Original practice · fully worked
Original variant: variance after unit conversion and calibration A chemical reading X has variance 16 in laboratory units. The published reading is Y=2.5X+7. What is the variance of Y?
C B A D E Variant answer in brief The additive calibration does not affect variance, while multiplication by 2.5 contributes a factor of 6.25, giving 100.
Setup
Setup The published reading is the affine transformation Y=2.5X+7 of a reading whose variance is 16.
Var ( X ) = 16 \operatorname{Var}(X)=16 Var ( X ) = 16 Model
Model The additive calibration 7 does not affect dispersion, while the multiplicative coefficient 2.5 enters the variance as its square.
Var ( Y ) = ( 2.5 ) 2 Var ( X ) \operatorname{Var}(Y)=(2.5)^2\operatorname{Var}(X) Var ( Y ) = ( 2.5 ) 2 Var ( X ) Compute
Compute Since 2.5 squared is 6.25, the transformed variance is 6.25 × 16, or 100.
Var ( Y ) = 6.25 ( 16 ) = 100 \operatorname{Var}(Y)=6.25(16)=100 Var ( Y ) = 6.25 ( 16 ) = 100 Answer
Answer Thus Var(Y)=100 squared laboratory units, corresponding to choice A.
100 units 2 (A) \boxed{100\ \text{units}^2\quad\text{(A)}} 100 units 2 (A) ✓ Variant verification record
Computation passed
Computed value 100
Main method Compute the variant by the affine-transformation variance rule.
Check value 100
Check method The original standard deviation is 4; the transformed standard deviation is 10, whose square is 100.
Answer key A
Agreement Pass Recorded 2026-07-31T00:00:00Z
