Question

A standardized exam's scores are normally distributed. In a recent year, the mean test score was 1467 and the standard deviation was 315. The test scores of four students selected at random are 1880, 1220, 2170, and 1350. Find the z-scores that correspond to each value and determine whether any of the values are unusual

Answer 1

To find the z-scores for each value, use the formula z=(x-mean)/standard deviation. The z-scores for the given test scores are 1.32, -0.78, 2.22, and -0.37. Based on the z-scores, the scores of 1880 and 2170 are considered unusual.

Step-by-step explanation:

To find the z-scores for each value, we can use the formula: z = (x - mean) / standard deviation. For the given test scores of 1880, 1220, 2170, and 1350, we can calculate the z-scores as follows:

For 1880: z = (1880 - 1467) / 315 ≈ 1.32

For 1220: z = (1220 - 1467) / 315 ≈ -0.78

For 2170: z = (2170 - 1467) / 315 ≈ 2.22

For 1350: z = (1350 - 1467) / 315 ≈ -0.37

To determine whether any of the values are unusual, we can consider the range within which most values fall.

Typically, z-scores between -2 and 2 are considered to be within a normal range, while z-scores outside of this range may be considered unusual.

Based on this, the z-scores of 1.32, -0.78, 2.22, and -0.37 indicate that the scores of 1880 and 2170 are unusual, as they fall outside of the normal range.