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

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=(X-\mu)/(\sigma)`

For X = 2470

`Z=(2470-2500)/(20)\\\\\Rightarrow\ Z=(-30)/(20)\\\\\Rightarrow\ Z=(-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.