If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes. A) 0.3085 B) 0.2674 C) 0.3551 D) 0.1915

Step 1 ：We are given that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute.

Step 2 ：We are asked to find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes.

Step 3 ：We can use the z-score formula to find the z-score for 3 minutes. The z-score formula 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 ：Substituting the given values into the z-score formula, we get \(Z = \frac{3 - 3.5}{1} = -0.5\).

Step 5 ：We can then use the standard normal distribution table or a function to find the probability that the z-score is less than -0.5. This gives us a probability of approximately 0.3085.

Step 6 ：Final Answer: The probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes is \(\boxed{0.3085}\).