Question

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

Answer 1

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.