Question 1168175
Let $\mu$ be the population mean length of unemployment, so $\mu = 30$ weeks.
Let $\sigma$ be the population standard deviation, so $\sigma = 2.4$ weeks.

**Find the probability that a single randomly selected value is greater than 29.7.**

For a single randomly selected individual ($n=1$), the distribution is normal with mean $\mu = 30$ and standard deviation $\sigma = 2.4$. We want to find $P(X > 29.7)$.
First, we calculate the z-score for $x = 29.7$:
$z = \frac{x - \mu}{\sigma} = \frac{29.7 - 30}{2.4} = \frac{-0.3}{2.4} = -0.125$

Now, we find the probability $P(Z > -0.125)$, where $Z$ is a standard normal random variable.
$P(Z > -0.125) = 1 - P(Z \le -0.125)$
Looking up the value in a standard normal distribution table or using a calculator, we find:
$P(Z \le -0.125) \approx 0.4502$

So, $P(Z > -0.125) = 1 - 0.4502 = 0.5498$.

$P(X > 29.7) = \boxed{0.5498}$

**Find the probability that a sample of size n=59 is randomly selected with a mean greater than 29.7.**

For a sample of size $n=59$, the sampling distribution of the sample mean $\bar{X}$ is approximately normal with mean $\mu_{\bar{X}} = \mu = 30$ and standard deviation $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{2.4}{\sqrt{59}} \approx \frac{2.4}{7.6811} \approx 0.31245$.

Now, we calculate the z-score for the sample mean $\bar{X} = 29.7$:
$z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{29.7 - 30}{0.31245} = \frac{-0.3}{0.31245} \approx -0.9602$

Now, we find the probability $P(Z > -0.9602)$, where $Z$ is a standard normal random variable.
$P(Z > -0.9602) = 1 - P(Z \le -0.9602)$
Looking up the value in a standard normal distribution table or using a calculator, we find:
$P(Z \le -0.9602) \approx 0.1685$

So, $P(Z > -0.9602) = 1 - 0.1685 = 0.8315$.

$P(M > 29.7) = \boxed{0.8315}$

Based on your previous answers:
a) 0.4483 is close to $P(Z \le -0.125) \approx 0.4502$, which is not the required probability for the first part.
b) 0.1685 is close to $P(Z \le -0.9602) \approx 0.1685$, which is $1 - P(M > 29.7)$.

Your calculations likely had a sign error or you were looking for the complementary probability.

Final Answer: The final answer is $\boxed{P(X > 29.7) = 0.5498, P(M > 29.7) = 0.8315}$