The probability that the sample mean is greater than 50200 is 0.976.
The standard error (SE) of the mean is calculated using the formula:
$$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$$
Substituting the given values:
$$\sigma_{\bar{X}} = \frac{4850}{\sqrt{36}} = 808.333$$
The Z-score is calculated using the formula:
$$Z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}}$$
Substituting the given values:
$$Z = \frac{48800 - 51803}{808.333} = -3.715$$
Since we are looking for \(P(\bar{X} < k)\), we use \(P(Z)\):
Substituting the Z-score into the standard normal distribution:
$$P = 0.0001$$
The probability that the sample mean is less than 48800 is 0.0001.