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# MATH_2040_Chapter_6_Notes Math 2040

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This 6 page Class Notes was uploaded by Ang Judd on Saturday February 13, 2016. The Class Notes belongs to Math 2040 at Southern Utah University taught by Said Bahi in Winter 2016. Since its upload, it has received 15 views. For similar materials see Business Statistics in Math at Southern Utah University.

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Date Created: 02/13/16

Chapter6 6.1 Random Variables Random Variable: A numerical value assigned to outcomes. Probiblity Distribution: A model that describes a specific kind of process Discrete Random Variable: A random variable which has a countable number of possible outcomes (no decimals) WHEN DISCRIBING DISCRETE RANDOM VARIABLES A={HHT, HTH, THH} In 3 tosses of a coin 1)State the Variable X= count of heads in 3 tosses *X refers to Event A* 2)List all possible values of the variable X taken values= 0,1,2,3 X=0 {TTT} X=1 {HTT, THT, TTH} X=2 {HHT, HTH, THH} X=3 {HHH} 3)Determine the probibilities of these values we want: P(X=0) P(0) P(X=1) P(1) P(X=2) P(2) P(X=3) P(3) *With small problems a table is fine* X P(X) 0 1/8 P(x)= 1 1 3/8 n(S) 2 3/8 3 1/8 Counting Principle 1st 2nd 3rd 2 x2 x2 =8 23 =8 6.2 Properties of Discrete Probability Distributions 1) ∑ P(x)=1 2) 0 ≤ P(x)≤ 1 3) P(x=xior x=j)=P(x=i) + P(x=j) P(A U B)=P(A)+P(B)-P(A ∩ B) P(A U B)=P(A)+P(B) mutually exclusive ex) toss a coin 3 times P{x=1 or x=2} P(1)= 3/8 P(2)=3/8 P(1 and 2)= 6/8 or 3/4 EX) DBS_6_2 slide 8 x= count of sales per day x= 0, 1, 2, 3, 4 a) P(x≥2)= P{x=2 or x=3 or x=4} P(2) + P(3)+ P(4) 60/200+ 40/200+ 40/200= 140/200=14/20= 7/10 0.3+0.2+0.2= 0.7 OR USE THE COMPLEMENT P(x≥2)=1-(Px<2)=1-P(x<1)=1-{ P(0)P(1)}= 1-(40/200+20/200=60/200= 0.7 b) P( at most 2 sales) P(x≤2) *From History* P(0)+P(1)+P(2) 40/200+20/200+60/200=120/200= 3/5 or 0.6 EX) DBS_6_2 slide 16 2 P(X=x)= x , if x= 1, 2, 3, 4 { 30 0 otherwise x P(x) work shown 0 0 *Theoretically from the frequency, We don't 1 1/30 1 /30 see the experiment, we're just given a 2 2 4/30 2 /30 formula* 3 9/30 3 /30 2 4 16/30 4 /30 5 0 6.3 Expected Value, Variance, and Standard Deviation of Random Variables Expected Value mean/μ mean of x or expected value of x μ=∑(x)= ∑ xP(x) Variance Ợ =V(x)=∑ (x-μ) =P(x) Standard Deviation Ợ=SD(x)=√Ợ 2 6.4 Discrete Uniform Distribution Uniform: Every outcome has the same probability. ex) Roll a die X P(x) P(X=x)= 1 1 1/6 n 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 P(X)= 1/6 6.5 Binomial Model Properties 1) There are only 2 possible outcomes (success and failure) 2) Each trial is independent of the others 3) A fixed number of trials (n) 4) P(success) is fixed (p) P(failure)=1-p 5) The binomial random variable X= count of successes X= 0, 1, 2, 3, …, n Want: P(0) P(1) P(2) P(3)….P(n) ex) A doctor has a 85% rate of success in surgeries. In any given day he will perform 5 surgeries. P(3 successes) X= Count of successes in a given day (binomial) X= 0, 1, 2, 3, 4, 5 P(X=3) SSSFF P(A and B)= P(AlB)xP(B) P(A)xP(B) S S S F F (0.85) (0.85) (0.85) (0.15) (0.15) = (0.85) (0.15) *Independet not mutually exclusive* *P(A) or P(B) is addition* S F F S S 3 2 (0.85) (0.15) (0.15) (0.85) (0.85) = (0.85) (0.15) F S S S F (0.15) (0.85) (0.85) (0.85) (0.15) = (0.85) (0.15) x n-x *How to compute Binomial Probabilities: X with n, p then P(X=x)=nCx*P *(1-p) Mean of Binomial μ=∑(x)=np same as expected value ex) same doctor n=s p=0.85 e(x)=s(0.85) 4.25 Varience Ợ =√(x)=np(1-p) ex) same doctor n=5 p=0.85 5(0.85)(0.15) 0.6375 Standard Deviation 2 √Ợ Ợ=√0.6375 0.798 EX) DBS_5_2 slide 6 toss coin 10 times P=1/2 n=10 P(6 heads) x= count of heads in 10 tosses 6 4 P(X=6)= 10 6/2) (1/2) 10! x 1 *It doesn't matter which one you simplify with- use the one that gets rid of the most numbers 6!4! 2^10 10x9x8x7 = 5x3x2x7 x 1 = 0.2051 4x3x2x1 1 2^10 On calculator 2nd-vars P(X=5)--> pdf (A) n,p,x P(X≤5)-->cdf (B) n,p,x ex) P(at most 6 heads in 10 trials) *cdf P(X≤6) P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6) Examples For All the Situations *Use the calculator 2nd-->vars-->binompdf or binomcdf X=6 (10, .3) P(X=7) binompdf (10, .3, 7) P(X≤3) binomcdf(10, .3, 3) P(X>6) 1-P(X≤6) or 1-binomcdf (10, .3, 6) P(X<4) P(X≤3)=P(0)+P(1)+P(2)+P(3) P(X≥6) 1-P(X≤5) or 1-binomcdf (10, .3, 5) TEST GO OVER Question 1) Toss a coin 4 times P(3 heads in 4 tosses) 3 HHHT (1/2) (1/2) *From 5.8, but this is binomial THHH (1/2) (1/2) P(X=x)=nC xp *(1-p)(n-x) 3 HTHH (1/2) (1/2) 4C3 P(3 heads)= 4C3(1/2) (1/2) 6.6 The Poisson Model Success/ Failure X= count of success (occurrances, arrivals) in an interval of time or space Binomial Poisson n= 100 tosses for 1 hour x= count of heads in 0=tcsuenst of heads P(X=30) per unit of time x f(x)=2 P(x= 5 heads every 10 minutes) f(x)=ex X= Poisson Random Variable * on calculator 2nd-->LN-->exponential λ= Average successes per time * on calculator 2nd-->vars-->poissonpdf/cdf P(X=x)= e λx x! ex)At an ER the average amount of arrivals per hour is 2.7. What is P(X=5)? -2.7 P(X=5)= e (2.7)5 5! 0.0803≈8% EX) DBS_6_6 slide 4 a) P(X=0) X= arrivals per 15 minutes time λ= 20 = 5 per 15 minutes 4 -5 0 -5 -5 P(X=0)= e (5) e (1) = e ≈0.0067 0! -1 Or poissonpdf(5,0) b) P(X≥3) λ=5 P(X≥3)= 1-P(X≤2) 1-poissoncdf (5,2) ≈0.8753 *Use the same "formulas" as Binomial, just with poisson* Mean=μ=λ Varience=Ợ =λ Standard Deviation=Ợ=√λ 6.7 Hypergeometric Distribution P(X=x)= (AC x) N-ACn-x) NCn Are counting the number of successes Success/ Failure Dependent ex)Katie has 52 cards, she selects a 5-card hand with out replacement. What is the probability of her having 2 hearts? X= Count of hearts in 5 trials (cards) P(X=2) P(A)= n(A) Hearts Non-Hearts n(S) 13 39 A={X=2} 13C2 39C3 (13C2)(39C3) = 78*9139 = 712842 ≈0.27 52C5 2598960 2598960

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