The commuting time for a student to travel from home to a college campus is normally distributed with a mean of 30 minutes and a standard deviation of 5 minutes. if the student leaves home at 8: 25 a. m., what is the probability that the student will arrive at the college campus later than 9 a. m.?

0.1587

Step-by-step explanation:

Let X be the commuting time for the student. We know that . Then, the normal probability density function for the random variable X is given by

. We are seeking the probability P(X>35) because the student leaves home at 8:25 A.M., we want to know the probability that the student will arrive at the college campus later than 9 A.M. and between 8:25 A.M. and 9 A.M. there are 35 minutes of difference. So,

= 0.1587

To find this probability you can use either a table from a book or a programming language. We have used the R statistical programming language an the instruction pnorm(35, mean = 30, sd = 5, lower.tail = F)

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."

with btw u said 100pts but it clearly says

i answerd and just got 10pts

no hate