Answer :
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]
[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]
