Continuous Random Variables and Probability Distributions
Every CRV has a probability density function associated with it,
pdf.
If X is a CRV call the pdf f(x).
Properties of any pdf:
f(x) ≥ 0 for all real x

 f ( x)dx  1

For CRV’s we associate probability with the area under the pdf (or
density curve) more specifically, if X is a CRV with pdf f(x) then
b
P(a  X  b)   f ( x)dx
a
Note that this implies some pretty important results:
Let y be any number: P(X = y) = 0
For CRV’s we never consider the probability that X is equal to any
number.
We always talk about P(X < y) or P(X > y) or P (a < X < b).
Note because the P(X = y) = 0 means that
P(X < y) = P(X ≤ y) and P(X > y) = P(X ≥ y)
Typically CRV’s are used for measurements and not counting.
However, because DRV’s can get difficulty to work with when
dealing with many values, CRV’s are used to approximate the
DRV’s.
Don’t not worry we will not have to integrate in this class. Most
pdf’s are not integrable anyway. Most of the important pdf’s have
been tableized or have functions which are simple functions.
The simplest CRV is the Uniform distribution.
X ~ Unif (a, b)
f(x) = 1 / (b – a)
for a ≤ x ≤ b
f(x) = 0
otherwise
μ = (a + b) / 2
σ2 = (b – a) 2 / 12
Since a and b are constants, the pdf is just a horizontal line, at one
over the length of the interval (a – b).
Finding probabilities just involves finding the area of rectangles.
Ex
Let X ~ Unif (0, 2)
f (x) = ½
if 0 ≤ x ≤ 2
0
otherwise
Draw Picture.
P(X < 1.5) = 1.5 * ½ = .75
P(X > .25) = 1.75 * ½ = .875
P (.25 < X < 1.5) = 1.25 * ½ = .625 = P(X < 1.5) – P(X < .25)
P(X = 1) = 0
Ex.
The amount of time a bus is late is uniformly distributed
with parameters 0 and 10. Find probability that a bus is less than 7
minutes late? Find probability that a bus is more than 2 minutes
late? Find probability that a bus is between 2 and 7 minutes late?
f(x) = x/10 = 0.1x
for 0 < X < 10
P(X < 7) = .1 * 7 = .7
P(X > 2) = .1 * 8 = .8
P (2 < X < 7) = .1 * 5 = .5 = P(X < 7) – P(X < 2)
Ex. Continued.
Given that the bus is 5 minutes late, Find the probability that is
more than 7 minutes late?
Find the 75th percentile.
Assuming that the bus’ arrivals are independent, find the
probability that the bus will be more than 8 minutes late the next
2 days?
In class practice:
Section 5.8 (page 238/248) 2, 4, 6, 8
Homework: 1, 3, 5, 7
The Exponential Distribution.
Called the waiting time or memory-less distribution.
X ~ Exp(μ) for μ > 0
1 −1𝑥
𝜇
𝑖𝑓 𝑥 ≥ 0
𝑓(𝑥) = 𝜇 𝑒
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
E(X) = μ and σ2 = μ2 σ = μ
note that f(x) > 0
Let m = 1/μ and let m be the parameter of the exponential
distribution, so we can rewrite
𝑚𝑒 −𝑚𝑥
𝑖𝑓 𝑥 ≥ 0
𝑓(𝑥) =
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
After integrating we find that:
P(X > a) = e –ma = A
P(X < a) = 1 – e –ma = 1 – A
Ex. Let X ~ Exp(m = 1/10)
P(X > 15) = e –15/10 = e –1.5 = .2231
P(X < 20) = 1 - e –20/10 = 1- e –2 = .8647
P(15 < X < 20) = P(X < 20) – P(X < 15) = .8647 - .7761 = .0886
E(X) = µ = 10
V(X) = σ2 = 100
σ = 10
Ex. Suppose that the length of time in hours between emergency
arrivals at a certain hospital is modeled as an exponential
distribution with μ = 2. Find the probability that:
a. More than 5 hours pass without an emergency.
b. Between 2.5 and 5 hours pass without an emergency.
c. More than 10 hours pass without an emergency given
more than 5 hours pass without an emergency.
In class examples: 12, 14
Homework : 9, 11, 13