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Samenvatting Statistiek hoorcollege 2
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Statistiek HC2Voorbeeld 1 – normale verdeling
Wat is de kans dat een aselect gekozen persoon een IQ heeft dat hoger is dan 105 als
IQ in de populatie normaal verdeeld is met µ = 100 en σ = 15?
X = IQ
Z = X – µ = 105 – 100 = 0,33 (grenswaarde)
σ 15
Tabel C.1 -> Z = 0,33 -> oppervlakte onder de grafiek / area beyond = 0,3707
P(Z>0,33) = kans dat Z-waarde groter is dan 0,33 = 0,3707
Dus kans dat aselect gekozen persoon IQ heeft dat hoger is dan 105 is 0,3707 (37,07%)
Waar ligt de grenswaarde in de populatie van de hoogste 10% scores?
Tabel C.2 : alleen positieve Z-waardes
Geeft het grootste gebied weer -> de linkerkant
Grootste 10% -> 0.10
Larger area -> 0.90 -> tabel C.2
Z-waarde (grenswaarde) = 1,2816
X = µ + Z x σ = 100 + 1,2816 x 15 = 119,2
Dus de grenswaarde voor de hoogste 10% scores is een IQ van 119,2
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