This Exam P sample reference tests Covariance and Correlation. Writing the common marginal variance as v and the covariance as c gives v+c=5 and 5v-4c=16. Solving yields v=4 and c=1, so choice D is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis changes 25-9c=16 to 25+9c=16 during substitution, producing c=-1.
BThis fails to square the coefficient 2 on the second marginal variance, using 3v-4c=16. Together with v+c=5, it gives c=-1/7, nearest to -0.1.
CThis is a decimal-place error after the correct elimination: dividing the residual 9 by 90 instead of by 9 gives 0.1.
EThis drops the covariance cross term from the second variance, obtaining 5v=16 and then c=5-3.2=1.8, whose nearest option is 2.0.
Original practice · fully worked
Original variant: covariance from a calibrated difference score
Two centered calibration readings U and V have variances 9 and 16. A diagnostic score is D=2U-V, and repeated trials show Var(D)=36. Calculate Cov(U,V).
A -4
B 1/3
C 4
D 8
E 16
Variant answer in brief
Expanding Var(2U-V) gives 4(9)+16-4Cov(U,V)=36. Therefore Cov(U,V)=4, so choice C is correct.
Setup
Setup
Let c denote the covariance of the two calibration readings.
c=Cov(U,V)
Model
Model
Expand the variance of the weighted difference, including its cross term.
Var(2U−V)=4Var(U)+Var(V)−4c
Compute
Compute
Insert the three supplied variances and isolate c.
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