Question

When traveling 40 mph (miles per hour), the distance that it takes Fred’s car to stop varies evenly between 120 and 155 feet. (This includes the reaction distance and the braking distance.) All of the questions are related to the stopping distance when Fred is traveling 40 mph.

a) Let S be the distance it takes for Fred’s car to stop at when traveling 40 mph. Find the distribution, parameter(s), and support of S.

b) What is the probability that it takes between 115 and 138 feet for the car to stop?

c) Find the expected distance it will take Fred to stop and the standard deviation of the stopping distance.

d) What values represent the middle 60% of Fred’s stopping distances?

e) Suppose a squirrel darts into the road as Fred is driving, and when Fred finally sees the squirrel and applies the brakes, the squirrel is 131 feet away. What is the probability that the squirrel survives its encounter with Fred (i.e. that Fred stops before he hits the squirrel)?

f) Fred knows that when it rains, it will take a minimum of 127 feet to stop. Fred is out driving while it is raining. If there is a stop sign that is 135 feet away, what is the probability that Fred stops in time?

Answer 1

a)
S follows a uniform distribution represented by U(120, 155)

b)
P(115 < X < 138)
= P(X < 138) - P(X < 115)
= (138 - 120)/(155 - 120) - (115 - 120)/(155 - 120)
= 0.6571

c)
expected value of S = 1/2*(120 + 155) = 137.5
variance = (1/12)*(155 - 120)^2 = 102.0833

d)
P(X < x1) = 0.2
(x1 - 120)/(155 - 120) = 0.2
x1 - 120 = 0.2*35
x1 = 120 + 0.2*35
x1 = 127

P(X > x2) = 0.2
(155 - x2)/(155 - 120) = 0.2
155 - x2 = 0.2*35
x2 = 155 - 0.2*35
x2 = 148

P(x1 < X < x2) = 0.6

e)
P(X < 131)
= (131 - 120)/(155 - 120)
= 0.3143

f)
here a = 127
P(X < 135)
= (135 - 127)/(155 - 127)
= 0.2857