This Exam P sample reference tests Exponential Distribution. Normalization identifies an exponential rate of 0.004. Its median is 173.29, below the benefit cap, so official choice C applies.
How to solve this Exponential Distribution question
Setup
Setup
Let X be exponential with rate 0.004 and let the reported benefit be the smaller of X and 250. First determine whether the uncapped median reaches the cap.
X∼Exp(0.004)
B=min(X,250)
Model
Model
Below 250 the capped and uncapped cumulative probabilities coincide, so solve the ordinary exponential median equation and then verify the resulting value is below 250.
P(B≤m)=P(X≤m)=1−e−0.004mfor m<250
Compute
Compute
Setting the cumulative probability to one half gives 173.2868. Because this is less than 250, capping does not alter the median.
1−e−0.004m=0.5
m=−0.004log(0.5)=173.2868<250
Answer
Answer
The median capped benefit rounds to 173, which is choice C.
173(C)
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EThe value 250 is the cap. It would be a median only if the uncapped median were at least 250, whereas it is 173.2868.
Original practice · fully worked
Original variant: median of a capped Weibull lifetime
A filter lifetime T has survival function exp(-(t/100)²) for t above zero. A dashboard reports B=min(T,120) hours. Determine the median displayed lifetime.
A 69.31 hours
B 83.26 hours
C 100.00 hours
D 117.74 hours
E 120.00 hours
Variant answer in brief
Solving exp(-(m/100)²)=0.5 gives m=100 √(log 2)=83.26, which is below the display cap.
Setup
Setup
The displayed lifetime is the smaller of the Weibull lifetime T and 120 hours. Check the ordinary Weibull median against that cap.
ST(t)=e−(t/100)2
B=min(T,120)
Model
Model
At a median the survival probability is one half. Invert the squared Weibull exponent, then compare the result with 120.
ST(m)=0.5when m<120
Compute
Compute
The uncapped median is 100 times the square root of log 2, or 83.2555 hours, which is below the dashboard cap.
m=100−log(0.5)=83.2555<120
Answer
Answer
Thus the median displayed lifetime is 83.26 hours, corresponding to choice B.
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