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 $\bigsqcup$.
(Round to two decimal places as needed.)
The $z$-score for 22 is $\square$.
(Round to two decimal places as needed.)
The $z$-score for 8 is $\square$.
(Round to two decimal places as needed.)
The $z$-score for 36 is $\square$.
Answer
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 \(\boxed{\text{None of the scores are unusual}}\).
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 \(\boxed{\text{None of the scores are unusual}}\).
