intermediate 30 minutes

Continuous Probability Distributions

Explore continuous distributions beyond the normal: exponential, uniform, and more. Learn probability density functions and their applications.

On This Page
Advertisement

Continuous vs Discrete Distributions

AspectDiscreteContinuous
ValuesCountable (0, 1, 2, …)Uncountable (any value in range)
FunctionPMF: P(X = x)PDF: f(x)
P(X = specific value)Can be positiveAlways 0
ProbabilitiesSum upIntegrate (area under curve)

Probability Density Function (PDF)

The probability density function f(x) describes the relative likelihood of values. Probability = area under the curve.

PDF Properties
  1. f(x)0f(x) \geq 0 for all x
  2. Total area under curve = 1
  3. P(aXb)=abf(x)dxP(a \leq X \leq b) = \int_a^b f(x) \, dx (area between a and b)
Mean and Variance (Continuous)

μ=E(X)=xf(x)dx\mu = E(X) = \int_{-\infty}^{\infty} x \cdot f(x) \, dx

σ2=Var(X)=(xμ)2f(x)dx\sigma^2 = Var(X) = \int_{-\infty}^{\infty} (x-\mu)^2 \cdot f(x) \, dx


1. Uniform Distribution

The uniform distribution assigns equal probability to all values in an interval [a, b].

Continuous Uniform Distribution

XUniform(a,b)X \sim \text{Uniform}(a, b)

f(x)=1bafor axbf(x) = \frac{1}{b-a} \quad \text{for } a \leq x \leq b

Mean: μ=a+b2\mu = \frac{a+b}{2}

Variance: σ2=(ba)212\sigma^2 = \frac{(b-a)^2}{12}

Random Number Generator

A random number is generated uniformly between 0 and 10.

X ~ Uniform(0, 10)

P(X < 3)? P(X<3)=30100=0.3P(X < 3) = \frac{3-0}{10-0} = 0.3

P(2 < X < 7)? P(2<X<7)=72100=0.5P(2 < X < 7) = \frac{7-2}{10-0} = 0.5

Mean and SD:

  • μ = (0+10)/2 = 5
  • σ² = (10-0)²/12 = 100/12 = 8.33
  • σ = 2.89
Waiting for a Bus

Buses arrive every 15 minutes. You arrive at a random time.

Wait time X ~ Uniform(0, 15)

P(wait less than 5 minutes)? P(X<5)=515=13=0.333P(X < 5) = \frac{5}{15} = \frac{1}{3} = 0.333

Expected wait time: μ=0+152=7.5 minutes\mu = \frac{0+15}{2} = 7.5 \text{ minutes}


2. Exponential Distribution

The exponential distribution models time between events in a Poisson process. It’s the continuous counterpart to the geometric distribution.

Exponential Distribution

XExponential(λ)X \sim \text{Exponential}(\lambda)

f(x)=λeλxfor x0f(x) = \lambda e^{-\lambda x} \quad \text{for } x \geq 0

Mean: μ=1λ\mu = \frac{1}{\lambda}

Variance: σ2=1λ2\sigma^2 = \frac{1}{\lambda^2}

CDF: P(Xx)=1eλxP(X \leq x) = 1 - e^{-\lambda x}

Customer Arrivals

Customers arrive at rate λ = 4 per hour (Poisson process).

Time between arrivals X ~ Exponential(4)

Mean time between customers: μ=14=0.25 hours=15 minutes\mu = \frac{1}{4} = 0.25 \text{ hours} = 15 \text{ minutes}

P(next customer arrives within 10 minutes = 1/6 hour)? P(X1/6)=1e4×1/6=1e0.667=10.513=0.487P(X \leq 1/6) = 1 - e^{-4 \times 1/6} = 1 - e^{-0.667} = 1 - 0.513 = 0.487

P(wait more than 30 minutes = 0.5 hours)? P(X>0.5)=e4×0.5=e2=0.135P(X > 0.5) = e^{-4 \times 0.5} = e^{-2} = 0.135

Memoryless Property

The exponential distribution is memoryless: the probability of waiting t more time units doesn’t depend on how long you’ve already waited.

P(X>s+tX>s)=P(X>t)P(X > s + t \,|\, X > s) = P(X > t)

Memoryless Property

Light bulb lifetime ~ Exponential with mean 1000 hours.

If a bulb has lasted 500 hours, P(lasting 200 more hours)?

Same as P(new bulb lasting 200 hours): P(X>200)=e200/1000=e0.2=0.819P(X > 200) = e^{-200/1000} = e^{-0.2} = 0.819

The bulb doesn’t “remember” it’s been on for 500 hours!


3. Chi-Square Distribution

The chi-square distribution arises when you sum squared standard normal variables. It’s crucial for hypothesis testing.

Chi-Square Distribution

If Z1,Z2,,ZkZ_1, Z_2, \ldots, Z_k are independent standard normal variables:

X=Z12+Z22++Zk2χ2(k)X = Z_1^2 + Z_2^2 + \cdots + Z_k^2 \sim \chi^2(k)

Mean: μ=k\mu = k (degrees of freedom)

Variance: σ2=2k\sigma^2 = 2k

Properties

  • Only takes positive values (x ≥ 0)
  • Right-skewed (especially for small df)
  • Approaches normal as df increases
  • Used in chi-square tests, confidence intervals for variance
Chi-Square Critical Values

Common critical values for chi-square tests:

dfχ²₀.₉₅ (right tail = 0.05)χ²₀.₀₅ (left tail = 0.05)
13.840.004
511.071.15
1018.313.94
2031.4110.85

4. t-Distribution

The t-distribution (Student’s t) is similar to the normal but with heavier tails. It’s used when estimating means with unknown population standard deviation.

t-Distribution

Tt(df)T \sim t(df)

Mean: μ = 0 (for df > 1)

Variance: σ2=dfdf2\sigma^2 = \frac{df}{df-2} (for df > 2)

Properties

  • Symmetric and bell-shaped like normal
  • Heavier tails than normal (more probability in extremes)
  • Approaches standard normal as df → ∞
  • At df ≈ 30, very close to normal
t vs z Critical Values
Confidencez (normal)t (df=10)t (df=30)
90%1.6451.8121.697
95%1.962.2282.042
99%2.5763.1692.750

With small samples, you need to go further from the mean for the same confidence!


5. F-Distribution

The F-distribution is the ratio of two chi-square variables. It’s used in ANOVA and comparing variances.

F-Distribution

If X1χ2(df1)X_1 \sim \chi^2(df_1) and X2χ2(df2)X_2 \sim \chi^2(df_2) are independent:

F=X1/df1X2/df2F(df1,df2)F = \frac{X_1/df_1}{X_2/df_2} \sim F(df_1, df_2)

Mean: μ=df2df22\mu = \frac{df_2}{df_2-2} (for df₂ > 2)

Properties

  • Only takes positive values
  • Right-skewed
  • Has TWO degrees of freedom parameters
  • Used in F-tests, ANOVA, regression analysis

6. Beta Distribution

The beta distribution is flexible for modeling proportions and probabilities.

Beta Distribution

XBeta(α,β)X \sim \text{Beta}(\alpha, \beta)

Values in [0, 1]

Mean: μ=αα+β\mu = \frac{\alpha}{\alpha + \beta}

Different α and β values create different shapes:

  • α = β = 1: Uniform distribution
  • α = β > 1: Symmetric, bell-shaped
  • α > β: Left-skewed
  • α < β: Right-skewed

Comparison of Continuous Distributions

DistributionSupportParametersCommon Use
Normal(-∞, ∞)μ, σNatural phenomena
Uniform[a, b]a, bRandom number generation
Exponential[0, ∞)λWait times
Chi-square[0, ∞)dfVariance tests
t(-∞, ∞)dfSmall sample means
F[0, ∞)df₁, df₂ANOVA, variance comparison
Beta[0, 1]α, βProportions

Choosing the Right Distribution

Identifying Distributions

Scenario 1: Time until next earthquake → Exponential (waiting time for rare event)

Scenario 2: Test scores of large population → Normal (central limit theorem)

Scenario 3: Bus arrival time within a 10-minute window → Uniform (any time equally likely)

Scenario 4: Sample variance of normal data → Chi-square (sum of squared normals)

Scenario 5: Batting average, proportion successful → Beta (values between 0 and 1)

Relationships Between Distributions

Many distributions are related:

  1. Normal → Chi-square: Z2χ2(1)Z^2 \sim \chi^2(1)
  2. Chi-square + Chi-square → F: Ratio of chi-squares
  3. Normal + Chi-square → t: Z divided by sqrt(χ²/df)
  4. Poisson → Exponential: Time between Poisson events
  5. Binomial → Normal: For large n (CLT)
  6. Exponential → Gamma: Sum of exponentials

Summary

In this lesson, you learned:

  • Continuous distributions use PDFs; probability = area under curve
  • Uniform: Equal probability over an interval
  • Exponential: Time between events (memoryless)
  • Chi-square: Sum of squared normals (variance tests)
  • t-distribution: Like normal but heavier tails (small samples)
  • F-distribution: Ratio of chi-squares (ANOVA)
  • Beta: Flexible distribution for proportions
  • Distributions are interconnected through mathematical relationships

Practice Problems

1. X ~ Uniform(5, 15). Find: a) P(X < 8) b) P(7 < X < 12) c) Mean and standard deviation

2. Phone calls arrive at rate 2 per hour (Poisson). Time between calls is exponential. a) Mean time between calls? b) P(more than 1 hour between calls)? c) P(call arrives in next 15 minutes)?

3. A chi-square distribution has df = 8. What is: a) The mean? b) The variance?

4. For a t-distribution with df = 15, the critical value for a two-tailed test at α = 0.05 is 2.131. What does this mean?

Click to see answers

1. a) P(X < 8) = (8-5)/(15-5) = 3/10 = 0.3 b) P(7 < X < 12) = (12-7)/(15-5) = 5/10 = 0.5 c) μ = (5+15)/2 = 10 σ² = (15-5)²/12 = 100/12 = 8.33 σ = 2.89

2. λ = 2 per hour a) Mean = 1/λ = 0.5 hours = 30 minutes b) P(X > 1) = e^(-2×1) = e^(-2) ≈ 0.135 c) P(X < 0.25) = 1 - e^(-2×0.25) = 1 - e^(-0.5) = 1 - 0.607 ≈ 0.393

3. a) Mean = df = 8 b) Variance = 2×df = 2×8 = 16

4. If the test statistic |t| > 2.131, we reject the null hypothesis. This leaves 2.5% probability in each tail of the distribution.

Next Steps

Continue building your statistical knowledge:

Advertisement

Was this lesson helpful?

Help us improve by sharing your feedback or spreading the word.