Summary
Samenvatting alle opdrachten uit de meetings uitgewerkt
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de opdrachten uit de meetings zijn hierin uitgewerkt
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By: JaleesaJocelyn • 5 year ago
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Opdrachten group meeting 21.9 employee application data
a) werknemers als individu
b)
Labels Quantitative variable Categorical variable
- Employee ID - Salary - Department
number - Number of years - Education
- Last name with the company
- First name - Age
- Middle initial
c)
ID- Last First Middle Education Department Age Years with Salary
number name name initial company
101 Visser Suzan De Accountancy Financiën 41 17
102 Baker Sophie Marketing Verkoop 44 13
103 Vaart Jack Van der Corporate Administratie 29 4
governance
104 Smit Lisa Commerciele Aankoop 32 8
economie
105 Wolf Eric de marketing Directie 52 23
1.11 survey of students
a, b)
- Age: quantitative, je kunt het tellen16 – 23 jaar (?)
- Like to sing: categoricalje deelt het in in ja of nee
- Play musical instrument: categorialje deelt het in op not at all, a little, pretty well
- Spend on food: quantitative: je kunt het bijhouden$0-$100 (?)
- Height: quantitative: je kunt het meten2-9 feet
1.44 describe the IQ scores
Shape: de vorm heeft en piek bij IQ-scores van 120-130.
Spread: de laagste was 72, de hoogste was 136.
Midpoint: duidelijk boven de 100, er zijn veel meer hogere scores dan lagere scores
Outliers: 136?
Diagram:
7 l 2479
8 l 69
9 l 01336778
10 l 0022333344555666777789
11 l 0000111112222333444455688999
12 l 003344677888
13 l 026
,1.45 describe the self-concept scores
2| 018
3| 05679
4| 023446799
5| 1111223344444556668899
6| 0000123334444455666677777899
7| 0000111223
8| 0
Je ziet dat 20,21 en 80 outliers zijn.
Center rond 60, rest tussen 35-73
Skewed on the left.
1.61- gosset’s data on double stout sales
a) the mean is (94+112+78+68+46+111+66+140+428+190+118+24)/12=122,92
b) the median is:
24-46-66-68-78-94-111-112-118-140-190-428
N is gelijk, dus (12+1)/2=6,5 dus median tussen 6 en 7e cijfer
De median is: (94+111)/2=102,5
c) De median, want er is een outlier, namelijk de 428 van Londen.
1.62- measures of spread for the double stout data
24-46-66-68-78-94-111-112-118-140-190-428
a) standard deviatie:
variance, s2= ((24-122,92)2+(46-122,92)2+(66-122,92)2+(68-122,92)2+(78-122,92)2+(94-
122,92)2+(111-122,92)2+(112-122,92)2+(118-122,92)2+(140-122,92)2+(190-122,92)2+(428-
122,92)2)/(12-1)
s2=32437540,36
s=5695,40
b) quartiles:
24-46-66-68-78-94-111-112-118-140-190-428
mean=122,92
Q1= 67
Q2= 102,5
Q3= 129
Q4=428
c)
1.63- are there outliers in the double stout data?
a) IQR = Q3-Q1
= 129-67
= 62
b) 1,5 * 62 = 93
outliers zijn dus onder -26 of boven 222. Londen is dus een outlier.
c) boxplot (papier los)
d) modified boxplot
e) stemplot
f) de stemplot is niet handig, want de data liggen te veel uit elkaar.
De boxplot is ook niet erg handig, want er is een enorme uitschieter.
De modified boxplot is het best.
, 1.91-create a data set
12-13-13-14-14-15-16-18-22-25-42
Mean: (12+13+13+14+14+15+16+18+22+25+42)/11=18,55
Median=15
Q3=20
1.93- a standard deviation contest
a) 5-5-5-5
b) 0,0,20,20
c)bij a kan elk cijfer gekozen worden, als het maar vier keer hetzelfde is. Bij b is er maar 1
optie, anders wordt de deviatie kleiner.
1.121- assign more grades
a) scores op een rij: 54-55-62-64-71-73-76-80-93-98
b)
x−μ
z=
σ
μ=72
σ=10
x−72
vb. berekening F-1,645 -1,645 =
10
x = -1,645 * 10 + 72
x = 55,55, dus F is onder 55,55 want lager kan niet.
c)
1.123-use a different range for the uniform distribution
a) height: 1/5=0,2
b) 0,2
c) 2/5=0,4
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