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A time for the 100 meter sprint of 14.9 seconds at a school where the mean time for the 100 meter sprint is 17.6 seconds and the standard deviation is 2.1 seconds. 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 or greater than 2.00.

Respuesta :

Answer:

Z-score = -1.285

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 17.6 seconds

Standard Deviation, σ = 2.1 seconds

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

For x = 14.9, we have:

[tex]z_{score} = \displaystyle\frac{14.9-17.6}{2.1} = -1.285[/tex]

A z-score to be unusual if its z-score is less than -2.00 or greater than 2.00.

Thus, the z-score is not unusual as it lies in the range [-2.00, 2.00]

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