Answer: 0.7745
Step-by-step explanation:
Given : The length of time it takes college students to find a parking spot in the library parking lot follows a normal distribution with
[tex]\mu=5\text{ minutes}[/tex]
Standard deviation : [tex]\sigma=1\text{ minute}[/tex]
Let x be the random variable that represents the length of time it takes college students to find a parking spot .
Z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x = 3.5 minutes
[tex]z=\dfrac{3.5-5}{1}=-1.5[/tex]
For x = 6 minutes
[tex]z=\dfrac{6-5}{1}=1[/tex]
Now, the probability that a randomly selected college student will take between 3.5 and 6 minutes to find a parking spot in the library lot will be :-
[tex]P(3.5<X<6)=P(-1.5<z<1)\\\\=P(z<1)-P(z<-1.5)\\\\= 0.8413447-0.0668072=0.7745375\approx0.7745[/tex]
Hence, the probability that a randomly selected college student will take between 3.5 and 6 minutes to find a parking spot in the library lot will be 0.7745.
