Midterm 1

Midterm 1 - I.05 method for kth%ile Returns item place in...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: I+.05 method: for kth %ile. Returns item place in array of ordered set. K as integer. N+1 method: . K in decimal form. For interpolators (e.g. 3.5), ipart + fpart(larger-smaller). Sample mean is just average Sample variance PDFs: P(a<x<b) = 1) F(x) >= 0 i.e. 2) CDF: F(x) = P(X<x) = . Shows amount of probability to that point. f(x) = some PDF µ=E[x]= Normal Distribution: X:n(µ, σ 2 ) -> P(Z>z)=P(Z<-z) PMFs: 1) F(x) >= 0 for all x 2) =1 Binomial distribution (discrete): Deals with some # of trials with one of two outcomes (success or failure in general) E[x]=n-p V(x)=np(1-p) CDF of: Poisson Distribution (discrete): P(X=x)= X:p(x, ), and x is # of some event in some time/distance λ Poisson is for some defined interval with events occurring at random Assure events are random through the interval, and can be split into subintervals so that: 1) Probability of more than one count in a subinterval is 0 2) Prob. Of one count of subinterval is same for all, and proportional to length 3) Count in each sub. Is independent of other subintervals. Exponential Distribution (continuous): Timed, so not binomial/poisson Time between successive events of a Poisson process PDF: for Good for slow wear-out Has lack of memory E(X) = , V(x)= e.g. X= distance until first crack, and need to find probability of no cracks in 10-mile stretch P(X> 10)= If there are a # of cracks in 10 miles, use Poisson Joint PDFs: F(x,y)-> P(a< x<b, c< Y< d) = Let Y = x+ c E[Y]= E(X+ c), E[x]= µ = E[x]+ E[c]-> µ+c V(Y)= V(X)+ V(c)= V(X)= σ 2 Y=cx 1 , E[Y]=cE[x]=cµ V(Y)=c 2 V(X) = c 2 σ 2 Y=c 1 x 1 +c 2 x 2 +…c n x n E(Y)=c 1 µ 1 +c 2 µ 2 +…+c n µ n V(Y)=c 1 2 σ 1 2 +c 2 2 σ 2 2 … Normal Prob. Plots 1) Rank data in ascending order 2) Plot each pt with i-.5 method (P=(i-0.5)/n E(A-B) = E(A)-E(B) V(A+-B) = V(A)+V(B) e.g. A:n(stuff) B:n(stuff) C:n(stuff) D=A-B-C E(D) = µ A +µ B +µ C V(D) = σ A 2 + σ B 2 + σ C 2 Central Limit Theorem: If underlying population is unknown, X1,x2…xn R.S. (mu, sigma squared), and if x-bar is sampling mean, Then is n(0, 1) as n-> infinity Type I error: Rejecting H 0 when it is true. Type II error: Failing to reject H when it is false....
View Full Document

This note was uploaded on 09/17/2008 for the course IEE 380 taught by Professor Anderson-rowland during the Spring '06 term at ASU.

Page1 / 8

Midterm 1 - I.05 method for kth%ile Returns item place in...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online