A car topped and drove 500 hundred miles from home near river, the sample mean was found to be 50724 hrs and the standard deviation of distance is 18112 km. interval of 95% confidence.

Since the sample size $n=500$ is large ($n \ge 30$), we use the z-distribution for the critical value.
For a 99% confidence interval, $\alpha = 1 - 0.99 = 0.01$.
We need $z_{\alpha/2} = z_{0.01/2} = z_{0.005}$.
From the standard normal distribution table, the critical z-value is $z_{0.005} = 2.576$.

The formula for the margin of error is:

The confidence interval is given by $\bar{x} \pm ME$.
Lower bound: $50724 - 2087.90 = 48636.10$ km
Upper bound: $50724 + 2087.90 = 52811.90$ km

The 99% confidence interval for the mean distance driven is: