .6 .
b) P (.2 < X < .6) = ? Say
it
in words: ?
c) For what x is there probability .4 of being greater than
x
?
(In notation: P(X > x) = .4. Find x)
B. Y = (number you get from) the sum of two spinners.
("Triangular")
This is the same random variable as Y in 10.14, p. 259!
a) The probability that the sum is a number less than .6 =
P( ?
) =.18
P(Y > .6) = ?
b) P(Y <
1.6)
= ? P (Y <
1) = ?
P( 1 < Y < 1.6) = ?
c) P(Y > x) = .92 Find x: ?
(Hint: P(Y<x) = .08)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
- - - - - -
Our most important probability model: NORMAL DISTRIBUTION
family.
Same techniques as before, only we ask "probability that one chosen at
random..." instead of "proportion of all..." Review Normal
techniques: Day 7 Day
8, Day
9
Take a random sample of size 1 from a population which is
N(110, 25).
(Give an individual, chosen at random, the "Classic IQ test", which
has a normal distribution, mean 110, s.d. 25. X is the
score
on the test.)
Find P(100 < X < 140), prob. that individual gets between
100 and 140. Work is on Day
9, what proportion.
Next: How does sample mean behave?
( pp.275-86)
Sample Chosen
from a Population
(varies)
(fixed, but usually unknown)
Calculate Numerical
summary:
Statistic
estimatingParameter
xbar
µ
We take a simple random sample of size n, find the sample mean xbar.
It will be different depending on the sample, so we have a random
phenomenon.
We measure the outcome as a number, the sample mean, so we have a
random variable X bar.
Law of Large Numbers (p. 273-4, "LLN") Take
observations
at
random from a population with population mean µ.
Then as the number of observations n increases, the sample mean xbar
gets
closer and closer to µ.
(Even if the population is infinitely large! Note--we don't say how
big n needs
to be for how close here.)
OR Let the sample size n get bigger. Then the xbars
will eventually get very close to the population mean µ.
OR As the sample size increases, the sample mean gets closer
to the population mean µ.
OR For a very large sample, the sample mean will (almost
certainly)
be very close to the population mean.
e.g. the bigger my statistics class, the closer their mean height
should be to the U.S. mean height for women.
Applet:
http://www.whfreeman.com/bps4e
"Law of Large Numbers" Roll a single die. X = number.
µ=
3.5. (Think X is number of spaces you can move in a board game.
Average per roll is 3.5.)
My result Oct.27: x
= 5 n=1 Xbar = 5/1 = 5.
Roll again, x = 2. n=2, Xbar = (5+2)/2 = 3.5
Again, x = 1 n=3. Xbar =
(5+2+1)/3 = 2.67 Again...
... Xbar for large
n; close to 3.5.
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