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

Question

2 years ago

14

nearest tenth?
0.8

−0.9

0.9

−0.8

2 answers:

Likurg_2 [28] · Answer 1 · 2023-09-25T04:23:43+03:00

6 0

x - mean

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

std. dev.

Here we have:

45 - 61.2

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

21.4

Ksju [112] · Answer 2 · 2023-09-25T05:49:04+03:00

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