To find the [tex]$z$[/tex]-score for the value 64 in a normal distribution with a mean of 90 and a standard deviation of 18, we can use the z-score formula:
[tex]\[
Z = \frac{X - \mu}{\sigma}
\][/tex]
Where:
- [tex]\( Z \)[/tex] is the z-score,
- [tex]\( X \)[/tex] is the value for which we are finding the z-score,
- [tex]\( \mu \)[/tex] (mu) is the mean of the distribution,
- [tex]\( \sigma \)[/tex] (sigma) is the standard deviation of the distribution.
Given the values:
- Mean ([tex]\( \mu \)[/tex]) = 90,
- Standard Deviation ([tex]\( \sigma \)[/tex]) = 18,
- Value ([tex]\( X \)[/tex]) = 64,
we can plug these values into the formula.
[tex]\[
Z = \frac{64 - 90}{18}
\][/tex]
Subtract the mean from the value:
[tex]\[
64 - 90 = -26
\][/tex]
Now, divide by the standard deviation:
[tex]\[
Z = \frac{-26}{18} \approx -1.4444444444444444
\][/tex]
So, the z-score for the value 64 is approximately:
[tex]\[
Z \approx -1.4
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{-1.4}
\][/tex]
