In: Math
Avoiding an accident while driving can depend on reaction time.
That time, measured from the time the driver first sees the danger
until the driver gets his/her foot on the brake pedal, can be
described by a normal model with mean 1.9 seconds and standard
deviation 0.13 seconds. Use the 68-95-99.7 rule
(NOT a z table) to answer the following questions. The pictures of
the 68-95-99.7 rule at this link might help.
http://www.oswego.edu/~srp/stats/6895997.htm
What percentage of drivers have a reaction time more than 2.16 seconds?
________%
What percentage of drivers have a reaction time less than 1.77
seconds?
________%
What percentage of drivers have a reaction time less than 2.03
seconds?
________%
The 68-95-99.7 Rule tells us about the approximate probability that is found within a certain number of standard deviations from the population mean.
• First, the 68-95-99.7 Rule says that the probability within 1 standard deviation from the mean is approximately 68%.
This means that:
P(μ−σ≤X≤μ+σ) ≈ 0.68
In this case, we have that:
μ−σ = 1.9−0.13 = 1.77\
μ+σ = 1.9+0.13=2.03
so then
Pr(1.77 ≤X ≤ 2.03) ≈ 0.68
Then, since the probability inside of the interval (1.77, 2.03) is 0.68, then using the Law of Complement, the probability OUTSIDE of 1.77,2.03 is 1 - 0.68 = 0.32. Also, since the normal distribution is symmetric, then we also conclude that half of that probability (0.32/2 = 0.16) goes to each of the two tails. This means that
P(X≤1.77) ≈ 0.16
Thus, percentage of drivers having reaction time less than 1.77 seconds is 16%
P(X≤2.03) ≈ 1−0.16 = 0.84
Thus, percentage of drivers having reaction time less than 2.03 seconds is 84%
• Similarly, the 68-95-99.7 Rule says that the probability within 2 standard deviations from the mean is approximately 95%. This means that:
(μ−2σ ≤ X ≤ μ+2σ) ≈ 0.95
In this case, we have that:
μ−2σ = 1.9−2×0.13 = 1.64
μ+2σ = 1.9+2×0.13 = 2.16
so then
Pr(1.64≤X≤2.16) ≈ 0.95
Then, since the probability inside of the interval (1.64,2.16) is 0.95, then using the Law of Complement, the probability OUTSIDE of 1.64,2.16 is 1 - 0.95 = 0.05. Also, since the normal distribution is symmetric, then we also conclude that half of that probability (0.05/2 = 0.025) goes to each of the two tails. This means that
P(X≤1.64) ≈ 0.025
P(X>2.16) ≈ 0.025
Thus, percentage of drivers having reaction time more than 2.16 seconds is 2.5%