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Reader View Available The amount of time to complete a physical activity in a PE class is approximately normally distributed with a mean of 37.6 seconds and a standard deviation of 6.5 seconds. a) What is the probability that a randomly chosen student completes the activity in less than 32.8 seconds? (Round answer to three decimal places) b) What is the probability that a randomly chosen student completes the activity in more than 42.9 seconds? (Round answer to three decimal places) c) What proportion of students take between 33.3 and 42.5 seconds to complete the activity? (Round answer to three decimal places) d) 80% of all students finish the activity in less than seconds. (Round answer to one decimal places) Submit Question

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Final answer:

To solve these probability problems, we use the z-score formula to standardize the values. The probability of completing the activity in less than 32.8 seconds is 0.229. The probability of completing the activity in more than 42.9 seconds is 0.207. The proportion of students who take between 33.3 and 42.5 seconds to complete the activity is 0.371. Lastly, 80% of students finish the activity in less than 43.1 seconds.

Step-by-step explanation:

To solve these probability problems, we first need to standardize the values using the z-score formula:

z = (x - mean) / standard deviation

a) To find the probability that a randomly chosen student completes the activity in less than 32.8 seconds:

z = (32.8 - 37.6) / 6.5 = -0.738

Using a z-table or calculator, we find that the probability is approximately 0.229.

b) To find the probability that a randomly chosen student completes the activity in more than 42.9 seconds:

z = (42.9 - 37.6) / 6.5 = 0.815

Again, using a z-table or calculator, we find that the probability is approximately 0.207.

c) To find the proportion of students who take between 33.3 and 42.5 seconds to complete the activity:

First, we calculate the z-scores for both values:

z1 = (33.3 - 37.6) / 6.5 = -0.662

z2 = (42.5 - 37.6) / 6.5 = 0.754

Using a z-table or calculator, we find the area under the curve between these two z-scores is approximately 0.371.

d) To find the value that separates the bottom 80% of students from the top 20%, we need to find the z-score that corresponds to the cumulative probability of 0.8:

Using a z-table or calculator, we find that the z-score is approximately 0.842.

We can then solve for x using the z-score formula:

0.842 = (x - 37.6) / 6.5

x - 37.6 = 0.842 * 6.5

x - 37.6 = 5.477x = 43.077

Therefore, 80% of students finish the activity in less than 43.1 seconds.

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