Study Guides (248,073)
Canada (121,281)
Administration (1,201)
ADM2303 (24)

Stats Memory Aid.doc

2 Pages
415 Views
Unlock Document

Department
Administration
Course
ADM2303
Professor
David Wright
Semester
Fall

Description
Jaclyn Ebert 6221545 Probability time segments. Per – purr – cat – cat eats fish – poisson. Describes the # of events in a given interval of time or space. Marginal (straight): P(x) -λ x P(x∩y) = P(x AND y) = P(x)*P(y) P(x) = e (λ )/x! P(0 OR 1 OR 2 OR 3) = e (1/1 + λ /1! + λ /2! + λ /3!) P(xuy) = P(x OR y) = P(x) + P(y) – P(x AND y) P(>3) = 1 – P(0 OR 1 OR 2 OR 3) Mutually exclusive (disjoint): if the occurrence of one event Mean of poisson = μ = λ do rate*units… ex: rate 0.6, takes 1.7 precludes the occurrence of another event. hours, so 0.6*1.7=1.02 and that is your lamda Then P(x OR y) = P(x) + P(y) SD of poisson = σ = √λ Conditional: P(x|y) = P(x given that y has occurred) P(xANDy) Exponential: Interval of time or space between events. Probability = /P(y) where P(y) > 0 density function. Independent: when P(y|x) = P(y) -λx f(x) = λe where x ≥ 0 or P(x AND y) = P(x)*P(y) vs. Disjoint: when P(x AND y) = 0 or when A and B overlap and Standard Normal Distrubution when both events can occur together. Bayes Theorem: I know P(x|y), and I want to know P(y|x). I have an initial estimate of P(y), I do a test and find that x is true, so I update my initial estimate to get P(y|x). Use tree diagrams, then multiple along the branches to get PAND y). Then to find P(y|x), use the conditional formula, and the denominator will be all the P(y)s μ x added together. z = x-μσ where: x = any point on the horizontal axis, σ = SD of P(D|+) = P(+|D)*P(D) / [ P(+|D)*P(D) + P(+|W)*P(W) ] Where: + = positive test result, D = patient has diabetes, W the normal distribution, μ = population mean. Then use table. = patient is well. Backwards: z = find P in middle of table somewhere, then: Random Variables x = μ + zσ Random variables (RVs) ‘X’ and ‘Y’ represent annual incomes of well E(x) = EV = mean = μ = ∑x*P(x) established unmarried young professional men and women. These two RVs are normally distributed and are statistically independent. Var(x) = σ² = ∑(x – EV)²[P(x)] = ∑[x – x*P(x)]² [P(x)] SD(x) = σ = √Var(x) = √∑[x – x*P(x)]² [P(x)] The mean values of ‘X’ and ‘Y’ are $100K and $80K with standard deviations of $20K and $15K respectively. Properties: Find P(The woman has a higher income than that of the man). E(x+y) = E(x) + E(y) P(W>M) = P(W-M>0) P(W-M>0) μ SD Var Var(x+y) = Var(x) + Var(y) W 80 15 ² => 225 ↓ SD(x+y) = √[Var(x) + Var(y)] M 100 20 ² => 400 + Discrete Probability Distrubutions W-M = -20 25 <= √ 625  Uniform: equal probabilities for all possible outcomes. Ex: roll z = (x-μ)/σ = (0+20)/25 = 0.8 P = 0.7881 from table P(W>M) = 1 – 0.7881 = 0.21 dice. Use random variable table. Combinations: Select x out of n with result good or bad. Correlation Coefficient: n! C = / x!*(n – x!) r = ∑(x-)(y-) s x / ∑(x-)² = ∑x /in (n-1)(s )(s ) √ n-1 Binomial: x out of n, two possible outcomes with n identical and x y independent trials. P of success remains constant from trial to trial. Correlation between Random Variables: Corr(x,y) ∑(x-)(y-)p Sampling with replacement. Known probability of being special. P(x) = /n! x!*(n-x)!)(q )-x (sx)(s y where: n = sample size, x = number of successes, n-x = number Joint Probabilities Meter: $1 Meter: $2 of failures, p = probability of success, q = (1-p) = probability of Fine: $0 0.5 0.4 failure. Fine: $50 0.01 0.09 Mean of binomial = μ =xnp x P(x) Mean Variance xP(x) (x-)²P(x) SD of binomial = σ = √npq = √np(1-p) 1 0.5+0.01=0.51 1(0.51)=0.51 (1-1.49)²(0.51)=0.12245 Normal Approximation: If np>10 and n(1-p)>10, then if P(≥50), 2 0.4+0.09=0.49 2(0.49)=0.98 (2-1.49)²(0.49)=0.1274 =1.49 Variance=0.2498 subtract 0.5 from x in the z calculation, and if P(≤50), add 0.5. ex: μ = 10 for a normal distribution, σ = 2.83 s=√Var=0.5 P(10) = P(9.5≤x≤10.5) = P[(9.5-10)/2.83≤z≤(10.5-10)/2.83] = P(- y P(y) Mean Variance 0.177≤z≤0.177) = 0.140( - μ) N-n yP(y) (y-)²P(y) Finite population: z = /[σ√( / N-1 0 0.5+0.4=0.9 0(0.9)=0 (0-5)²(0.9)=22.5 50 0.01+0.09=0.1 50(0.1)=5 (50-5)²(0.1)=202.5 Poisson: Known arrival rate. Average # of events per segment is λ. =5 Variance=225 # of events is random, occurrence of one does not influence the chance an other. Probability that an event occurs in a given segment s=√Var=15 is the same for all segments. Ex: # customers returning products per Corr(x,y) = [(1-1.49)(0-5)0.5 + (2-1.49)(0-5)0.4 + month, # phone calls per min, # cracks per km of highway. Usually (1-1.49)(50-5)0.01 + (2-1.49)(50-5)0.09] / (0.5*15)Normal opulation Yes Normal Dist Yes Large Dist Sample? The answer will No be approximate c (4 marks You want to examine 20 market-size cod which have achieved a weight of over 4.25 kg without your new feeding program. You take a random sample of 250 market-sized cod. What is the probability that this sample will contain enough cod for your purposes? market-size, with a standard deviation of 1 month. The average weight of market size cod is 3.75 kg with a standard deviation of 0.38 kg, and we wil assume it is Normally distributed. [denived from: Change 2 30 30 or Sample Number 30 Limits For n X /R LCL Normal opulation Yes Normal Dist Yes Large Dist Sample? The answer will No be approximate c (4 marks You want to examine 20 market-size cod which have achieved a weight of over
More Less

Related notes for ADM2303

Log In


OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit