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

Step-by-step 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 , $z=\dfrac{x-\mu}{\sigma}$

z-score corresponds x = 2 , $z=\dfrac{2-3.5}{1}=-1.5$

z-score corresponds x = 4.5 , $z=\dfrac{4.5-3.5}{1}=1$

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