Question 1168176
Let $\mu$ be the mean annual salary for graduates 10 years after graduation, so $\mu = 159000$.
Let $\sigma$ be the standard deviation of the annual salary, so $\sigma = 40000$.

**Find the probability that a single randomly selected graduate has a mean value between $146815.8 and $156969.3 dollars.**

For a single randomly selected graduate (n=1), the distribution is normal with mean $\mu = 159000$ and standard deviation $\sigma = 40000$. We want to find $P(146815.8 < X < 156969.3)$.
First, we calculate the z-scores for the lower and upper bounds:
$z_1 = \frac{146815.8 - 159000}{40000} = \frac{-12184.2}{40000} \approx -0.3046$
$z_2 = \frac{156969.3 - 159000}{40000} = \frac{-2030.7}{40000} \approx -0.0508$

Now, we find the probabilities corresponding to these z-scores using a standard normal distribution table or a calculator:
$P(Z < -0.3046) \approx 0.3803$
$P(Z < -0.0508) \approx 0.4797$

The probability $P(146815.8 < X < 156969.3) = P(-0.3046 < Z < -0.0508) = P(Z < -0.0508) - P(Z < -0.3046) = 0.4797 - 0.3803 = 0.0994$.

$P(146815.8 < X < 156969.3) = \boxed{0.0994}$

**Find the probability that a random sample of size n=97 has a mean value between $146815.8 and $156969.3 dollars.**

For a sample of size $n=97$, the sampling distribution of the sample mean $\bar{X}$ is approximately normal with mean $\mu_{\bar{X}} = \mu = 159000$ and standard deviation $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{40000}{\sqrt{97}} \approx \frac{40000}{9.84885} \approx 4061.39$.

Now, we calculate the z-scores for the lower and upper bounds of the sample mean:
$z_1 = \frac{146815.8 - 159000}{4061.39} = \frac{-12184.2}{4061.39} \approx -3.0000$
$z_2 = \frac{156969.3 - 159000}{4061.39} = \frac{-2030.7}{4061.39} \approx -0.5000$

Now, we find the probabilities corresponding to these z-scores using a standard normal distribution table or a calculator:
$P(Z < -3.0000) \approx 0.0013$
$P(Z < -0.5000) \approx 0.3085$

The probability $P(146815.8 < \bar{X} < 156969.3) = P(-3.0000 < Z < -0.5000) = P(Z < -0.5000) - P(Z < -3.0000) = 0.3085 - 0.0013 = 0.3072$.

$P(146815.8 < M < 156969.3) = \boxed{0.3072}$