Question

For a car traveling at 30 miles per hour (mph), the distance required to brake to a stop is normally distributed with a mean of 50 feet and a standard deviation of 8 feet. Suppose you are traveling 30 mph in a residential area and a car moves abruptly into your path at a distance of 60 feet ahead of you.

If you apply your brakes, what is the probability it will take you between 44 and 56 feet to brake the car to a stop?

0.5468

0.4592 0

0.4532

For the problem situation presented, what is the probability it will take more than 60 feet to stop the car?

0.1056

0.9999

0.9032

0.8944

Answer 1

Solution :

Given that ,

mean = = 50

standard deviation = = 8

P(44 < x < 56) = P((44 - 50 / 8) < (x - ) / < (56 - 50) / 8) )

= P(-0.75 < z < 0.75)

= P(z < 0.75) - P(z < -0.75)

= 0.7734 - 0.2266 = 0.5468

Probability = 0.5468

P(x > 60) = 1 - P(x < 60)

= 1 - P((x - ) / < (60 - 50) / 8)

= 1 - P(z < 1.25)

= 1 - 0.8944   

= 0.1056

P(x > 60) = 0.1056

Probability = 0.1056