The length of time it takes college students to find a parking spot in the library parking lot follows a normal distribution with a mean of 5.0 minutes and a standard deviation of 2 minute. Find the proportion of college students that will take between 2.5 and 6.0 minutes to find a parking spot in the library lot. Round your solution to 4 decimals:

Step 1 ：We are given a normal distribution with a mean of 5.0 minutes and a standard deviation of 2 minutes. We are asked to find the proportion of college students that will take between 2.5 and 6.0 minutes to find a parking spot.

Step 2 ：To solve this, we can use the properties of the normal distribution. We can calculate the z-scores for 2.5 and 6.0 minutes, and then find the area under the curve between these two z-scores. This area represents the proportion of students that will take between 2.5 and 6.0 minutes to find a parking spot.

Step 3 ：The formula for the z-score is: \(z = \frac{X - \mu}{\sigma}\) where X is the value we are interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 4 ：After calculating the z-scores, we can use the cumulative distribution function (CDF) of the normal distribution to find the area under the curve. The CDF gives the probability that a random variable is less than or equal to a certain value.

Step 5 ：The proportion of students that will take between 2.5 and 6.0 minutes to find a parking spot is then given by: \(P(2.5 < X < 6.0) = P(X < 6.0) - P(X < 2.5)\) where \(P(X < x)\) is the CDF of the normal distribution at x.

Step 6 ：Calculating the z-scores for 2.5 and 6.0 minutes, we get \(z_{2.5} = -1.25\) and \(z_{6.0} = 0.5\).

Step 7 ：Using the CDF of the normal distribution, we find that \(P(X < 2.5) = 0.1056\) and \(P(X < 6.0) = 0.6915\).

Step 8 ：Subtracting these probabilities, we find that the proportion of students that will take between 2.5 and 6.0 minutes to find a parking spot is \(0.6915 - 0.1056 = 0.5858\).

Step 9 ：Final Answer: The proportion of college students that will take between 2.5 and 6.0 minutes to find a parking spot in the library lot is \(\boxed{0.5858}\).