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Tentamen Elementaire Statistiek VEMSTA11, collegejaar 2019-2020Opleiding Vastgoed & Makelaardij
antwoordmodel
Antwoordblad 1
Meerkeuzevragen
(totaal 40 punten, 4 punten per vraag)
Vraag 1 2 3 4 5 6 7 8 9 10
Antwoord B C A D C B B B A D
Eindantwoorden
(totaal 35 punten)
vraag antwoord vraag Antwoord
11a Klasse waarin de mediaan ligt: 13a 10*10*10*10 = 104
klasse 40 - <50 km =10.000 mogelijkheden
(2 punten) (mediaan = 36e waarneming) (3 punten)
11b 13b 5*4*3*2= 120
Sx=20,36 mogelijkheden
(3 punten) (3 punten)
11c 13c 4e positie: 2 mogelijkheden
Variatiecoëfficiënt= (alleen de 0 of 5)
(3 punten) stand.afw/gemiddelde (3 punten) 3e positie: 9 mogelijkheden
2e positie: 8 mogelijkheden
=0,4627 1e positie: 7 mogelijkheden
Dus 2*9*8*7=1008 mogelijkheden
12a 14a y=ax+b →y=2,29187*x+0,0597
E(x) = 55%*40 dus:
(3 punten) = 22 personen (3 punten) leegstand=
2,29187 *afstand + 0,0597
12b P(x = 20) = binomPdf(40,0.55,20) 14b r = 0,9827
= 0,1025 (of 10,25%)
(3 punten) (3 punten)
12c P(x > 20) = 1 - P(x < 19) 14c 4,5 invullen in formule →
= 1 - binomCdf(40, 0.55, 19)
=0,78696 ≈0,7870 (of 78,70%)
10,37 ≈ 10 maanden
(3 punten) (3 punten)
1
, Antwoordblad 2
Vraag Antwoord
15 𝟕×𝟏+𝟔×𝟐+𝟑×𝟐+𝟔×𝟑+𝟑×𝒙
(5 punten) = 𝟔, 𝟓
𝟏𝟏
𝟒𝟑+𝟑𝒙 𝟔,𝟓
→ 𝟏𝟏 = 𝟏
→ 𝟒𝟑 + 𝟑𝒙 = 𝟏𝟏 × 𝟔, 𝟓
→ 𝟑𝒙 = 𝟕𝟏, 𝟓 − 𝟒𝟑
→ 𝟑𝒙 = 𝟐𝟖, 𝟓
𝟐𝟖,𝟓
Dus: 𝒙 = 𝟑 = 𝟗, 𝟓
16a x: aantal km per week
(4 punten) X~N(μ = 250, σ = 75)
P(x < 175) = NormalCDF(0, 175, 250, 75) = 0,1582 (15,82%)
(of m.b.t. vuistregels: 16%)
16b 2 weken achter elkaar: 0,15822 = 0,0250 (2,5%)
(4 punten)
of m.b.b.t. vuistregels: 0,162= 0,0256 (2,6%)
16c Als voor grenswaarde g geldt: P(x > g)= 5%
(5 punten)
→ g = InvNorm(0.95 , 250, 75) =373,364
→ omhoog afronden → vanaf 374 km
16d IQR = Q3 - Q1
(7 punten)
= InvNorm(0.75, 250, 75) - InvNorm(0.25, 250, 75)
= 101,2 km
2
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