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

b

step-by-step explanation:

alright so i think for this one it would to draw it out. this may be hard for me to try and draw, but i will do my best on here:

        r

  /       |     \

notice how it forms 2 triangles inside one bigger triangle.

so if we know the angle spr is the same as psr that is saying the measure of angle p is the same as angle s. this means the big triangle psr is either isosceles or equilateral. in both cases, the triangle is perfectly symmetrical when split down the middle, which is in this case rq. because we know it is perfectly symmetrical on both sides, we know that the two triangles inside the bigger one are equal.