Question

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. Find the probability that a randomly selected college student will take between 2 and 6 minutes to find a parking spot in the main parking lot.

Answer 1

0.624 is the probability that a randomly selected college student will take between 2 and 6 minutes to find a parking spot in the main parking lot.

Explanation:

We are given the following information in the question:

Mean, μ = 5 minutes

Standard Deviation, σ = 2 minutes

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

Formula:

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

P(student will take between 2 and 6 minutes )

`P(2 \leq x \leq 6) = P(\displaystyle(2 - 5)/(2) \leq z \leq \displaystyle(6-5)/(2)) = P(-1.5 \leq z \leq 0.5)\\\\= P(z \leq 0.5) - P(z < -1.5)\\= 0.691 - 0.067 = 0.624 = 62.4\%`

`P(2 \leq x \leq 6) = 62.4\%`

0.624 is the probability that a randomly selected college student will take between 2 and 6 minutes to find a parking spot in the main parking lot.