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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 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 minutes.

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Answer:

The probability is the pvalue of
Z = X - \mu, in which X is the number of minutes we want to find the probability of time being less of, and
\mu is the mean.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the z-score of a measure X is given by:


Z = (X - \mu)/(\sigma)

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 p-value, we get the probability that the value of the measure is greater than X.

Standard deviation of 1 minute.

This means that
\sigma = 1

Find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than minutes.?

This is the pvalue of Z for X. So


Z = (X - \mu)/(\sigma)


Z = (X - \mu)/(1)


Z = X - \mu

The probability is the pvalue of
Z = X - \mu, in which X is the number of minutes we want to find the probability of time being less of, and
\mu is the mean.

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