Sure! To find the [tex]$z$[/tex]-score for a value in a normal distribution, use the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where:
- [tex]\( X \)[/tex] is the value you are interested in (64 in this case),
- [tex]\( \mu \)[/tex] is the mean of the distribution (90 in this case),
- [tex]\( \sigma \)[/tex] is the standard deviation (18 in this case).
Let's apply the values to the formula step-by-step:
1. Subtract the mean from the value:
[tex]\[ 64 - 90 = -26 \][/tex]
2. Divide this result by the standard deviation:
[tex]\[ \frac{-26}{18} = -1.4444444444444444 \][/tex]
So, the [tex]$z$[/tex]-score is approximately [tex]\(-1.4\)[/tex].
From the given options:
- [tex]$-3.6$[/tex]
- [tex]$-1.4$[/tex]
- [tex]$1.4$[/tex]
- [tex]$3.6$[/tex]
The approximate [tex]$z$[/tex]-score for the value 64 is [tex]\(-1.4\)[/tex].
