The Central Idea
Three continuous probability families appear repeatedly in probability, statistics, science, engineering, economics, and applied modeling:
- uniform distributions for equally dense outcomes across a bounded interval;
- normal distributions for symmetric variation concentrated near a mean;
- exponential distributions for waiting times associated with constant-rate Poisson processes.
Each family is defined by a probability density function and characterized by one or more parameters.
The Continuous Uniform Distribution
A continuous uniform distribution assigns constant probability density across an interval
\[ [a,b]. \]We write
\[ \boxed{X\sim\operatorname{Uniform}(a,b)}. \]Uniform Density Function
The density must have total area \(1\). Since the support has width
\[ b-a, \]the constant height must be
\[ \frac{1}{b-a}. \]Therefore,
\[ \boxed{ f_X(x)= \begin{cases} \frac{1}{b-a}, & a\le x\le b,\\ 0, & \text{otherwise}. \end{cases}} \]Why the Uniform Density Is Constant
Equal-length subintervals receive equal probabilities.
If two subintervals inside \([a,b]\) have the same width, then they have the same area under the density curve and therefore the same probability.
Uniform Interval Probabilities
If
\[ X\sim\operatorname{Uniform}(a,b) \]and
\[ a\le cTherefore,
\[ \boxed{ P(c\le X\le d)=\frac{d-c}{b-a} }. \]Example: Uniform Probability
Let
\[ X\sim\operatorname{Uniform}(0,10). \]Find
\[ P(2\le X\le5). \]The desired interval has width \(3\), while the full support has width \(10\).
\[ P(2\le X\le5) = \frac{3}{10}. \]Therefore,
\[ \boxed{P(2\le X\le5)=0.3}. \]Uniform Mean
The continuous uniform distribution is symmetric about the midpoint of its support.
Therefore,
\[ \boxed{ \mathbb{E}[X]=\frac{a+b}{2} }. \]Uniform Variance
For
\[ X\sim\operatorname{Uniform}(a,b), \]the variance is
\[ \boxed{ \operatorname{Var}(X)=\frac{(b-a)^2}{12} }. \]The standard deviation is
\[ \boxed{ \sigma_X=\frac{b-a}{\sqrt{12}} }. \]The Uniform CDF
The cumulative distribution function is
\[ \boxed{ F_X(x)= \begin{cases} 0, & xThe CDF rises linearly because the density is constant.
The Normal Distribution
The normal distribution is a continuous distribution with a symmetric bell-shaped density.
It is determined by two parameters:
- the mean \(\mu\), which controls location;
- the standard deviation \(\sigma>0\), which controls spread.
We write
\[ \boxed{ X\sim N(\mu,\sigma^2) }. \]The Normal Density Function
The density of
\[ X\sim N(\mu,\sigma^2) \]is
\[ \boxed{ f_X(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} }. \]Although the formula is important, normal probabilities are usually calculated through standardization, tables, calculators, or software rather than by integrating this density directly.
Shape of the Normal Distribution
A normal density is:
- symmetric about \(\mu\);
- highest at the mean;
- unimodal;
- spread over the entire real line;
- asymptotic to the horizontal axis in both tails.
The total area under the curve is \(1\).
Mean, Median, and Mode
Because a normal distribution is symmetric and unimodal,
\[ \boxed{ \text{mean}=\text{median}=\text{mode}=\mu }. \]The center of the distribution occurs at the same location under all three measures.
Effect of the Mean
Changing \(\mu\) shifts the entire normal distribution horizontally without changing its shape.
Larger \(\mu\) moves the distribution to the right. Smaller \(\mu\) moves it to the left.
Effect of the Standard Deviation
Increasing \(\sigma\) spreads the distribution out and lowers its peak.
Decreasing \(\sigma\) concentrates the distribution more tightly around the mean and raises the peak.
The total area remains \(1\).
The Standard Normal Distribution
The standard normal distribution has
\[ \mu=0 \]and
\[ \sigma=1. \]We write
\[ \boxed{Z\sim N(0,1)}. \]Standardization
If
\[ X\sim N(\mu,\sigma^2), \]then
\[ \boxed{ Z=\frac{X-\mu}{\sigma} } \]has the standard normal distribution.
This transformation converts values into numbers of standard deviations above or below the mean.
Interpreting a z-Score
A z-score tells how far a value lies from the mean in standard-deviation units.
- \(z=0\): exactly at the mean;
- \(z=1\): one standard deviation above the mean;
- \(z=-2\): two standard deviations below the mean.
Example: Compute a z-Score
Suppose
\[ X\sim N(70,8^2) \]and consider
\[ x=86. \]Then
\[ z = \frac{86-70}{8} = \boxed{2}. \]The value \(86\) lies two standard deviations above the mean.
Normal Probabilities
Normal probabilities are areas under the normal density curve.
For example,
\[ P(X\le x) \]is the area to the left of \(x\).
Standardization converts this to
\[ P\left( Z\le \frac{x-\mu}{\sigma} \right). \]The Standard Normal CDF
The standard normal cumulative distribution function is commonly denoted
\[ \Phi(z)=P(Z\le z). \]Therefore,
\[ \boxed{ P(X\le x) = \Phi\left(\frac{x-\mu}{\sigma}\right) }. \]Normal Interval Probabilities
For
\[ aThen
\[ \boxed{ P(a\le X\le b) = \Phi(z_b)-\Phi(z_a) }. \]Symmetry of the Standard Normal Distribution
The standard normal distribution is symmetric about \(0\).
Therefore,
\[ \boxed{ \Phi(-z)=1-\Phi(z) }. \]This identity helps relate upper-tail and lower-tail probabilities.
The Empirical Rule
For a normal distribution, approximately:
- \(68\%\) of values lie within \(1\) standard deviation of the mean;
- \(95\%\) lie within \(2\) standard deviations;
- \(99.7\%\) lie within \(3\) standard deviations.
This is often called the 68-95-99.7 rule.
Using the Empirical Rule
If
\[ X\sim N(100,15^2), \]then approximately \(95\%\) of values lie between
\[ 100-2(15)=70 \]and
\[ 100+2(15)=130. \]Thus,
\[ \boxed{P(70\le X\le130)\approx0.95}. \]Normal Quantiles
Sometimes the probability is known and the corresponding value must be found.
If
\[ P(Z\le z_p)=p, \]then \(z_p\) is the standard normal \(p\)-quantile.
For a general normal variable,
\[ \boxed{ x_p=\mu+\sigma z_p }. \]The Exponential Distribution
The exponential distribution models continuous waiting times associated with a constant-rate Poisson process.
If events occur at rate
\[ \lambda>0 \]per unit time or exposure, then the waiting time \(T\) until the next event may be modeled as
\[ \boxed{ T\sim\operatorname{Exponential}(\lambda) }. \]The Exponential Density
The exponential probability density function is
\[ \boxed{ f_T(t)= \begin{cases} \lambda e^{-\lambda t}, & t\ge0,\\ 0, & t<0. \end{cases}} \]The density is largest near \(0\) and decreases as \(t\) increases.
The Exponential CDF
For \(t\ge0\),
\[ F_T(t) = \int_0^t\lambda e^{-\lambda u}\,du. \]Therefore,
\[ \boxed{ F_T(t)=1-e^{-\lambda t} }. \]Exponential Survival Probability
The probability of waiting longer than \(t\) is
\[ P(T>t) = 1-F_T(t). \]Therefore,
\[ \boxed{ P(T>t)=e^{-\lambda t} }. \]This survival function is central to exponential waiting-time models.
Example: Waiting More Than Two Hours
Suppose
\[ T\sim\operatorname{Exponential}(0.5) \]where time is measured in hours.
Then
\[ P(T>2) = e^{-0.5(2)} = e^{-1}. \]Therefore,
\[ \boxed{P(T>2)\approx0.3679}. \]Exponential Mean
For
\[ T\sim\operatorname{Exponential}(\lambda), \]the expected waiting time is
\[ \boxed{ \mathbb{E}[T]=\frac1\lambda }. \]Higher event rates correspond to shorter expected waiting times.
Exponential Variance
The variance is
\[ \boxed{ \operatorname{Var}(T)=\frac{1}{\lambda^2} }. \]Therefore,
\[ \boxed{ \sigma_T=\frac1\lambda }. \]The exponential distribution has equal mean and standard deviation.
Exponential Memorylessness
Like the geometric distribution, the exponential distribution is memoryless.
For \(s,t\ge0\),
\[ \boxed{ P(T>s+t\mid T>s)=P(T>t) }. \]Given that no event has occurred by time \(s\), the remaining waiting-time distribution is the same as it was at time \(0\).
Why Exponential Memorylessness Occurs
In a constant-rate Poisson process, the future event rate does not depend on how long the process has already gone without an event.
The process has no probabilistic memory of the elapsed waiting time.
Poisson Counts and Exponential Waiting Times
The Poisson and exponential distributions describe two sides of the same constant-rate process.
- Poisson: how many events occur in a fixed interval?
- Exponential: how long until the next event occurs?
If the Poisson rate is \(\lambda\), then the corresponding exponential waiting-time rate is also \(\lambda\).
Example: Rate and Mean Waiting Time
Suppose events occur at an average rate of
\[ 4 \]per hour.
Then
\[ \lambda=4 \]and the expected waiting time is
\[ \mathbb{E}[T] = \frac14\text{ hour}. \]Since one quarter of an hour is \(15\) minutes,
\[ \boxed{\mathbb{E}[T]=15\text{ minutes}}. \]Uniform, Normal, and Exponential Compared
| Uniform | Normal | Exponential |
|---|---|---|
| Bounded support | Support \((-\infty,\infty)\) | Support \([0,\infty)\) |
| Constant density | Bell-shaped density | Decreasing density |
| Parameters \(a,b\) | Parameters \(\mu,\sigma\) | Parameter \(\lambda\) |
| Symmetric | Symmetric | Right-skewed |
| Equal density over support | Most density near mean | Most density near zero |
Recognizing a Uniform Model
Consider a uniform model when:
- the variable is continuous;
- it is restricted to a bounded interval;
- equal-length subintervals should receive equal probability;
- there is no reason to favor one location within the interval over another.
Recognizing a Normal Model
A normal model may be reasonable when:
- the variable is continuous;
- values cluster around a central mean;
- deviations above and below the center are approximately symmetric;
- extreme values become progressively less common;
- the underlying mechanism aggregates many small influences.
Recognizing an Exponential Model
Consider an exponential model when:
- the variable is a nonnegative waiting time;
- events arise from a roughly constant-rate Poisson process;
- the memoryless assumption is reasonable;
- short waits are more common than long waits.
Common Error: Treating Density as Point Probability
For all three continuous distributions,
\[ P(X=x)=0 \]at every exact point.
Probability is obtained from area over intervals, not from the density height at a single point.
Common Error: Forgetting the Uniform Support
A uniform density is zero outside
\[ [a,b]. \]Probability assigned outside the support is therefore zero.
Common Error: Using the Wrong Uniform Height
The density height is not
\[ \frac1b \]unless \(a=0\).
The correct height is
\[ \boxed{\frac{1}{b-a}}. \]Common Error: Confusing Variance and Standard Deviation in Normal Notation
In
\[ X\sim N(\mu,\sigma^2), \]the second parameter is the variance
\[ \sigma^2, \]not the standard deviation \(\sigma\).
Common Error: Reversing the z-Score Formula
The correct standardization is
\[ \boxed{ z=\frac{x-\mu}{\sigma} }. \]Subtract the mean first, then divide by the standard deviation.
Common Error: Treating the Empirical Rule as Exact
The \(68\%-95\%-99.7\%\) values are useful approximations for normal distributions.
They are not exact probability identities.
Common Error: Treating \(\lambda\) as Mean Waiting Time
In an exponential distribution,
\[ \lambda \]is the event rate.
The mean waiting time is
\[ \boxed{\frac1\lambda}. \]Common Error: Using a Negative Exponential Waiting Time
Exponential waiting times are nonnegative:
\[ T\ge0. \]The density is zero for negative values.
A Reliable Procedure for Uniform Problems
- Identify the interval \([a,b]\).
- Write the density height \(1/(b-a)\).
- Translate the event into a subinterval.
- Compute probability as subinterval width divided by total width.
- Use \((a+b)/2\) for the mean.
- Use \((b-a)^2/12\) for the variance.
A Reliable Procedure for Normal Problems
- Identify \(\mu\) and \(\sigma\).
- Translate the probability statement into an interval or tail.
- Standardize each boundary using \(z=(x-\mu)/\sigma\).
- Use the standard normal CDF, table, calculator, or software.
- Subtract CDF values for interval probabilities.
- Use symmetry or complements when they simplify the calculation.
- Interpret the probability in the original units.
A Reliable Procedure for Exponential Problems
- Identify the constant event rate \(\lambda\).
- Confirm that the waiting time is nonnegative.
- Use \(P(T>t)=e^{-\lambda t}\) for survival probabilities.
- Use \(P(T\le t)=1-e^{-\lambda t}\) for cumulative probabilities.
- Use \(1/\lambda\) for the mean waiting time.
- Use \(1/\lambda^2\) for the variance.
- Check that time units and rate units are compatible.
Practice 1: Uniform Probability
Let
\[ X\sim\operatorname{Uniform}(2,8). \]Find
\[ P(3\le X\le5). \]Practice 2: Uniform Mean and Variance
Using Practice 1, find \(\mathbb{E}[X]\) and \(\operatorname{Var}(X)\).
Practice 3: z-Score
Suppose
\[ X\sim N(50,10^2). \]Find the z-score corresponding to
\[ x=65. \]Practice 4: Interpret a z-Score
What does
\[ z=-1.5 \]mean relative to the mean?
Practice 5: Empirical Rule
Suppose
\[ X\sim N(100,20^2). \]Approximately what interval contains \(95\%\) of the distribution?
Practice 6: Exponential Survival Probability
Let
\[ T\sim\operatorname{Exponential}(0.25). \]Find
\[ P(T>4). \]Practice 7: Exponential Cumulative Probability
Using Practice 6, find
\[ P(T\le4). \]Practice 8: Exponential Mean and Variance
Let
\[ T\sim\operatorname{Exponential}(2). \]Find the mean and variance.
Practice 9: Choose the Model
Identify the most appropriate family.
- A random point is equally likely to occur anywhere between \(0\) and \(1\).
- A symmetric measurement varies around a central value with bell-shaped behavior.
- The waiting time until the next event in a constant-rate Poisson process.
Practice 10: Identify the Error
A student says that for
\[ X\sim N(40,9), \]the standard deviation is \(9\).
Explain the error.
Practice Answers
Practice 1
\[ P(3\le X\le5) = \frac{5-3}{8-2} = \frac{2}{6} = \boxed{\frac13}. \]Practice 2
\[ \mathbb{E}[X] = \frac{2+8}{2} = \boxed{5}. \] \[ \operatorname{Var}(X) = \frac{(8-2)^2}{12} = \frac{36}{12} = \boxed{3}. \]Practice 3
\[ z = \frac{65-50}{10} = \boxed{1.5}. \]Practice 4
The value lies
\[ \boxed{1.5} \]standard deviations below the mean.
Practice 5
Approximately \(95\%\) lies within two standard deviations:
\[ 100\pm2(20). \]Therefore,
\[ \boxed{60\le X\le140}. \]Practice 6
\[ P(T>4) = e^{-0.25(4)} = e^{-1}. \]Therefore,
\[ \boxed{P(T>4)\approx0.3679}. \]Practice 7
\[ P(T\le4) = 1-e^{-1}. \]Therefore,
\[ \boxed{P(T\le4)\approx0.6321}. \]Practice 8
\[ \mathbb{E}[T] = \frac12. \] \[ \operatorname{Var}(T) = \frac{1}{2^2} = \frac14. \]Therefore,
\[ \boxed{ \mathbb{E}[T]=\frac12,\qquad \operatorname{Var}(T)=\frac14 }. \]Practice 9
- Uniform.
- Normal.
- Exponential.
Practice 10
In the notation
\[ N(\mu,\sigma^2), \]the second parameter is the variance. Thus
\[ \sigma^2=9 \]and
\[ \boxed{\sigma=3}. \]What to Remember
For
\[ X\sim\operatorname{Uniform}(a,b), \]the density is
\[ \boxed{ f_X(x)=\frac{1}{b-a} } \]on \([a,b]\), with
\[ \boxed{ \mathbb{E}[X]=\frac{a+b}{2} } \]and
\[ \boxed{ \operatorname{Var}(X)=\frac{(b-a)^2}{12} }. \]For
\[ X\sim N(\mu,\sigma^2), \]standardization uses
\[ \boxed{ Z=\frac{X-\mu}{\sigma} }. \]For
\[ T\sim\operatorname{Exponential}(\lambda), \]the survival probability is
\[ \boxed{ P(T>t)=e^{-\lambda t} }. \]Its mean and variance are
\[ \boxed{ \mathbb{E}[T]=\frac1\lambda } \]and
\[ \boxed{ \operatorname{Var}(T)=\frac1{\lambda^2} }. \]The central modeling distinction is: uniform distributions spread density evenly across a bounded interval, normal distributions model symmetric bell-shaped variation, and exponential distributions model continuous waiting times in constant-rate processes.
Where We Go Next
The distributions studied so far describe one random variable at a time. Many applications involve several random quantities whose behavior must be studied together.
The next lesson, Joint Distributions, develops probability models for two or more random variables and introduces marginal, conditional, and dependence relationships.