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

Answer:

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

Step-by-step 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

$P(2.5 \leq x \leq 5) = P(\dfrac{2.5 - \mu}{\sigma} \leq \dfrac{X-\mu}{\sigma}\leq \dfrac{5- \mu}{\sigma})$

$P(2.5 \leq x \leq 5) = P(\dfrac{2.5 - 4}{1} \leq Z \leq \dfrac{5- 4}{1})$

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

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

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

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

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