Question

Assume that it takes a college student an average of 10 minutes to find a parking spot in the main parking lot. Assume also that this time is normally distributed with a standard deviation of 5 minutes. What time is exceeded by approximately 75% of the college students when trying to find a parking spot in the main parking lot?

Answer 1

75% of college students exceed 6.63 minutes when trying to find a parking spot.

Explanation:

We are given the following information in the question:

Mean, μ = 10 minutes

Standard Deviation, σ = 5 minutes

We are given that the distribution of time for parking is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

P(X < x) = 0.25

We have to find the value of x such that the probability is 0.25.

P(X < x)

`P( X < x) = P( z < \displaystyle(x - 10)/(5))=0.25`

Calculation the value from standard normal z table, we have,

`P(z<-0.674) = 0.25`

`\displaystyle(x - 10)/(5) = -0.674\\x = 6.63`

Hence, 75% of college students exceed 6.63 minutes when trying to find a parking spot.