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The burning time of a very large candle is normally distributed with mean of 2500 hours and standard deviation of 20 hours. Find the z-score thay correspondes to a lifespan of 2470 hours. A. 2.5 B.1.5 C.-1.5 D-2.4

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JeanaShupp

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.

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