A normally distributed assay has mean 50 and standard deviation 4. Using P(Z>1.5)=0.0668, the probability an assay exceeds 56 is most nearly:
- (A)0.0668
- (B)0.134
- (C)0.433
- (D)0.933
Show worked solution
Answer: (A)
The standardized value is (56-50)/4=1.5, so the requested upper-tail probability is 0.0668.
Standardization measures distance from the mean in standard deviations.
The supplied upper-tail area applies directly.
FE Reference Handbook — Probability: Normal Distribution
Why the other choices appear
- (B)Doubling the one-tail probability gives the two-tail probability outside plus-or-minus 1.5.
- (C)Subtracting the upper-tail area from one-half gives the area between mean and z=1.5.
- (D)Using the lower cumulative probability answers X less than 56.