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Leokris [45]
2 years ago
14

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
Mathematics
2 answers:
Likurg_2 [28]2 years ago
6 0

x - mean

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

std. dev.


Here we have:


45 - 61.2

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

21.4

Ksju [112]2 years ago
3 0

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

z=\frac{x-mean}{\sigma  } =\frac{x-61,2}{21.4}

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

X = \frac{45-61.2}{21.4} =0.0757\\=-0.076

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