Question

1. Assume the waiting time at the BMV is uniformly distributed from 10 to 60 minutes, i.e. X ∼ U ( 10 , 60 )X ∼ U ( 10 , 60 )

What is the expected time waited (mean), and standard deviation for the above uniform variable?  

1B) What is the probability that a person at the BMV waits longer than 45 minutes?

1C) What is the probability that an individual waits between 15 and 20 minutes, OR 35 and 40 minutes, i.e. P ( 15 ≤ X ≤ 20 ∪ 35 ≤ X ≤ 40 )?

Answer 1

a)
Here, the given values of lower limit, a = 10 and upper limit, b = 60
For Uniform distribution,
Mean = (a + b)/2
Mean = (10 + 60)/2 = 35

Standard Deviation = sqrt((b - a)^2/12)
Standard Deviation = sqrt((60 - 10)^2/12 = 14.4338

b)

Here, the given values of lower limit, a = 10 and upper limit, b = 60

For Uniform Distribution,
P(X >= x) = (b - x)/(b - a)
P(X >= 45) = (45 - 10)/(60 - 10)
P(X >= 45) = 0.3

c)

Here, the given values of lower limit, a = 10 and upper limit, b = 60

For Uniform Distribution,
P(x1 <= X <= x2) = (x2 - x1)/(b - a)
P(15 <= X <= 20) = (20 - 15)/(60 - 10)
P(15 <= X <= 20) = 0.1

Here, the given values of lower limit, a = 10 and upper limit, b = 60

For Uniform Distribution,
P(x1 <= X <= x2) = (x2 - x1)/(b - a)
P(35 <= X <= 40) = (40 - 35)/(60 - 10)
P(35 <= X <= 40) = 0.1

. P ( 15 ≤ X ≤ 20 ∪ 35 ≤ X ≤ 40 ) = 0.1 + 0.1 = 0.2