Question

# 1. Assume the waiting time at the BMV is uniformly distributed from 10 to 60 minutes,...

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 )?

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

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