To find the correct expression for the [tex]\( z \)[/tex]-score of Din's test score, we need to use the formula for the [tex]\( z \)[/tex]-score in a normally distributed data set. The formula is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where:
- [tex]\( X \)[/tex] is the value of the data point (Din's test score),
- [tex]\( \mu \)[/tex] is the mean of the data set, and
- [tex]\( \sigma \)[/tex] is the standard deviation.
Given:
- Din's test score [tex]\( X = 92 \)[/tex],
- Mean [tex]\( \mu = 83 \)[/tex],
- Standard deviation [tex]\( \sigma = 5 \)[/tex],
We substitute these values into the formula:
[tex]\[ z = \frac{92 - 83}{5} \][/tex]
So, the correct expression Din would write to find the [tex]\( z \)[/tex]-score of her test score is:
[tex]\[ z = \frac{92 - 83}{5} \][/tex]
From the given options, this matches the second answer choice:
[tex]\[ z = \frac{83 - 92}{5} \][/tex]
is incorrect as it reverses the numerator,
[tex]\[ 2-\frac{92-83}{5} \][/tex]
is incorrect because it adds an unnecessary subtraction,
[tex]\[ z-\frac{5-83}{92} \][/tex]
is incorrect due to the incorrect placement of values and operations.
Therefore, the correct answer choice is:
[tex]\[ z = \frac{92 - 83}{5} \][/tex]
