100% tevredenheidsgarantie Direct beschikbaar na je betaling Lees online óf als PDF Geen vaste maandelijkse kosten 4.2 TrustPilot
logo-home
Tentamen (uitwerkingen)

MA116 Midterm 1 | Questions And Answers | 100% Correct Answers Rated A+

Beoordeling
-
Verkocht
-
Pagina's
6
Cijfer
A+
Geüpload op
03-07-2024
Geschreven in
2023/2024

Skewed right- 1. Long ____ tail 2. Mean and median relationship 3. Box and whiskers plot - where is the median? - ANS1. Right 2. Mean>Median 3. Toward right of box Skewed left- 1. Long ____ tail 2. Mean and median relationship 3. Box and whiskers plot - where is the median? - ANS1. Left 2. Mean<Median 3. Toward left of box Independence rules (3) - ANS1. P(A∩B) = P(A) * P(B) 2. P (A∩B/B) = P(A) 3. P (A∩B/A) = P(B) True or false- Disjoint events are independent - ANSFALSE- not always independent Permutations rule- does order matter? replacement? - ANSYes. Yes. Permutations and combinations rule: - N= - n= - ANS-N= Total elements -n= # in group Permutations formula - ANS(N!) / (N-n)! Combinations rule- does order matter? replacement? - ANSNo. No Combinations formula - ANS(N!) / n! (N-n)! Random variables (Mean, Standard Deviation) - ANS(sample mean, σ/√n) PDF- total area under curve= - ANS1 What is the probability that a continuous random variable is equal to any single point? - ANS0 General normal v. standard normal distributions- relationship between mean, median, and mode - ANSGeneral normal: mean=median=mode Standard normal: mean=median=mode=0 Qualities of a binomial distribution (4) - ANS-N IDENTICAL trials -2 possible mutually exclusive outcomes -P(success) remains the same form trial to trial -Independent trials Binomial distribution - ANSX~Bin (n,p) Bernoulli distrbution - ANSBin(1,p) = Ber(p) Binomial distribution- mean - ANSnp Binomial distribution- standard deviation - ANS√npq Binomial distribution- shape - ANS- p<.5= right skewed -p=.5= symmetric -p>.5= left skewed Normal distribution- if z-score is greater or less than +- 3.49 - ANSP= 0 or 1 depending on problem Percentiles: formula - ANSP(Z≤zp)=p p= - ANSPercentile (ex. 80th percentile) How to get zp - ANSPlug .8 inside Z-table. Get Z-score. Then transform. Normal approximation to the binomial- what must you check? - ANSnpq>10 Normal approximation to the binomial- What is very important to remember? - ANSThe ±.5 issue Normal approximation to the binomial- What letter do we use? - ANSY Normal approximation to the binomial- P(X=A) - ANSP(A-.5<Y<A+.5) Normal approximation to the binomial- P(X≤A) - ANSP(Y<A+.5) Normal approximation to the binomial- P(X<A) - ANSP(Y<A-.5) Normal approximation to the binomial- P(X≥A) - ANSP(Y>A-.5) Normal approximation to the binomial- P(X>A) - ANSP(Y>A+.5) Normal approximation to the binomial- P(A≤X≤B) - ANSP(A-.5<X<B+.5) Normal approximation to the binomial- P(A<X<B) - ANSP(A+.5<X<B-.5) Normal approximation to the binomial- What is unusual? - ANS<.05 Chi squared properties (2) - ANS-Skewed right -Depends on degrees of freedom Chi squared distribution- relationship between degrees of freedom and symmtery - ANS↑ DF, ↑ symmetry T-distribution properties (3) - ANS-Symmetric, but fatter -Depends on degrees of freedom -Similar to Z when N>30 Z-transform for sampling mean (point estimation) - ANS(x-µ) / (σ/√n) Sampling distribution for mean - ANSX~N (µ,σ) Conditions for point estimation for proportion - ANS-SRS -NPQ>10 -Independence= n<N.05 Point estimation for proportion- test statistics - ANS(Phat - Po) / √PoQo/n Confidence interval in English - ANS-We are ___% confident that the true unknown value of (µ, p, etc) is between (___,___). -If we construct 100 intervals, 95 percent of them will contain (µ, p, etc). Confidence interval- variance/sd-- Relationship between N and CI - ANS↑N, ↓Width of CI Type 1 error - ANSHo is rejected when it is true Type 2 error - ANSHo is NOT rejected when it is false

Meer zien Lees minder
Instelling
MA116
Vak
MA116









Oeps! We kunnen je document nu niet laden. Probeer het nog eens of neem contact op met support.

Geschreven voor

Instelling
MA116
Vak
MA116

Documentinformatie

Geüpload op
3 juli 2024
Aantal pagina's
6
Geschreven in
2023/2024
Type
Tentamen (uitwerkingen)
Bevat
Vragen en antwoorden

Onderwerpen

€10,49
Krijg toegang tot het volledige document:

100% tevredenheidsgarantie
Direct beschikbaar na je betaling
Lees online óf als PDF
Geen vaste maandelijkse kosten

Maak kennis met de verkoper
Seller avatar
Kiranga
5,0
(1)

Maak kennis met de verkoper

Seller avatar
Kiranga John Hopkins University
Volgen Je moet ingelogd zijn om studenten of vakken te kunnen volgen
Verkocht
1
Lid sinds
1 jaar
Aantal volgers
0
Documenten
122
Laatst verkocht
1 jaar geleden

Here we offer revised study materials to elevate your educational outcomes. We have verified learning materials (Research, Exams Questions and answers, Assignments, notes etc) for different courses guaranteed to boost your academic results. We are dedicated to offering you the best services and you are encouraged to inquire further assistance from our end if need be. Having a wide knowledge in Nursing, trust us to take care of your Academic materials and your remaining duty will just be to Excel. Remember to give us a review, it is key for us to understand our clients satisfaction. We highly appreciate refferals given to us. Also clients who always come back for more of the study content we offer are extremely valued. All the best

Lees meer Lees minder
5,0

1 beoordelingen

5
1
4
0
3
0
2
0
1
0

Recent door jou bekeken

Waarom studenten kiezen voor Stuvia

Gemaakt door medestudenten, geverifieerd door reviews

Kwaliteit die je kunt vertrouwen: geschreven door studenten die slaagden en beoordeeld door anderen die dit document gebruikten.

Niet tevreden? Kies een ander document

Geen zorgen! Je kunt voor hetzelfde geld direct een ander document kiezen dat beter past bij wat je zoekt.

Betaal zoals je wilt, start meteen met leren

Geen abonnement, geen verplichtingen. Betaal zoals je gewend bent via Bancontact, iDeal of creditcard en download je PDF-document meteen.

Student with book image

“Gekocht, gedownload en geslaagd. Zo eenvoudig kan het zijn.”

Alisha Student

Veelgestelde vragen