Question

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

Answer 1

x - mean

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

std. dev.

Here we have:

45 - 61.2

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

21.4

Answer 2

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