My bus is scheduled to depart at noon. however, in reality, the departure time varies randomly, with expected departure times 12 o'clock noon and a standard deviation of 6 minutes. assume the departure time is normally distributed. if i get to the bus stop 5 minutes past noon, what is the probability that the bus has not yet departed?

0.2033

Step-by-step explanation:

Given : The departure time varies randomly, with expected departure times 12 o'clock noon and a standard deviation of 6 minutes.

Let

Let x be the random variable that represents the departure time .

We assume the departure time is normally distributed.

The formula to find the z-score :-

For x= 5 , we have

Now, the probability that the bus has not yet departed :-

Hence, the probability that the bus has not yet departed = 0.2033

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)