Answer:
b. 2.28%.
Step-by-step explanation:
Mean temperatue (μ) = 1000°F
Standard Deviation (σ) = 50 °F
For any temperature value, X, the z-score is given by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
For X= 900°F
[tex]z=\frac{900-1000}{50}\\z=-2.0[/tex]
A z-score of -2.0 corresponds to the 2.28-th percentile of a normal distribution. Therefore, the probability that X<900 is:
[tex]P(X<900) = 2.28\%[/tex]