Question

A normal distribution of data has a mean of 90 and a standard deviation of 18. what is the approximate z-score for the value 64? –3.6 –1.4 1.4 3.6

Answer 1

Answer:

-1.4

Step-by-step explanation:

Answer 2

Answer:

The approximate z-score for the value 64 is -1.4.

Step-by-step explanation:

Since, the z-score or standard score for the value x is,

$z=\frac{x-\mu}{\sigma}$

Where,

$\mu$ is the mean score,

$\sigma$ is the standard deviation,

Here, x = 64,

$\mu=90$

$\sigma=18$

Hence, the z-score for the value 64 is,

$z=\frac{64-90}{18}$

$=\frac{-26}{18}$

$=-1.44444444444\approx -1.4$

Second option is correct.