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 4 minutes and a standard deviation of 1 minute. Find the probability that a randomly selected college student will take between 2.5 and 5 minutes to find a parking spot in the library lot.

Answer 1

the probability that a randomly selected college student will take between 2.5 and 5 minutes to find a parking spot in the library lot is 0.77454

Explanation:

Given that:

mean = 4

standard deviation = 1

The objective is to find the probability that a randomly selected college student will take between 2.5 and 5 minutes to find a parking spot in the library lot.

i.e

`P(2.5 \leq x \leq 5) = P((2.5 - \mu)/(\sigma) \leq (X-\mu)/(\sigma)\leq (5- \mu)/(\sigma))`

`P(2.5 \leq x \leq 5) = P((2.5 - 4)/(1) \leq Z \leq (5- 4)/(1))`

`P(2.5 \leq x \leq 5) = P((-1.5)/(1) \leq Z \leq (1)/(1))`

`P(2.5 \leq x \leq 5) = P({-1.5}\leq Z \leq 1)`

`P(2.5 \leq x \leq 5) = P({Z < 1})- P(Z < -1.5)`

`P(2.5 \leq x \leq 5) = 0.84134- 0.06680`

`\mathbf{P(2.5 \leq x \leq 5) = 0.77454}`