standard deviation was 316 . The test scores of four students selected at random are 1950, 1290, 2250, and 1410 . Find the z-scores that correspond to each value and determine whether any of the values are unusual. The z-score for 1950 is (Round to two decimal places as needed.)

To find the z-score for a value, you can use the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \( X \) is the value, \( \mu \) is the mean of the data set, and \( \sigma \) is the standard deviation. In this case, we need to first calculate the mean of the test scores of the four students. The test scores are 1950, 1290, 2250, and 1410. 1. **Calculate the mean (\( \mu \))**: \[ \mu = \frac{(1950 + 1290 + 2250 + 1410)}{4} = \frac{6900}{4} = 1725 \] 2. **Using the standard deviation (\( \sigma \))**: Given \( \sigma = 316 \). 3. **Calculate the z-score for each value**: - For \( X = 1950 \): \[ z = \frac{(1950 - 1725)}{316} = \frac{225}{316} \approx 0.71 \] - For \( X = 1290 \): \[ z = \frac{(1290 - 1725)}{316} = \frac{-435}{316} \approx -1.38 \] - For \( X = 2250 \): \[ z = \frac{(2250 - 1725)}{316} = \frac{525}{316} \approx 1.66 \] - For \( X = 1410 \): \[ z = \frac{(1410 - 1725)}{316} = \frac{-315}{316} \approx -0.99 \] Now, rounding to two decimal places, the z-scores are: - 1950: \( z \approx 0.71 \) - 1290: \( z \approx -1.38 \) - 2250: \( z \approx 1.66 \) - 1410: \( z \approx -0.99 \) To determine if any values are unusual, typically z-scores above 2 or below -2 are considered unusual. Here, none of the z-scores exceed these limits, so all values are within the usual range.