Final answer:
The z-score for a student running 100m in 14.2 seconds, where the mean is 17.6 and the standard deviation is 2.1, is approximately -1.6, which is within two standard deviations of the mean. Hence, the time is not considered statistically unusual.
Step-by-step explanation:
The question pertains to a statistical analysis of sprint times using measures such as the mean, standard deviation, and z-score to determine if a time is unusual. In statistical terms, a score is typically considered unusual if it lies more than two standard deviations from the mean. In this context, if a student runs the 100m sprint in 14.2 seconds, we can calculate the z-score by subtracting the mean time (17.6 seconds) from the student's time and then dividing by the standard deviation (2.1 seconds). This is then used to determine whether the time is statistically unusual.
Z-Score Calculation Example
The z-score is calculated as:
Z = (Student's time - Mean time) / Standard deviation
Z = (14.2 - 17.6) / 2.1 = -3.4 / 2.1 = -1.619 (approximately -1.6)
Interpreting the Z-Score
Since -1.6 is within two standard deviations, the student's time is not considered unusual in a statistical sense. Therefore, based on the z-score, 14.2 seconds for the 100m sprint would be regarded as not unusual for the student.