Question

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 7.0 minutes and a standard deviation of 1 minute. Find the cut-off time which 75.8% of the college students exceed when trying to find a parking spot in the library parking lot. Please explain answer using TI-84 invnorm or normalcdf functions as we don't use the table in my class.

Answer 1

First we have find the z score corresponding to 0.758 probability then we will find the cut- off time (x) which 75.8% students exceed.

In Ti 84 calculator follow the instructions

Press 2ND> VARS> select DISTR > in DISTR select invNorm(0.758) >ENTER  

You will get Z = 0.78

Now we have to find x (cut-off time )

X = $\mu +\sigma *Z$

we are given $\mu$ = mean = 7.0 and standard deviation = $\sigma$ =1

X = 7 + (1*0.78)

X = 7.78  

This is required answer