This Exam P sample reference tests Mixture Distributions. For a positive median m, the mixture CDF is 0.20+0.80 Φ((m-1000)/400). Setting it to 0.50 requires a conditional normal percentile of 0.375, giving m=872.54 and choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AUsing the no-loss probability 0.20 itself as the percentile inside the normal component gives 1000+400 Phi⁻¹(0.20)=663.35, nearest 663.
BUsing 0.50-0.20=0.30 as the conditional percentile without dividing by the positive-loss probability gives 790.24, nearest 790.
DA value near 994 corresponds to a conditional percentile near 0.494 and makes the mixture CDF about 0.595. It rounds toward the positive-loss median without performing the required probability rescaling.
EThe value 1,000 is the median of the normal component alone. The 20% atom at zero pulls the unconditional median below it.
Original practice · fully worked
Original variant: median of two overlapping score models
A scoring engine is selected at random. Engine A is selected with probability 0.40 and produces a score uniformly distributed from 0 to 1. Engine B is selected otherwise and has conditional distribution function F_B(x)=x² on that same interval. Calculate the median of the unconditional score.
A 0.5000
B 0.6000
C 0.6243
D 0.6385
E 0.7071
Variant answer in brief
The mixture CDF is 0.40x+0.60x squared. Setting it to one half gives the positive root x=(-0.40+√(1.36))/1.20=0.6385, choice D.
Setup
Setup
Let M be the selected engine and X the resulting score. Average the two conditional CDFs.
FX(x)=0.40FX∣M=A(x)+0.60FX∣M=B(x)
FX(x)=0.40x+0.60x2,0≤x≤1
Model
Model
Set the unconditional CDF equal to one half and form the quadratic median equation.
0.40m+0.60m2=0.50
0.60m2+0.40m−0.50=0
Compute
Compute
Use the positive root because the score lies between zero and one.
m=2(0.60)−0.40+0.402+4(0.60)(0.50)
m=0.6384919825
Answer
Answer
The unconditional score median is approximately 0.6385.
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