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 take between 2 and 4.5 minutes to find a parking spot in the library parking lot.
To find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot, we need to calculate the area under the normal distribution curve between 2 and 4.5 minutes.
First, we need to calculate the z-scores for the two time values:
Z1 = (2 - 3.5) / 1 = -1.5
Z2 = (4.5 - 3.5) / 1 = 1
Next, we need to find the area under the normal curve between these two z-scores. We can use a standard normal distribution table or a statistical software to find the probabilities associated with these z-scores.
Using a standard normal table, we find:
P(Z < -1.5) = 0.0668
P(Z < 1) = 0.8413
Therefore, the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot is:
P(2 < X < 4.5) = P(-1.5 < Z < 1) = P(Z < 1) - P(Z < -1.5) = 0.8413 - 0.0668 = 0.7745
So, the probability is 0.7745 or 77.45%.