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.

Respuesta :

Answer:

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

Step-by-step explanation:

Normal Probability Distribution:

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\sigma = 1[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{X - \mu}{1}[/tex]

[tex]Z = X - \mu[/tex]

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