To solve the problem, we analyze the sampling distribution and calculate the probabilities. Let’s break it down step by step.

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(i) Sampling Distribution of $\bar{x}$

The sampling distribution of the sample mean $\bar{x}$ has the following properties:

1. Mean ($\mu_{\bar{x}}$):
   The mean of the sampling distribution is equal to the population mean:
   $\mu_{\bar{x}} = \mu = 31.5 \, \text{minutes}.$

2. Standard Error ($\sigma_{\bar{x}}$):
   The standard error of the sample mean is calculated using the formula:
   $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{50}} = \frac{12}{7.071} \approx 1.697 \, \text{minutes}.$

Thus, the sampling distribution of $\bar{x}$ is normally distributed with:
$\bar{x} \sim N(31.5, 1.697).$

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(ii) Probability that $\bar{x}$ is within 1 minute of the population mean

We are asked for the probability that $\bar{x}$ is within 1 minute of the population mean, i.e.,
$P(30.5 \leq \bar{x} \leq 32.5).$

We convert the values to standard $z$-scores using the formula:
$z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}.$

For $\bar{x} = 30.5$:
$z = \frac{30.5 - 31.5}{1.697} = \frac{-1}{1.697} \approx -0.589.$

For $\bar{x} = 32.5$:
$z = \frac{32.5 - 31.5}{1.697} = \frac{1}{1.697} \approx 0.589.$

The probability is:
$P(30.5 \leq \bar{x} \leq 32.5) = P(-0.589 \leq z \leq 0.589).$

Using standard normal distribution tables or a calculator, we find:
$P(-0.589 \leq z \leq 0.589) \approx 0.4433.$

Thus, the probability is approximately 44.33%.

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(iii) Probability that $\bar{x}$ is within 3 minutes of the population mean

Here, we calculate $P(28.5 \leq \bar{x} \leq 34.5)$:
$P(28.5 \leq \bar{x} \leq 34.5).$

For $\bar{x} = 28.5$:
$z = \frac{28.5 - 31.5}{1.697} = \frac{-3}{1.697} \approx -1.768.$

For $\bar{x} = 34.5$:
$z = \frac{34.5 - 31.5}{1.697} = \frac{3}{1.697} \approx 1.768.$

The probability is:
$P(28.5 \leq \bar{x} \leq 34.5) = P(-1.768 \leq z \leq 1.768).$

Using standard normal distribution tables or a calculator, we find:
$P(-1.768 \leq z \leq 1.768) \approx 0.9228.$

Thus, the probability is approximately 92.28%.

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Summary of Results

1. The sampling distribution of $\bar{x}$ is $N(31.5, 1.697)$.
2. The probability that $\bar{x}$ is within 1 minute of the population mean is 44.33%.
3. The probability that $\bar{x}$ is within 3 minutes of the population mean is 92.28%.

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Do you want a detailed breakdown of how to use the $z$-table or a plot of the normal distribution?

Relative Questions

1. How is the standard error of the mean derived, and why does it depend on the sample size?
2. Why does the probability increase when the range around the mean is widened?
3. What happens to the sampling distribution if the sample size increases to 100?
4. How would the probabilities change if the population standard deviation were larger, say 15 minutes?
5. Why is the normal distribution used for the sampling distribution of the mean?

Tip: Always confirm whether the sample size is large enough to approximate the sampling distribution as normal.