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The amount of time to complete a physical activity in a PE class is approximately normally normally distributed with a mean of 32.5 seconds and a standard deviation of 6.5 seconds. What is the probability that a randomly chosen student completes the activity in less than 28.4 seconds?

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Final answer:

The probability that a randomly chosen student completes the activity in less than 28.4 seconds is approximately 26.43%, which is obtained by calculating the z-score and then using the standard normal distribution table.

Step-by-step explanation:

To find the probability that a randomly chosen student completes the activity in less than 28.4 seconds, we will use the z-score formula for a normally distributed variable.

The z-score formula is given by:

z = (X - μ) / σ

where:

  • X is the value for which we are finding the probability (28.4 seconds)
  • μ (mu) is the mean of the distribution (32.5 seconds)
  • σ (sigma) is the standard deviation of the distribution (6.5 seconds)

So, the z-score for 28.4 seconds is:

z = (28.4 - 32.5) / 6.5

z = -4.1 / 6.5

z ≈ -0.63

Finally, we use the standard normal distribution table (or a calculator with statistical functions) to find the probability corresponding to a z-score of -0.63. This probability is approximately 0.2643, so there's a 26.43% chance that a randomly selected student will complete the activity in less than 28.4 seconds.

User Huzaifa Saifuddin
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