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.

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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.