Suppose 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.
what is the probability that a randomly selected college student will take between 2.5 and 4.5 minutes to find a parking spot in the library parking lot?

So the probability between 2  and 4.5 =  0.8413 – 0.0668

= 0.7745

Step-by-step explanation:

mean = 3.5

standard deviation = 1 minute

Apply  z statistic

for 4.5 minutes and 2 minutes

Calculate z for  4.5 minutes

z= (4.5-3.5)/1

 = 1.0

Calculate z for 2 minutes

(2-3.5)/1 = -1.5

z= 1 is 0.8413

z= - 1.5  is 0.0668

So the probability between 2  and 4.5 =  0.8413 – 0.0668

= 0.7745

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2