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The amount of time to complete a physical activity in a PE class is approximately normally distributed with a mean of 39.5 seconds and a standard deviation of 6.6 seconds.

What is the probability that a randomly chosen student completes the activity in less than 36.6 seconds?

Answer :

Final answer:

The probability that a randomly chosen student completes a PE activity in less than 36.6 seconds given a normal distribution with a mean of 39.5 seconds and a standard deviation of 6.6 seconds is approximately 33.00%.

Explanation:

The probability that a randomly chosen student completes a PE activity in less than 36.6 seconds, given that the activity completion time follows a normal distribution with a mean (μ) of 39.5 seconds and a standard deviation (σ) of 6.6 seconds, can be calculated using the Z-score formula:

Z = (X - μ) / σ

Where X is the value for which we want to find the probability.

First, let's find the Z-score for X = 36.6 seconds:

Z = (36.6 - 39.5) / 6.6 = -2.9 / 6.6 ≈ -0.4394

Then, we look up the Z-score in the standard normal distribution table, which provides the probability that a value is less than the given Z-score.

The probability associated with a Z-score of -0.4394 is approximately 0.3300 (33.00%). Therefore, there is a 33.00% chance that a randomly chosen student will complete the activity in less than 36.6 seconds.

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