Question

Find the z-score corresponding to the given value and use the z-score to determine whether the value is unusual. Consider a score to be unusual if its z-score is less than -2.00 and greater than 2.00. Round the z-score to the nearest tenth, if necessary.

A time for the 100 meter sprint of 15.0 seconds at a school where the mean time for the 100 meter sprint is 17.5 seconds and the standard deviation is 2.1 seconds.

a. -1.2; not unusual

b. 1.2; not unusual

c. -2.5; unusual

d. -1.2; unusual

Answer 1

The z-score for a 15.0 second 100m sprint, given a mean of 17.5 seconds and a standard deviation of 2.1 seconds, is approximately -1.2. Since the z-score is between -2 and 2, the time of 15.0 seconds is not considered unusual.

Step-by-step explanation:

To find the z-score corresponding to a 100 meter sprint time of 15.0 seconds, we use the formula z = (x - μ) / σ, where x is the value (15.0 seconds), μ is the mean (17.5 seconds), and σ is the standard deviation (2.1 seconds). Plugging in the numbers, we get:

z = (15.0 - 17.5) / 2.1 = -2.5 / 2.1 ≈ -1.2

Since the z-score of -1.2 is between -2 and 2, it is not considered unusual according to the given criteria. Therefore, the correct choice is a. -1.2; not unusual.