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Studentka2010 [4]
2 years ago
14

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
Mathematics
1 answer:
dezoksy [38]2 years ago
7 0

Answer: C.-1.5

Step-by-step explanation:

Given: The burning time of a very large candle is normally distributed with mean(\mu) of 2500 hours and standard deviation(\sigma) of 20 hours.

Let X be a random variable that represent the burning time of a very large candle.

Formula: Z=\dfrac{X-\mu}{\sigma}

For X = 2470

Z=\dfrac{2470-2500}{20}\\\\\Rightarrow\ Z=\dfrac{-30}{20}\\\\\Rightarrow\ Z=\dfrac{-3}{2}\\\\\Rightarrow\ Z=-1.5

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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