Assume that it takes a college student an average of 5 minutes to find a parking spot in the main parking lot. Assume also that this time is normally distributed with a standard deviation of 2 minutes. What time is exceeded by approximately 75% of the college students when trying to find a parking spot in the main parking lot?

Respuesta :

Answer:

A time of 3.65 minutes is exceeded by approximately 75% of the college students when trying to find a parking spot in the main parking lot

Step-by-step explanation:

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

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 5, \sigma = 2[/tex]

What time is exceeded by approximately 75% of the college students when trying to find a parking spot in the main parking lot?

This is 100-75 = 25th percentile, which is X when Z has a pvalue of 0.25. So it is X when Z = -0.675.

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

[tex]-0.675 = \frac{X - 5}{2}[/tex]

[tex]X - 5 = -0.675*2[/tex]

[tex]X = 3.65[/tex]

A time of 3.65 minutes is exceeded by approximately 75% of the college students when trying to find a parking spot in the main parking lot