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Studentka2010 [4]

3 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

1 answer:

dezoksy [38]3 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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Check the picture below.

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Step-by-step explanation:

Step 1:

The three basic formula needed to solve these questions are:

$sin\theta = \frac{oppositeside}{hypotenuse} , cos\theta = \frac{adjacentside}{hypotenuse}, tan\theta= \frac{opposite side}{adjacent side}.$

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