A standardized exam's scores are normally distributed. In a recent year, the mean test score was 20.7 and the standard deviation was 5.4 . The test scores of four students selected at random are 14, 22, 8, and 36 . Find the $z$-scores that correspond to each value and determine whether anv of the values are unusual.

The $z$-score for 14 is $⨆$.
(Round to two decimal places as needed.)
The $z$-score for 22 is $◻$.
(Round to two decimal places as needed.)
The $z$-score for 8 is $◻$.
(Round to two decimal places as needed.)
The $z$-score for 36 is $◻$.

Step 1 ：Calculate the z-score for each student using the formula $z=\frac{X-\mu }{\sigma }$, where $X$ is the student's score, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Step 2 ：For the student with a score of 14, the z-score is calculated as $z=\frac{14-20.7}{5.4}=-1.24$ (rounded to two decimal places).

Step 3 ：For the student with a score of 22, the z-score is calculated as $z=\frac{22-20.7}{5.4}=0.24$ (rounded to two decimal places).

Step 4 ：For the student with a score of 8, the z-score is calculated as $z=\frac{8-20.7}{5.4}=-2.35$ (rounded to two decimal places).

Step 5 ：For the student with a score of 36, the z-score is calculated as $z=\frac{36-20.7}{5.4}=2.83$ (rounded to two decimal places).

Step 6 ：In a normal distribution, any z-score greater than 3 or less than -3 is considered unusual. Therefore, none of the scores are unusual, which is the final answer $\boxed{{\displaystyle \text{None of the scores are unusual}}}$.