Answer:
To solve this problem, we'll use the properties of the sampling distribution of the sample mean.
a.) Find \( \mu \):
The mean of the sampling distribution of the sample mean (\( \mu \)) is equal to the population mean (\( \mu_x \)). So, \( \mu = 145 \) minutes.
b.) Find \( \sigma \):
The standard deviation of the sampling distribution of the sample mean (\( \sigma \)) is calculated using the formula:
\[ \sigma = \frac{\sigma_x}{\sqrt{n}} \]
Where:
- \( \sigma_x \) = population standard deviation (14 minutes)
- \( n \) = sample size (49 races)
\[ \sigma = \frac{14}{\sqrt{49}} = \frac{14}{7} = 2 \]
c.) Find the probability that the runner will average between 143 and 148 minutes in these 49 marathons.
To find this probability, we need to convert the individual times to z-scores and then use the standard normal distribution table.
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
- \( X \) = individual time (143 and 148 minutes in this case)
- \( \mu \) = population mean (145 minutes)
- \( \sigma \) = standard deviation of the sampling distribution of the sample mean (2 minutes)
For 143 minutes:
\[ Z_1 = \frac{143 - 145}{2} = -1 \]
For 148 minutes:
\[ Z_2 = \frac{148 - 145}{2} = \frac{3}{2} = 1.5 \]
Now, we'll use the standard normal distribution table to find the probabilities associated with these z-scores:
For \( Z = -1 \), \( P(Z < -1) \approx 0.1587 \)
For \( Z = 1.5 \), \( P(Z < 1.5) \approx 0.9332 \)
Now, to find the probability between these two values, we subtract the cumulative probability at \( Z = -1 \) from the cumulative probability at \( Z = 1.5 \):
\[ P(-1 < Z < 1.5) = P(Z < 1.5) - P(Z < -1) = 0.9332 - 0.1587 = 0.7745 \]
So, the probability that the runner will average between 143 and 148 minutes in these 49 marathons is approximately 0.7745.
