hs 67bps chapter 101 chapter 10 introducing probability
TRANSCRIPT
HS 67 BPS Chapter 10 1
Chapter 10
Introducing Probability
HS 67 BPS Chapter 10 2
Idea of Probability• Probability is the
science of chance behavior
• Chance behavior is unpredictable in the short run, but is predictable in the long run
• The probability of an event is its expected proportion in an infinite series of repetitions
The probability of any outcome of a random variable is an expected (not observed) proportion
HS 67 BPS Chapter 10 3
How Probability BehavesCoin Toss Example
Eventually, the proportion of
heads approaches 0.5
HS 67 BPS Chapter 10 4
How Probability Behaves“Random number table example”
The probability of a “0” in Table B is 1 in 10 (.10)
Q: What proportion of the first 50 digits in Table B is a “0”?
A: 3 of 50, or 0.06
Q: Shouldn’t it be 0.10?
A: No. The run is too short to determine probability. (Probability is the proportion in an infinite series.)
HS 67 BPS Chapter 10 5
Probability models consist of two parts:1) Sample Space (S) = the set of all possible
outcomes of a random process. 2) Probabilities for each possible outcome in
sample space S are listed.
Probability Models
Probability Model “toss a fair coin”
S = {Head, Tail}
Pr(heads) = 0.5
Pr(tails) = 0.5
HS 67 BPS Chapter 10 6
Rules of Probability
HS 67 BPS Chapter 10 7
Rule 1 (Possible Probabilities)Let A ≡ event A
0 ≤ Pr(A) ≤ 1
Probabilities are always between 0 and 1.
Examples:
Pr(A) = 0 means A never occurs
Pr(A) = 1 means A always occurs
Pr(A) = .25 means A occurs 25% of the time
HS 67 BPS Chapter 10 8
Rule 2 (Sample Space)Let S ≡ the entire Sample Space
Pr(S) = 1All probabilities in the sample space together must sum to 1
exactly.
Example: Probability Model “toss a fair coin”, shows that Pr(heads) + Pr(tails) = 0.5 + 0.5 = 1.0
HS 67 BPS Chapter 10 9
Rule 3 (Complements)Let Ā ≡ the complement of event A
Pr(Ā) = 1 – Pr(A)
A complement of an event is its opposite
For example:
Let A ≡ survival then Ā ≡ death
If Pr(A) = 0.95, then
Pr(Ā) = 1 – 0.95 = 0.05
HS 67 BPS Chapter 10 10
Events A and B are disjoint if they are mutually exclusive. When events are
disjoint
Pr(A or B) = Pr(A) + Pr(B)
Age of mother at first birth
(A) under 20: 25%
(B) 20-24: 33%
(C) 25+: 42% } Pr(B or C) = 33% + 42% = 75%
Rule 4 (Disjoint events)
HS 67 BPS Chapter 10 11
Discrete Random Variables
Example: A couple wants three children. Let X ≡ the number of girls they will haveThis probability model is discrete:
Discrete random variables address outcomes that take on only discrete (integer) values
HS 67 BPS Chapter 10 12
• Example Generate random number between 0 and 1 infinite possibilities.
• To assign probabilities for continuous random variables density models (recall Ch 3)
Continuous Random VariablesContinuous random variables form a continuum of possible outcomes.
This is the density model for random numbers between 0 and 1
HS 67 BPS Chapter 10 13
Area Under Curve (AUC)The AUC concept (Chapter 3) is essential to working with continuous random variables.
Example: Select a number between 0 and 1 at random.
Let X ≡ the random value.
Pr(X < .5) = .5
Pr(X > 0.8) = .2
HS 67 BPS Chapter 10 14
Normal Density Curves
→
z
x
♀ Height X~N(64.5, 2.5)
Z Scores
Introduced in Ch 3: X~N(µ, ).
Standardized Z~N(0, 1)
zx
HS 67 BPS Chapter 10 15
If I select a woman at random a 99.7% chance she is between 57" and 72"
68-95-99.7 Rule• Let X ≡ ♀ height
(inches)• X ~ N (64.5, 2.5) • Use 68-95-99.7 rule
to determine heights for 99.7% of ♀
• μ ± 3σ = 64.5 ± 3(2.5)
= 64.5 ± 7.5 = 57 to 72
HS 67 BPS Chapter 10 16
Calculating Normal Probabilities when 68-95-99.7 rule does not apply
Recall 4 step procedure (Ch 3)
A: State
B: Standardize
C: Sketch
D: Table A
HS 67 BPS Chapter 10 17
Illustration: Normal Probabilities
What is the probability a woman is between 68” and 70” tall? Recall X ~ N (64.5, 2.5)
4.15.2
)5.6468(
z
z(70 64.5)2.5
2.2
A: State: We are looking for Pr(68 < X < 70)
B: Standardize
Thus, Pr(68 < X < 70) = Pr(1.4 < Z < 2.2)
HS 67 BPS Chapter 10 18
Illustration (cont.)
C: Sketch
D: Table A: Pr(1.4 < Z < 2.2) = Pr(Z < 2.2) − Pr(Z < 1.4)= 0.9861 − 0.9192 = 0.0669