Answer: C.-1.5
Step-by-step explanation:
Given: The burning time of a very large candle is normally distributed with mean[tex](\mu)[/tex] of 2500 hours and standard deviation[tex](\sigma)[/tex] of 20 hours.
Let X be a random variable that represent the burning time of a very large candle.
Formula: [tex]Z=\dfrac{X-\mu}{\sigma}[/tex]
For X = 2470
[tex]Z=\dfrac{2470-2500}{20}\\\\\Rightarrow\ Z=\dfrac{-30}{20}\\\\\Rightarrow\ Z=\dfrac{-3}{2}\\\\\Rightarrow\ Z=-1.5[/tex]
So, the z-score they corresponds to a lifespan of 2470 hours. =-1.5
Hence, the correct option is C.-1.5.
