The amount of time to complete a physical activity in a PE class is normally distributed with a mean of 36.8 seconds and a standard deviation of 5.9 seconds. Round answers to 4 decimal places.
a) What is the probability that a randomly chosen student completes the activity in less than 32 seconds? 
b) What is the probability that a randomly chosen student completes the activity in more than 40.1 seconds? 
c) What proportion of students take between 34.4 and 40.3 seconds to complete the activity? 
d) 90% of all students finish the activity in less than  seconds.

To solve these problems, we will use the standard normal distribution formula. The formula for converting a value \( X \) to its corresponding \( Z \)-score is:

\[
Z = \frac{X - \mu}{\sigma}
\]
where:
- \( X \) is the value we are converting,
- \( \mu \) is the mean of the distribution,
- \( \sigma \) is the standard deviation.

The normal distribution has a mean \( \mu = 36.8 \) seconds and a standard deviation \( \sigma = 5.9 \) seconds.

### (a) Probability that a randomly chosen student completes the activity in less than 32 seconds:

1. Calculate the \( Z \)-score for \( X = 32 \):

\[
Z = \frac{32 - 36.8}{5.9} = \frac{-4.8}{5.9} \approx -0.8136
\]

2. Using a standard normal distribution table or calculator, find the probability corresponding to \( Z = -0.8136 \). The cumulative probability for this \( Z \)-score is approximately:

\[
P(Z < -0.8136) \approx 0.2081
\]

Thus, the probability that a randomly chosen student completes the activity in less than 32 seconds is **0.2081**.

### (b) Probability that a randomly chosen student completes the activity in more than 40.1 seconds:

1. Calculate the \( Z \)-score for \( X = 40.1 \):

\[
Z = \frac{40.1 - 36.8}{5.9} = \frac{3.3}{5.9} \approx 0.5593
\]

2. Using a standard normal distribution table or calculator, find the cumulative probability for \( Z = 0.5593 \), which is:

\[
P(Z < 0.5593) \approx 0.7123
\]

To find the probability of more than 40.1 seconds:

\[
P(X > 40.1) = 1 - P(Z < 0.5593) = 1 - 0.7123 = 0.2877
\]

Thus, the probability that a student takes more than 40.1 seconds is **0.2877**.

### (c) Proportion of students who take between 34.4 and 40.3 seconds:

1. Calculate the \( Z \)-score for \( X = 34.4 \):

\[
Z = \frac{34.4 - 36.8}{5.9} = \frac{-2.4}{5.9} \approx -0.4068
\]

2. Calculate the \( Z \)-score for \( X = 40.3 \):

\[
Z = \frac{40.3 - 36.8}{5.9} = \frac{3.5}{5.9} \approx 0.5932
\]

3. Using the standard normal table:
   - \( P(Z < -0.4068) \approx 0.3420 \)
   - \( P(Z < 0.5932) \approx 0.7236 \)

4. The proportion of students taking between 34.4 and 40.3 seconds is:

\[
P(34.4 < X < 40.3) = P(Z < 0.5932) - P(Z < -0.4068) = 0.7236 - 0.3420 = 0.3816
\]

Thus, the proportion of students taking between 34.4 and 40.3 seconds is **0.3816**.

### (d) 90% of all students finish the activity in less than how many seconds?

We need to find \( X \) such that \( P(X < X_{90}) = 0.90 \). The \( Z \)-score corresponding to 90% is approximately \( Z = 1.2816 \).

Now, convert this \( Z \)-score back to \( X \) using the formula:

\[
X_{90} = \mu + Z \cdot \sigma = 36.8 + 1.2816 \cdot 5.9 \approx 36.8 + 7.5626 = 44.3626
\]

Thus, 90% of all students finish the activity in less than **44.3626 seconds**.

---

Would you like further details or clarifications on any of these? Here are some additional related questions:

1. How would the probability change if the standard deviation was larger?
2. What would happen if the mean time were shifted by 5 seconds?
3. How do we interpret the Z-scores in terms of physical activity performance?
4. How would you calculate the median time in this scenario?
5. What is the probability of a student finishing exactly in 36.8 seconds?

**Tip:** In a normal distribution, the probability of observing any *exact* value is technically zero, since continuous distributions measure ranges, not single points.

The amount of time to complete a physical activity in a PE class is normally distributed with a mean of 36.8 seconds and a standard deviation of 5.9 seconds. Round answers to 4 decimal places.
 a) What is the probability that a randomly chosen student completes the activity in less than 32 seconds?
 b) What is the probability that a randomly chosen student completes the activity in more than 40.1 seconds?
 c) What proportion of students take between 34.4 and 40.3 seconds to complete the activity?
 d) 90% of all students finish the activity in less than how many seconds?

Solve a normal distribution problem in a PE class where students' activity times are normally distributed with a mean of 36.8 seconds and a standard deviation of 5.9 seconds. Learn to calculate probabilities for times less than 32 seconds, more than 40.1 seconds, and between 34.4 and 40.3 seconds. Determine the time for 90% completion.

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