Rule of probability. Let A and B be two events (sets of elementary events). 11. If P (AB) = P (A)P (B), then A and B are independent.

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1 Percetile: the αth percetile of a populatio is the value x 0, such that P (X x 0 ) α% For example the 5th is the x 0, such that P (X x 0 ) 5% 05 Rule of probability Let A ad B be two evets (sets of elemetary evets) 1 A probability must be a value betwee 0 ad 1, icludig 0 or 1, ie, 0 P (A) 1 The sum of probabilities of all simple evets is 1 P (e i ) 1 3 A B : A or B or both happe 4 AB : both A ad B happe 5 Ā : A does ot happe 6 P (A B) + P (AB) P (A) + P (B) 7 P (Ā) 1 P (A) 8 If AB, the P (A B) P (A) + P (B) It is called A ad B are mutually exclusive 9 P (A B) meas give B happes, the probability that A happes 10 P (A B) P (AB), P (A B)P (B) P (AB) P (B) 11 If P (AB) P (A)P (B), the A ad B are idepedet Example: Let A be the evet a perso has O type blood, ad B be the evet a perso is a Asia Suppose 5% Asia populatio ad 30% o-asia populatio have O type blood Ad 60% world populatio is Asia 1 What is the probability a perso has O type blood What is the probability that a perso is a Asia with O-type blood 3 what is the probability that a Asia perso has O-type blood 4 If a perso has O-type blood, what is the probability he is a Asia 5 What is the probability that a perso is a Asia or has O-type blood, or both 6 If a perso has O type blood, what is the probability that he/she is a Asia? Exercise 43, 45, 445, 447, 459, 460, 465, 473, 479, 495 about coditioal probability: 465, 468, 4134 Radom selectio from multigroups Suppose there are 10 red balls, 6 blue balls ad 7 gree balls i a box Radomly pick 10 balls from the box 1

2 1 What is the probability that 4 red, 3 blue ad gree balls are picked What is the probability that o blue balls is picked 3 What is the probability that exactly 10 red or blue balls are picked Sample mea, variace ad stadard deviatio Give a sample (a set of observatios) {x 1,, x } Sample mea x x x Sample variace s (xi x) 1 (xi x) Sample stadard deviatio s 1 Expected value, populatio mea, variace ad stadard deviatio Suppose the sample space of a radom discrete variable X is {x 1,, x }, ad f(x i ) P (X x i ) X f(x) xf(x) x f(x) x 1 f(x 1 ) x 1 f(x 1 ) x 1f(x 1 ) x f(x ) x f(x ) x f(x ) Total 1 µ E(x) x i f(x i ) x i f(x i ) 1 µ E(X) x i f(x i ) σ V ar(x) (x i µ) f(x i ) x i f(x i ) µ 3 σ sd(x) V ar(x) Exercise 51, 59, 593 Table of Distributios Biomial Normal Studet s t χ Biomial(, p) N(µ, σ) T (df) χ (df) Parameters, p µ,σ degree of freedom degree of freedom type discrete cotiuous cotiuous cotiuous Mea Expected Value E(X) p E(X) µ N/A N/A Stadard deviatio sd(x) p(1 p) sd(x) σ N/A N/A

3 How Biomial(, p) distributio arises If the proportio of object A i a populatio is p, ad X is the radom umber of object A i a radom sample of size, the X Biomial(, p) Exercise: 553, 565 How N(µ, σ) distributio arises (Cetral Limit Theorem) Regardless of distributio, if a cotiuous radom umber X has mea E(X) µ, ad sd(x) σ, ad X is the sample mea of sample of size > 30, the X N(µ, σ/ ) approximately (Normal Approximatio of biomial distributio) If X Biomial(, p) ad 0 ad p is ot close to 0 ad 1, the approximately X N(p, p(1 p)) Exercises: 61, 618, 69, 639, 76 How T (df) distributio arises If X N(µ, σ), the X X X use the sample stadard deviatio N(µ, σ/ ) Suppose σ is ot give, we to approximate σ s (Xi X) 1 If is large ( 30), the If is small ( < 30), the Z X µ s/ T X µ s/ N(0, 1) t( 1) How χ (df) distributio arises If X N(µ, σ), the χ (Xi X) ( 1)s χ ( 1) σ σ Cosequetly, if Z 1, Z are iid stadard ormal radom variables, ie, the χ Z i N(0, 1), i1 3 Z i χ ()

4 Drawig Iferece Iferece related to Normal distributio Poit estimate Distributio p µ ( 30) µ 1 µ ( 1, 30) umber of objects i ˆp X X i X Ȳ ( ) p(1 p) ˆp N p, X N(µ, σ/ σ ) X 1 Ȳ N(µ 1 µ, + σ ) 1 Stadard ˆp(1 ˆp) Error SE(ˆp) SE( s/ SE( X Ȳ ) s 1 + s 1 Error Margi Z α/ SE(ˆp) Z α/ SE( X) Z α/ SE( X Ȳ ) CI ˆp ± Z α/ SE(ˆp) X ± Zα/ SE( X) X Ȳ ± Z α/ SE( X Ȳ ) Test of Hypothesis of α sigificace level Null Hypothesis H 0 : p p 0 H 0 : µ µ 0 H 0 : µ 1 µ δ 0 Test Statistics ˆp p 0 Z p0 (1 p 0 )/ Z X µ 0 s/ X Ȳ δ 0 H 1 : RR Z Z α/ p-value: P (Z Z obs ) H 1 :> RR: Z Z α p-value: P (Z Z obs ) H 1 :< RR: Z Z α p-value: P (Z Z obs ) s s Iferece related to No-Normal Distributio 4

5 µ ( < 30) σ Poit Estimate X s Distributio X N(µ, σ/ ) χ ( 1)s σ χ ( 1) Stadard Error s/ N/A Error Margi t α/ s/ N/A cofidece Iterval X ± tα/ s/ ( ( 1)s, χ α/ ) ( 1)s χ 1 α/ Null Hypothesis H 0 : µ µ 0 H 0 : σ σ0 Test Statistics t X µ 0 s/ χ ( 1)s σ0 H 1 : RR t t α/ χ χ α/ or χ χ 1 α/ H 1 :< RR t t α χ χ 1 α H 1 :> RR t t α χ χ α ( ) ( 1)s Note: the cofidece iterval of σ is ( 1)s, χ α/ χ 1 α/ Exercises: 887, 891, 895, 8100, , 946, 968, 971, 974(a)