Question

If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1​ minute, find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking​ lot, rounding to four decimal places.

Answer 1

Answer: 0.7745

Explanation:

Given : The length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with
`\mu=3.5` minutes and
`\sigma=1` minute.

Let x represents the time to find a parking spot in the library parking​ lot.

Since ,
`z=(x-\mu)/(\sigma)`

z-score corresponds x = 2 ,
`z=(2-3.5)/(1)=-1.5`

z-score corresponds x = 4.5 ,
`z=(4.5-3.5)/(1)=1`

Required probability :

`\text{P-value }: P(2<x<4.5)=P(-1.5<z<1)\\\\=P(z<1)-P(z<-1.5)\\\\=P(z<1)-(1-P(z<1.5))\\\\0.8413447-(1-0.9331927)\\\\=0.7745374\approx0.7745`

[using z-value table.]

Hence, the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking​ lot, (rounding to four decimal place )= 0.7745