Answer:
b. 2.28%.
Explanation:
Mean temperatue (μ) = 1000°F
Standard Deviation (σ) = 50 °F
For any temperature value, X, the z-score is given by:
For X= 900°F

A z-score of -2.0 corresponds to the 2.28-th percentile of a normal distribution. Therefore, the probability that X<900 is:
