The amount of time to complete a physical activity in a PE class is approximately normally distributed with a mean of 38 seconds and a standard deviation of 6.8 seconds. Enter decimal answers rounded to 4 decimal places for a-c, 1 place for part d. a) What is the probability that a randomly chosen student completes the activity in less than 33 seconds? b) What is the probability that a randomly chosen student completes the activity in more than 42.3 seconds? c) What proportion of students take between 33.8 and 42.8 seconds to complete the activity? d) \( 70 \% \) of all students finish the activity in less than

To tackle this problem, we can use the properties of the normal distribution to find the required probabilities and proportions. a) To find the probability that a student completes the activity in less than 33 seconds, standardize the value using the formula: \( z = \frac{(X - \mu)}{\sigma} \). Plugging in, we get \( z = \frac{(33 - 38)}{6.8} \approx -0.7353 \). Using the z-table, we find that \( P(Z < -0.7353) \approx 0.23 \) or 0.2300 when rounded to 4 decimal places. b) Similarly, for more than 42.3 seconds, first find the z-score: \( z = \frac{(42.3 - 38)}{6.8} \approx 0.6324 \). From the z-table, \( P(Z > 0.6324) \approx 0.2650 \) when rounded to 4 decimal places. c) For students completing the activity between 33.8 and 42.8 seconds, calculate the z-scores: z(33.8) is approximately -0.6471 and z(42.8) is about 0.5882. Look up these values and find \( P(33.8 < X < 42.8) \approx P(Z < 0.5882) - P(Z < -0.6471) \approx 0.2771 \). So this comes out to approximately 0.2771 (rounded to 4 decimal places). d) To find the time at which 70% of students finish the activity, we look for the z-score that corresponds to 0.70, which is approximately 0.524. We can then use the formula \( X = \mu + z \cdot \sigma = 38 + (0.524 \cdot 6.8) \), yielding \( X \approx 43.6 \) (rounded to 1 decimal place). So the final answers are: a) 0.2300 b) 0.2650 c) 0.2771 d) 43.6