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Summary Statistiek (SQMB1): Business Statistics in Practice

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Summary of "Business Statistics in Practice" by Bowerman, O'Connell, Murphree, for the subject Statistics, aka SQMB1. This achievement rating: 8.0

Institution
Module

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Hfd 1-3 – Inleiding / basiskennis (niet specifiek




onderdeel van tentamen)
- Data Types:
Categorical Numerical Nominal => geen volgorde
(or Qualitative) Ordinal => volgorde
Non-numerical => ja/nee, man/vrouw, etc.
Nominal
Ordinal
Quantitative => waarnemingen vertegenwoordigen een waarde
Numerical Ratio => waarnemingen waarbij ratio/waarde
relevant zijn én er een 0-waarde aanwezig is.
Interval => ratio/waarde niet relevant en geen
0-waarde aanwezig (bijvoorbeeld temperatuur,
want “0 graden” betekent niet “geen
temperatuur”)
- Graphs / Pie Charts (Taartdiagram) / Histograms (Kolomdiagram)
- Pareto chart => kolommen vertegenwoordigen telling van waarnemingen, de lijn




cumulatief
- Mean (Gemiddelde)
- Median (Mediaan) => middelste waarneming in serie gesorteerde waarnemingen (1,
3, 3, 3, 4). Bij even aantal: gemiddelde van middelste twee waarnemingen.
- Mode (Modus) => meest voorkomende waarneming (1, 3, 3, 3, 4)
- Variance (Variantie) => som van gekwadrateerd verschil tussen waarnemingen en
gemiddelde
- Standard Deviation (Standaard Deviatie) => wortel van de variantie
- Coefficient of Variaton (Variatiecoëfficiënt) => relatieve spreidingsmaat, de
spreiding gemeten ten opzichte van het gemiddelde. Formule: Standaard Deviatie /
Gemiddelde.
- Covariance (Covariantie) => in welke mate twee toevalsvariabelen (lineair) met
elkaar samenhangen. Covariantie geeft aan of, en indirect in welke mate, de waarden
van de ene variabele toe- dan wel afnemen bij toenemende waarden van de andere.

- Expected Value => mogelijke uitkomsten vermenigvuldigd met hun kans
(voorbeeld van de dobbelsteen: 1/6 * 1 + 1/6 * 2 + …. + 1/6 * 6 = 3,5)

Steekproe Populatie
f

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264676428

, Gemiddelde
Standaardafwijking
Variatiecoëfficiënt



Z-score (relatieve locatie
waarneming binnen data
set)
Proportion p P


Hfd 4 – Probability
Kans (P) waarde tussen 0 en 1

Experiment elk proces dat goed-gedefinieerde uitkomsten oplevert

Sample space set van alle mogelijke uitkomsten (dobbelsteen: 1, 2, 3, 4, 5, 6)



Zes basisregels

1. Complement of an Event P(AC) = 1 – P(A)

2. Union of Two Events P(A  B) → A óf B, totaal oppervlak van twee cirkels
Mutually exclusive P(A  B) = P(A) + P(A) → oftewel, cirkels overlappen niet,
som van P’s
Not mutually exclusive P(A  B) = P(A) + P(A) - P(A B) → oftewel,
cirkels overlappen (deels),
en dus moet de dubbeltelling er uit gehaald worden.

3. Intersection of Two Events P(A B) → A én B
Multiplication Law P(A B) = P(A|B) * P(B) = P(B|A) * P(A)
Reden voor gebruik A|B is de eventuele afhankelijkheid,
zie regel 6.
Indien onafhankelijk, dan volstaat P(A) * P(B), de
productregel.

4. Mutually Exclusive Events P(A B) = 0 → geen overlappende cirkels

5. Conditional Probability P(A|B) → kans op A wanneer (nadat) B is
opgetreden



6. Multiplication P(A B) = P(A|B) * P(B) = P(B|A) * P(A)
Independent events if: P(A|B) = P(A) or P(B|A) = P(B) → oftewel, indien
de kans op A niet
verandert als gevolg van kans op A wanneer B optreedt
en vice versa.

Counting Rules
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264676428

, - Multistep experiments (n1) * (n2) * … * (nk)
→ voorbeeld Markley Oil (4) en Collings Minning (2) = 8
combi’s (4*2)

- Combinations N! → voorbeeld 5! = 5 * 4 * 3 * 2 * 1 = 120
“zonder terugleggen”, en “volgorde is van belang”

- Select n objects from set N


- Count outcomes of “select n from N”




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264676428

, Hfd 4 – Probability


1. The card game of Euchre employs a deck that consists of all four



of each of the aces, kings, queens, jacks, tens, and nines (one of each suit—clubs, diamonds, spades, and hearts).



(a)
Find the probability that a randomly selected card from a Euchre deck is a jack (J).

4 1
P(J) = =
24 6

(b)
Find the probability that a randomly selected card from a Euchre deck is a spade (S).

6 1
P(S) = =
24 4

(c)
Find the probability that a randomly selected card from a Euchre deck is a jack or an ace (A).

1 1 1
P(J U A) = + – 0 =
6 6 3

(d)
Find the probability that a randomly selected card from a Euchre deck is a jack or a spade.

1 1 1 3
P(J U S) = + – =
6 4 24 8



(e) Are the events J and A mutually exclusive? J and S? Why or why not?


Yes for J and A, No for J and S; A jack and an ace
cannot occur in a single draw, where a jack and a spade
can occur simultaneously.


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264676428

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Uploaded on
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Number of pages
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Written in
2014/2015
Type
SUMMARY

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Showing 7 of 11 reviews
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5 year ago

Bye, Kayleigh. Too bad you're not satisfied. Others rate this summary with 4 stars. And I got an 8.0 as an exam grade with this (and with livestock practice).

5 year ago

5 year ago

Thank you for your nice review, Caroline. Hope for you a nice study result.

6 year ago

6 year ago

Too bad you're not satisfied, Jamal. What are you missing in the summary? I also got 8.0 from this base in combination with fiks practice.

6 year ago

6 year ago

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