To calculate the approximate \( z \)-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. The formula for the \( z \)-score is given by:
[tex]\[
z = \frac{{X - \mu}}{{\sigma}}
\][/tex]
where:
- \( X \) is the value in question (64 in this case),
- \( \mu \) is the mean of the distribution (90),
- \( \sigma \) is the standard deviation of the distribution (18).
Let's substitute the values into the formula and solve for \( z \):
[tex]\[
z = \frac{{64 - 90}}{{18}}
\][/tex]
First, calculate the numerator:
[tex]\[
64 - 90 = -26
\][/tex]
Now, divide by the standard deviation:
[tex]\[
z = \frac{{-26}}{{18}}
\][/tex]
Performing the division gives:
[tex]\[
z \approx -1.444
\][/tex]
We can approximate \( z \) as -1.4 when rounded to one decimal place. Therefore, the approximate \( z \)-score for the value 64 is:
[tex]\[
\boxed{-1.4}
\][/tex]
