Continuous Random Variables &
Probability
Density Functions
Applied Statistics
MTH-361A | Spring 2026 | University of Portland
Objectives
- Develop an understanding of the exponential random
variable
- Develop an understanding of the normal random
variable
- Know how to compute probabilities of the exponential and
normal random variables
Rate to “Success” (Discrete)
Geometric R.V. (Revisited)
- Independent Bernoulli trials with “success” probability \(p\)
- Count the number of “fail” trials before the first “success”
- The probability of getting \(k\)
number of “fail” before a “success” follows the Geometric PMF
- The expected number of fails before a “success” is \(\frac{1-p}{p}\)
\(\star\) In a time-based
interpretation, if events occur in discrete time steps, the geometric
r.v. represents the number of time steps required until an event of
interest happens.
Assessment Retakes
A professor allows students to take a short assessment quiz, and if
they do not pass, they can revise their answers and retake the quiz in
the next session. The probability that a student passes on any given
attempt is \(p=0.40\), and attempts
continue until the student passes.
Let \(X\) be the number of “fail”
attempts before the student get a “pass”.
Information Given:
- R.V. is \(X \sim
\text{Geom}(p)\)
- Expected value is \(\text{E}(X) =
\frac{1-p}{p}\)
- What is the expected number of “fail” until a student get a “pass”?
\[\text{E}(X) = \frac{1-0.40}{0.40} =
1.5\]
\(\star\) On average, the student
would “fail” on their 1st attempt before they get a “pass”. The chances
of a “pass” with \(1\) “fail” attempt
is \(0.64\).
Computing Probabilities:
- Geometric PMF is \(P(X=k) = (1-p)^k
p\)
- What is the probability that a student get a “pass” with at most
\(1\) “fail” attempt? \[
\begin{aligned}
P(X \le 1) & = \sum_{k=0}^1 P(X=k) \\
& = (0.60)^0 (0.40) + (0.60)^1 (0.40) \\
P(X \le 1) & = 0.64
\end{aligned}
\]
Using R:
## [1] 0.64
\(\star\) Note that the R function
pgeom is the Geometric CDF.
Rate to “Success” (Continuous)
Exponential R.V.
- Unit length until an event occurs
- “success” event happens at a constant rate
\(\star\) In a time-based
interpretation, if events occur in continuous time, the exponential r.v.
represents the length of time required until an event of interest
happens.
The Exponential R.V.
The exponential r.v. is a continuous r.v. that
models the time until an event occurs, given that the event happens at a
constant rate over time called \(\lambda\): \[X
\sim \text{Exp}(\lambda)\]
Sample Space:
- \(x \in [0,\infty)\) because the
length of time until an event occurs can be any positive unit
length
Rate Parameter
- \(\lambda\) is reciprocal to the
average time of an event occurs.
- This means that \(\frac{1}{\lambda}\) is the average time
until a “success” occurs.
The Exponential R.V.: PDF
The exponential r.v. \(X \sim
\text{Exp}(\lambda)\) has infinite possible outcomes (or infinite
sized sample space) where \(\lambda >
0\) is the rate of “success” with PDF given as \[f(x) = \lambda e^{-\lambda x}, \quad x \ge
0\]
\(\star\) The exponential r.v.
models the unit length until an event happens.
Assessment Completion Length
A class of students is taking a quiz, and the time it takes for
students to finish the quiz follows an exponential r.v., assuming
unlimited quiz time allocation. On average, a student takes 15 minutes
to complete the quiz.
Let \(X\) represent the time to
finish the quiz.
Information Given:
- The time it takes for a student to finish a quiz follows an
exponential distribution: \[X \sim
\text{Exp}(\lambda)\] where \(\lambda\) is the rate parameter.
- \(\lambda = \frac{1}{15}\), which
can be interpreted in two ways:
- on average, one student finishes the quiz in 15 minutes
- on average, a student finishes \(\frac{1}{15}\) part of the quiz per
minute
Computing Probabilities:
- The exponential PDF is \[f(x) =
\frac{1}{15} e^{-\frac{1}{15} x}\]
- What is the probability that a student finishes a quiz less than or
equal to 15 minutes? \[
\begin{aligned}
P(X \le 15) & = \int_0^{15} f(x) \ dx \\
& = \int_0^{15} \frac{1}{15} e^{-\frac{1}{15} x} \ dx \\
P(X \le 15) & \approx 0.632
\end{aligned}
\]
Using R:
## [1] 0.6321206
\(\star\) Note that the R function
pexp is the Exponential CDF.
Probabilities of PDFs
A Probability Density Function (PDF) \(f(x)\) describes the likelihood of a
continuous r.v. taking a specific value from a continuous interval.
The probability of a single point is zero for continuous
distributions: \[P(X = x) = 0, \ \text{for
any } x\] because continuous distributions are defined over an
infinite number of possible values, and the probability at a single
point is infinitesimally small.
Instead, we calculate probabilities over intervals using
integration:
- Cumulative \(\longrightarrow \displaystyle
P(X \le x) = \int_{-\infty}^x f(t) \ dt\)
- Interval \(\longrightarrow \displaystyle
P(a \le X \le b) = \int_a^b f(t) \ dt\)
\(\star\)
- PMF (for discrete r.v.): \(P(X=k)\)
is a meaningful probability
- PDF (for continuous r.v.): \(P(X=x)=0\), but density matters for
interval probabilities.
Flipping \(n\) Coins
Suppose we conduct an experiment of flipping \(n\) fair coins. The sample space \(S\) contains all possible outcomes, where
the number of outcomes is \(2^n\).
Visualizing the possible outcomes using Pascal’s
triangle
\(\star\) Pascal’s Triangle helps us
visualize the total possible sequences of “success” (\(H\)) outcomes given \(n\) independent trials.
Counting the Number of \(H\)
outcomes (Discrete)
Let \(X\) be the r.v. that counts
the number of \(H\) outcomes in \(n\) trials.
Pascal’s triangle helps us count:
\(\star\) The binomial coefficient
tells you how many ways \(k\) “success”
outcomes can occur in \(n\) trials,
corresponding to the \(k\)-th column in
the \(n\)-th row of Pascal’s
Triangle.
The Binomial Coefficient
Pascal’s Triangle and Combinations This formula
calculates the number of ways to choose \(k\) elements from a set of \(n\). Each number in Pascal’s Triangle
corresponds to a combination. Also known as the binomial
coefficient.
Probability of Observing \(H\)
outcomes in \(n\) Trials
Suppose we want to compute the probability of observing a certain
number of “success” (\(H\)) outcomes in
\(n\) trials. Note that for a fair coin
the probability of “success” is \(\frac{1}{2}\).
Pascal’s triangle helps us compute these
probabilities:
Flipping “enough” \(n\) Coins
(Continuous)
To best illustrate this idea, here is a video.
\(\star\) The video explains how
random events can produce a predictable pattern, specifically, how many
random outcomes together form a normal “bell-curve” distribution.
Normal Approximation to the Binomial
\(\star\) The Binomial distribution
is approximately the normal distribution given large enough samples
because of the Law of Large Numbers.
The Normal R.V.
A normal r.v. is a type of continuous r.v. whose
probability distribution follows the normal distribution, also
known as the Gaussian distribution. The normal distribution is
characterized by two parameters, \(\mu\) as the mean and \(\sigma^2\) as the variance: \[X \sim \text{N}(\mu,\sigma^2)\]
Sample Space:
- \(x \in (-\infty,\infty)\) because
the normal r.v. can take any value from the entire real number line and
it is a continuous random variable.
Parameters
- \(\mu\) is the mean (center) of the
distribution.
- \(\sigma^2\) is the variance, which
measures the spread of the distribution.
The Normal R.V.: PDF
The normal r.v. \(X \sim
\text{N}(\mu,\sigma^2)\) has infinite possible outcomes (or
infinite sized sample space) where \(\mu\) is the mean and \(\sigma^2\) is the variance with PDF given
as \[f(x) = \frac{1}{\sigma \sqrt{2 \pi}}
\exp{\left(-\frac{1}{2} \left( \frac{x - \mu}{\sigma} \right)^2\right)},
\ -\infty < x < \infty\] where the term \(\exp{(\cdot)}\) is the exponential function
\(e^{(\cdot)}\). We write it
this way because the function the exponent term is complicated.
\(\star\) The normal r.v. often
approximates the distribution of many types of data, especially when
there are large numbers of independent factors contributing to the
outcome.
The “z-score”
A standardized score is a measure on how many
standard deviations a value is from the mean. This computed by the
z-score formula: \[z =
\frac{x - \mu}{\sigma}.\]
Using the z-score transformation, the normal PDF reduces to the
standard normal PDF, \[f(z)
= \frac{1}{\sqrt{2 \pi}} \exp{\left(-\frac{z^2}{2}\right)}, \ -\infty
< z < \infty.\]
\(\star\) The standardized score is
used to compare two normal distributions with different means and
variances.
Normal Approximation to the Binomial
When can we use it?
The binomial distribution \(X \sim
\text{Binom}(n,p)\) can be approximated by a normal distribution
when:
- \(np\) (expected “success”) and
\(np(1-p)\) (expected “fail”) are
sufficiently large.
Approximation Formula
- If \(X \sim \text{Binom}(n,p)\),
then for large \(n\), \[X \approx \text{N}(\mu,\sigma^2)\] where
\(\mu = np\) and \(\sigma^2 = np(1-p)\).
\(\star\) The normal approximation
simplifies binomial probability calculations for large \(n\).
Percentiles (1/2)
A large introductory statistics course has a final exam consisting of
many independent multiple-choice questions. Each question has a
probability of “success” \(p = 0.70\)
(a typical student answers correctly with probability \(0.7\)). The exam has \(n = 100\) questions.
Let \(X\) represent the number of
correct answers on the exam.
Information Given:
The number of correct answers follows a Binomial distribution:
\[X \sim \text{Binom}(n,p)\] where
\(n = 100\) and \(p = 0.70\).
Mean and standard deviation:
\[
\begin{aligned}
\mu & = np \\
& = 100(0.70) = 70 \\
\sigma^2 & = \sqrt{np(1-p)} \\
& = \sqrt{100(0.70)(1-0.70)} \approx 4.583
\end{aligned}
\]
Normal Approximation:
- Because the sample size is large and both
\[
\begin{aligned}
np & = 100(0.70) = 70 \\
n(1-p) & = 100(1-0.70) = 30
\end{aligned}
\] are greater than 10, the Binomial distribution can be
approximated by a Normal distribution: \[X
\approx \text{N}(np,np(1-p)).\]
Percentiles (2/2)
The \(90\)th percentile of the
standard normal distribution is \(\displaystyle z_{0.90} \approx 1.283\).
That is interpreted as \(1.283\)
standard deviations from the mean.
Computing Percentiles:
- Convert the standardized score to the exam score using the “z-score”
formula:
\[
\begin{aligned}
x & = \mu + z \sigma \\
x & = 70 + (1.283)(4.583) \\
x & \approx 75.88
\end{aligned}
\]
- This suggests a score of about \(76\) correct answers is roughly the \(90\)th percentile.
Interpretation:
- A student who scores about \(\frac{76}{100}\) performed better than
approximately \(90\)% of the
class.
Using R:
qnorm(0.90, 100*0.7, sqrt(100*0.7*0.3))
## [1] 75.87281
\(\star\) The R function
qnorm is the inverse CDF of the Normal distribution.
