Continuous Probability Distributions

A continuous uniform distribution is a probability distribution in which all values between two specified limits (\(a\) and \(b\)) are equally likely. The probability is uniformly spread over the interval \([a, b]\).

Probability Density Function (PDF): \[ f(x) = \begin{cases} \frac{1}{b-a}, & a \leq x \leq b \\ 0, & \text{otherwise} \end{cases} \]

Mean (Expected Value): \[ E(X) = \frac{a + b}{2} \]

Variance: \[ \text{Var}(X) = \frac{(b - a)^2}{12} \]

Cumulative Distribution Function (CDF): \[ F(x) = \begin{cases} 0, & x < a \\ \frac{x - a}{b - a}, & a \leq x \leq b \\ 1, & x > b \end{cases} \]

• The variable is continuous and can take any real value within a specified range.

• Every value between \(a\) and \(b\) is equally likely.

• The problem may mention a flat or constant probability density.

• Usually used for time, distance, or measurement-based outcomes.

Keywords to look for:

• “equally likely between a and b”

• “randomly chosen between two continuous values”

• “uniformly distributed over an interval”

A commuter arriving at a bus stop finds that the time between successive buss is uniformly distributed between 0 and 20 minutes. Let \(X\) be the waiting time (in minutes) from the commuter's arrival until the next bus. Find:

• [i.] The PDF of \(X\).

• [ii.] The probability that the commuter waits more than 7 minutes.

• [iii.] The expected waiting time and the variance.

Solution:

Since interarrival times are uniform on \([0,20]\), we have \(X\sim U(0,20)\).

• [i.] The PDF of a continuous uniform on \([a,b]\) is \[ f_X(x)=\begin{cases} \dfrac{1}{b-a}, & a\le x\le b,\\[4pt] 0, & \text{otherwise.} \end{cases} \] Here \(a=0,b=20\), so \[ f_X(x)=\frac{1}{20},\quad 0\le x\le 20. \]

• [ii.] For \(X\sim U(0,20)\), \[ P(X>7)=\frac{b-7}{b-a}=\frac{20-7}{20-0}=\frac{13}{20}=0.65. \] (Alternatively integrate: \(\int_{7}^{20}\tfrac{1}{20}\,dx=\tfrac{13}{20}\).)

• [iii.] The mean and variance of \(U(a,b)\) are \[ E[X]=\frac{a+b}{2},\qquad \mathrm{Var}(X)=\frac{(b-a)^2}{12}. \] Thus \[ E[X]=\frac{0+20}{2}=10,\qquad \mathrm{Var}(X)=\frac{20^2}{12}=\frac{400}{12}=\frac{100}{3}\approx 33.3333. \]

A laboratory thermostat produces temperatures uniformly between \(10^\circ\)C and \(50^\circ\)C during a calibration process. Let \(X\) denote the recorded temperature. Compute:

• [i.] The mean and variance of \(X\).

• [ii.] The probability that \(X<30^\circ\)C.

Solution:

Here \(X\sim U(10,50)\) with \(a=10\), \(b=50\).

• [i.] Mean: \[ E[X]=\frac{a+b}{2}=\frac{10+50}{2}=30. \] Variance: \[ \mathrm{Var}(X)=\frac{(b-a)^2}{12}=\frac{(50-10)^2}{12}=\frac{40^2}{12}=\frac{1600}{12}=\frac{400}{3}\approx 133.3333. \]

• [ii.] Since the distribution is uniform, \[ P(X<30)=\frac{30-10}{50-10}=\frac{20}{40}=\frac{1}{2}=0.5. \] (Equivalently, integrate \(f_X(x)=1/40\) from \(10\) to \(30\): \(\int_{10}^{30}\tfrac{1}{40}\,dx=\tfrac{20}{40}\).)

A customer support call lasts uniformly between 2 and 12 minutes. Let \(X\) denote the call duration (minutes). Find the probability that a randomly selected call lasts between 4 and 9 minutes. Also give the PDF and expected call duration.
Solution:

\(X\sim U(2,12)\) so \(a=2\), \(b=12\).

• PDF: \[ f_X(x)=\frac{1}{b-a}=\frac{1}{10},\quad 2\le x\le 12. \]

• Probability: \[ P(4\le X\le 9)=\frac{9-4}{12-2}=\frac{5}{10}=0.5. \] (Integration: \(\int_4^9 \tfrac{1}{10}\,dx=\tfrac{5}{10}\).)

• Expected value: \[ E[X]=\frac{2+12}{2}=7\ \text{minutes}. \] (Variance for completeness: \(\mathrm{Var}(X)=\dfrac{10^2}{12}=\dfrac{100}{12}=\dfrac{25}{3}\approx 8.3333\).)

Let \(X \sim U(0,4)\) and \(Y \sim U(0,6)\), independent random variables. Find:

• [i.] The probability that \(X + Y < 5\).

• [ii.] The probability that \(X > Y\).

Solution: Both \(X\) and \(Y\) are uniform on \([0,4]\times[0,6]\), so total area \(=24\).

(i) \(P(X+Y<5)\):

Area \(=\int_0^4 (5-x)\,dx =12 \Rightarrow P(X+Y<5)=\tfrac{12}{24}=\tfrac12.\)

(ii) \(P(X>Y)\):

Favourable region: triangle below \(x=4\) and above \(y=x\). Area \(=\int_0^4(4-y)\,dy=8 \Rightarrow P(X>Y)=\tfrac{8}{24}=\tfrac13.\)

Final answers: (i) \(\tfrac{1}{2}\), (ii) \(\tfrac{1}{3}\).

Let \(X\) be uniformly distributed on \([-2,\,5]\). Find the cumulative distribution function \(F_X(x)\). Solution:

For \(X\sim U(a,b)\) with \(a=-2\), \(b=5\), the CDF is \[ F_X(x)=\begin{cases} 0, & x<a=-2,\\[4pt] \dfrac{x-a}{b-a}=\dfrac{x+2}{7}, & -2\le x\le 5,\\[8pt] 1, & x>b=5. \end{cases} \] This piecewise formula gives, for example, \(F_X(0)=\dfrac{0+2}{7}=\dfrac{2}{7}\).

Median: For a continuous uniform the median equals the mean: \[ \text{median}=\frac{a+b}{2}=\frac{-2+5}{2}=\frac{3}{2}=1.5. \]

The Exponential Distribution is a continuous probability distribution used to model the time between successive events in a Poisson process. It is memoryless, meaning the probability of an event occurring in the future is independent of the past.

Probability Density Function (PDF): \[ f(x; \lambda) = \lambda e^{-\lambda x}, \quad x \ge 0, \lambda > 0 \]

Cumulative Distribution Function (CDF): \[ F(x) = 1 - e^{-\lambda x} \]

Mean (Expected Value): \[ E(X) = \frac{1}{\lambda} \]

Variance: \[ Var(X) = \frac{1}{\lambda^2} \]

Memoryless Property: \[ P(X > s + t \mid X > s) = P(X > t) \]

Look for the following keywords in the problem statement:

• "Time between arrivals"

• "Waiting time until the next event"

• "Lifetimes of devices or components"

• "Memoryless property"

• Events that happen independently and continuously over time

Such problems are often tied to Poisson processes, where the inter-arrival times are exponentially distributed.

The PDF is highest at $x = 0$ and decreases exponentially.

The average time between calls at a call center is 3 minutes. What is the probability that the next call comes in more than 5 minutes?

Solution: $\lambda = \frac{1}{3}$.
We use: $P(X > 5) = e^{-\lambda x} = e^{-5/3} \approx 0.1889$.

The time to failure of a light bulb follows an exponential distribution with $\lambda = 0.02$. What is the expected life of the bulb?

Solution: $E(X) = \frac{1}{\lambda} = \frac{1}{0.02} = 50$ hours.

If a system fails with a mean time of 10 hours, find the probability that it fails before 4 hours.

Solution: $\lambda = 1/10$.
$P(X < 4) = 1 - e^{-\lambda x} = 1 - e^{-0.4} \approx 0.3297$.

A machine breaks down on average every 15 days. What is the variance in the time between breakdowns?

Solution: $\lambda = 1/15$, so: $Var(X) = \frac{1}{\lambda^2} = 225$ days².

A component lasts more than 8 years with probability 0.2. What is the failure rate $\lambda$?

Solution: $P(X > 8) = e^{-\lambda \cdot 8} = 0.2$.
Taking $\ln$ on both sides: $-8\lambda = \ln(0.2) \Rightarrow \lambda = -\frac{\ln(0.2)}{8} \approx 0.2012$.

The Normal Distribution is a continuous probability distribution that is symmetric about its mean. It models many natural phenomena such as height, test scores, and measurement errors.

Probability Density Function (PDF): \[ f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{ -\frac{(x - \mu)^2}{2\sigma^2} } \]

Where:

• $\mu$ = mean

• $\sigma$ = standard deviation

• $x$ = continuous random variable

Properties:

• Symmetric about the mean

• Mean = Median = Mode

• Total area under the curve = 1

• Empirical Rule:

  • 68% data within 1$\sigma$

  • 95% data within 2$\sigma$

  • 99.7% data within 3$\sigma$

A problem likely involves a Normal Distribution if:

• It involves continuous data (e.g., heights, weights, time).

• You are given mean and standard deviation.

• The data is symmetric or bell-shaped.

• It mentions that data follows a "normal distribution" or "Gaussian distribution".

• It asks for probability between two values or more/less than a certain value.

• It involves a standardization step using Z-scores.

The Standard Normal Distribution is a special case of the Normal Distribution where the mean ($\mu$) is 0 and the standard deviation ($\sigma$) is 1.

A random variable $Z$ follows a standard normal distribution if it has the following probability density function (pdf): \[ f(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}, \quad -\infty < z < \infty \]

This distribution is symmetric about $z=0$.

To convert a normal variable $X \sim N(\mu, \sigma^2)$ to a standard normal variable $Z$, we use: \[ Z = \frac{X - \mu}{\sigma} \] The Z-score tells us how many standard deviations a value is from the mean.

Use the Standard Normal Distribution when:

• You are dealing with a normally distributed variable and have been given the mean and standard deviation.

• You are asked to compute probabilities or quantiles using Z-scores.

• The problem involves symmetry around the mean.

• Values are standardized (i.e., Z-scores are used).

The weights of apples in an orchard are normally distributed with a mean of 150 g and a standard deviation of 12 g. What is the probability that a randomly chosen apple weighs more than 165 g?

\[ Z = \frac{165 - 150}{12} = 1.25 \] \[ P(X > 165) = P(Z > 1.25) = 1 - 0.8944 = 0.1056 \] Answer: 10.56%

The lifespan of LED bulbs follows a normal distribution with mean 2000 hours and standard deviation 200 hours. What is the probability that a bulb lasts between 1800 and 2200 hours?

\[ Z_1 = \frac{1800 - 2000}{200} = -1,\quad Z_2 = \frac{2200 - 2000}{200} = 1 \] \[ P(1800 < X < 2200) = P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6826 \] Answer: 68.26%

The volume of milk in a carton is normally distributed with mean 1000 ml and standard deviation 8 ml. What percentage of cartons will contain more than 1015 ml?

\[ Z = \frac{1015 - 1000}{8} = 1.875 \Rightarrow P(Z > 1.875) = 1 - 0.9699 = 0.0301 \] Answer: 3.01%

Let $X \sim N(60, 16)$. Find the value $x$ such that $P(X < x) = 0.975$.

From Z-tables, $P(Z < 1.96) = 0.975$ \[ x = \mu + Z\sigma = 60 + 1.96 \cdot 4 = 60 + 7.84 = 67.84 \] Answer: 67.84

The exam scores in a statistics class are normally distributed with mean 70 and standard deviation 6. What is the probability that a student scores between 65 and 80?

\[ Z_1 = \frac{65 - 70}{6} = -0.83,\quad Z_2 = \frac{80 - 70}{6} = 1.67 \Rightarrow P = P(-0.83 < Z < 1.67) \] \[ = 0.9525 - 0.2033 = 0.7492 \] Answer: 74.92%

If $X \sim N(45, 9)$, find $P(42 < X < 50)$.

\[ Z_1 = \frac{42 - 45}{3} = -1,\quad Z_2 = \frac{50 - 45}{3} = 1.67 \] \[ P = P(-1 < Z < 1.67) = 0.9525 - 0.1587 = 0.7938 \] Answer: 79.38%

Find $x$ such that $P(X < x) = 0.20$ for $X \sim N(100, 25)$.

\[ P(Z < z) = 0.20 \Rightarrow z = -0.84 \] \[ x = \mu + z\sigma = 100 - 0.84 \cdot 5 = 95.8 \] Answer: 95.8

$X \sim N(500, 100^2)$. Find $P(X > 650)$.

\[ Z = \frac{650 - 500}{100} = 1.5 \Rightarrow P(Z > 1.5) = 1 - 0.9332 = 0.0668 \] Answer: 6.68%

For $X \sim N(0,1)$, find $P(|X| < 1.5)$.

\[ P(|X| < 1.5) = P(-1.5 < X < 1.5) = 0.9332 - 0.0668 = 0.8664 \] Answer: 86.64%

The battery life of a smartphone is normally distributed with mean 8 hours and standard deviation 1.2 hours. What is the probability that a phone lasts more than 10 hours?

\[ Z = \frac{10 - 8}{1.2} = 1.67 \Rightarrow P(Z > 1.67) = 1 - 0.9525 = 0.0475 \] Answer: 4.75%

The Student's t-distribution is a probability distribution used when we want to estimate population parameters, particularly the mean, from a small sample, or when the population standard deviation is unknown.

Key Points:

• It is symmetric and bell-shaped like the normal distribution.

• It has heavier tails than the normal distribution, which accounts for more variability in small samples.

• As the sample size increases, the t-distribution approaches the standard normal distribution.

Degrees of Freedom (df): - For a sample of size $n$, df = $n - 1$. - It reflects the number of independent values that can vary in your calculation.

Uses:

• One-sample t-test: Compare a sample mean to a population mean.

• Two-sample t-test: Compare the means of two independent samples.

• Paired t-test: Compare means from two related samples.

• Confidence intervals: Estimate population mean when standard deviation is unknown.

The Chi-Square ($\chi^2$) distribution is a continuous distribution commonly used to analyze variance and categorical data.

Key Points:

• Skewed to the right, especially for small degrees of freedom.

• Becomes more symmetric as df increases, approaching normality.

• Degrees of freedom depend on sample size and the number of constraints (e.g., df = n - 1 for variance tests).

Uses:

• Goodness-of-fit tests.

• Test of independence in contingency tables.

• Confidence intervals for variance of a normally distributed population.

• Part of ANOVA test statistics.