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

Answer:

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

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(X < x) = 0.25

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

P(X < x)  

$P( X < x) = P( z < \displaystyle\frac{x - 10}{5})=0.25$  

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

$P(z$

$\displaystyle\frac{x - 10}{5} = -0.674\\x = 6.63$  

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