Assume that it takes a college student an average of 5 minutes to find a parking spot in the main parking lot. Assume also that this time is normally distributed with a standard deviation of 2 minutes. What time is exceeded by approximately 75% of the college students when trying to find a parking spot in the main parking lot?

Look up 0.7500 in your Z table. That is -.674 std from the mean.

Now scale and shift it for your problem:

5 - .675(2) = 3.65

A handy place to play around with this stuff is

http://davidmlane.com/hyperstat/z_table.html

To find the time exceeded by approximately 75% of the college students when trying to find a parking spot, we need to determine the z-score corresponding to the 75th percentile of the normal distribution.

Step 1: Calculate the z-score corresponding to the 75th percentile.
The 75th percentile corresponds to a cumulative probability of 0.75. We can find the corresponding z-score using a standard normal distribution table or calculator.

Step 2: Once we have the z-score, we can calculate the corresponding time using the formula:
X = μ + (z * σ)
where X is the time (in minutes), μ is the mean (average) time, z is the z-score, and σ is the standard deviation.

In this case, the mean (μ) is 5 minutes and the standard deviation (σ) is 2 minutes.

Step 1: Finding the z-score
Using a standard normal distribution table or calculator, we can find that the z-score corresponding to a cumulative probability of 0.75 is approximately 0.6745.

Step 2: Calculating the time
X = 5 + (0.6745 * 2)
X ≈ 5 + 1.349
X ≈ 6.349

Therefore, approximately 75% of college students will exceed a time of 6.349 minutes when trying to find a parking spot in the main parking lot.