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 3.5 minutes and a standard deviation of 1 minute. Find the probability that a randomly selected college student will take between 2.0 and 4.5 minutes to find a parking spot in the library lot. Report the probability to 3 decimal places.

0.774 = 77.40% probability that a randomly selected college student will take between 2.0 and 4.5 minutes to find a parking spot in the library lot.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Find the probability that a randomly selected college student will take between 2.0 and 4.5 minutes to find a parking spot in the library lot.

pvalue of Z when X = 4.5 subtracted by the pvalue of Z when X = 2. So

X = 4.5

has a pvalue of 0.841

X = 2

has a pvalue of 0.067

0.841 - 0.067 = 0.774

0.774 = 77.40% probability that a randomly selected college student will take between 2.0 and 4.5 minutes to find a parking spot in the library lot.

50x5=$250

step-by-step explanation: