Question

a standardized exams scores are normally distributed. in a recent year, the mean test score was 20.8 and the standard deviation was 5.7. the test scores of four students selected at random are 14, 21, 9, and 37. find the z - scores that correspond to each value and determine whether any of the values are unusual.
the z - score for 14 is - 1.19. (round to two decimal places as needed.)
the z - score for 21 is 0.04. (round to two decimal places as needed.)
the z - score for 9 is . (round to two decimal places as needed.)

Explanation:

Step1: Recall z - score formula

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the data - point, $\mu$ is the mean, and $\sigma$ is the standard deviation. Here, $\mu = 20.8$ and $\sigma=5.7$.

Step2: Calculate z - score for $x = 9$

Substitute $x = 9$, $\mu = 20.8$, and $\sigma = 5.7$ into the formula: $z=\frac{9 - 20.8}{5.7}=\frac{-11.8}{5.7}\approx - 2.07$.

Step3: Determine if values are unusual

A value is considered unusual if its z - score is less than - 2 or greater than 2. The z - score for 9 is approximately - 2.07, so the score of 9 is unusual. The z - score for 14 is - 1.19 (not unusual), the z - score for 21 is 0.04 (not unusual). For the fourth value, if we calculate the z - score for $x = 37$: $z=\frac{37-20.8}{5.7}=\frac{16.2}{5.7}\approx2.84$. Since $2.84>2$, the score of 37 is unusual.

Answer:

The z - score for 9 is approximately - 2.07. The unusual values are 9 and 37.