This Exam P sample reference tests Continuous Random Variables. This problem asks for the point where a bounded continuous density is largest. Differentiating its polynomial kernel leaves the sole admissible maximizing critical point at 2 minus the square root of 3/2, approximately 0.775, so choice B.
How to solve this Continuous Random Variables question
Setup
Setup
The positive constant multiplying the density does not affect its mode. Shift the variable two units to center the remaining polynomial, then maximize it on the stated support.
g(x)=3(x−2)2−(x−2)4+4
g′(x)=6(x−2)−4(x−2)3
Model
Model
Factor the derivative and retain only stationary points inside the support.
g′(x)=2(x−2)(3−2(x−2)2)=0
x=2,x=2±23
2+23>3
Compute
Compute
Compare the admissible stationary points with the boundary limits. The lower critical point gives the largest density height.
x↓0limf(x)=0,f(2)=185=0.277778
f(2−23)=288125=0.434028
x↑3limf(x)=125=0.416667
Answer
Answer
The density reaches its unique maximum at approximately 0.775.
2−23=0.775255…(B)
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AThe lower support boundary has limiting density zero. Selecting 0.000 confuses an endpoint with the interior maximum.
CAlthough the derivative vanishes at 2, the density there is 5/18 and the derivative changes from negative to positive; this point is a local minimum.
DThe upper-boundary limit is 5/12, below 125/288 at the maximizing critical point. The boundary itself is also excluded from the support.
EThe upper algebraic maximizer is 2 plus the square root of 3/2, or 3.224745, which lies outside the interval. The admissible lower root is listed in choice B.
Original practice · fully worked
Original variant: most likely instrument recovery time
For a positive recovery time t, an instrument has density equal to one sixth of the product of one plus t and an exponential decay factor with exponent negative t/2. The density is zero otherwise. Calculate the mode.
A 0
B 0.5
C 1
D 2
E 10/3
Variant answer in brief
The log-density derivative is the reciprocal of one plus t, minus one half. It vanishes at time 1 and changes from positive to negative there, so the recovery-time mode is 1 and choice C.
Setup
Setup
The normalizing factor is positive and constant, so maximize the density kernel.
h(t)=(1+t)e−t/2,t>0
logh(t)=log(1+t)−2t
Model
Model
Differentiate the log-density; it has the same maximizing locations as the density.
dtdlogh(t)=1+t1−21
1+t1=21⟹t=1
Compute
Compute
The derivative is positive before one and negative after one, proving a unique maximum.
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