criley225
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will give brainliest. Given a mean of 61.2 and a standard deviation of 21.4, what is the z-score of the value 45 rounded to the nearest tenth?


0.8


−0.9


0.9


−0.8

Respuesta :

altavistard

x - mean

z-score = ------------------

std. dev.


Here we have:


45 - 61.2

z-score = ------------------ = -0.757

21.4

Answer:

-0.076 is answer.

Step-by-step explanation:

Let X be the corresponding random variable.

Given that X is Normal with mean = 61.2 and std deviation = 21.4

We know that x to z conversion is as follows

[tex]z=\frac{x-mean}{\sigma  } =\frac{x-61,2}{21.4}[/tex]

When x =45 we find correspond z score by substituting for z

X = [tex]\frac{45-61.2}{21.4} =0.0757\\=-0.076[/tex]

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