Elementary Distributions

This chapter lays the foundation for understanding probabilistic phenomena by introducing elementary discrete and continuous distributions. Mastery of these distributions is critical for comprehending advanced probabilistic models and is frequently assessed in examinations through both direct application and theoretical analysis.

Chapter Contents

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| Topic |

|---|-------| | 1 | Discrete Distributions | | 2 | Continuous Distributions |

We begin with Discrete Distributions.

Part 1: Discrete Distributions

Discrete distributions model random variables that can take on a finite or countably infinite number of values. We use them to analyze outcomes such as the number of successes in trials or the count of events in a specific interval.

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Core Concepts

1. Bernoulli Distribution

The Bernoulli distribution describes the outcome of a single trial with exactly two possible results: "success" (with probability $p$) or "failure" (with probability $1-p$).

Worked Example:

Consider a component that functions correctly with a probability of $0.9$. We want to find the probability that a single randomly selected component functions correctly.

Step 1: Define the random variable and parameters.

> Let $X=1$ if the component functions correctly, and $X=0$ otherwise. This is a Bernoulli trial.
> The probability of success is $p = 0.9$.

Step 2: Apply the Bernoulli PMF for $x=1$.

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Answer: The probability that a single component functions correctly is $0.9$.

:::question type="MCQ" question="A quality control inspector checks a product for defects. The probability of a product being defective is $0.05$. What is the probability that a randomly selected product is not defective?" options=["0.05","0.95","0.50","0.10"] answer="0.95" hint="Identify success and failure, and their probabilities." solution="Let $X=1$ if the product is defective (success) and $X=0$ if it is not defective (failure).
The probability of success is $p = 0.05$.
We want to find the probability that the product is not defective, which corresponds to $X=0$.

Step 1: Identify the probability of failure.
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Answer: The probability that a randomly selected product is not defective is $0.95$."
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2. Binomial Distribution

The Binomial distribution models the number of successes in a fixed number of independent Bernoulli trials. Each trial has the same probability of success, $p$.

Worked Example:

A fair coin is tossed 10 times. We want to find the probability of getting exactly 7 heads.

Step 1: Define the random variable and parameters.

> Let $X$ be the number of heads in 10 tosses. This follows a Binomial distribution.
> Number of trials, $n = 10$.
> Probability of success (getting a head), $p = 0.5$.
> Number of desired successes, $k = 7$.

Step 2: Apply the Binomial PMF.

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Answer: The probability of getting exactly 7 heads in 10 tosses is approximately $0.117$.

:::question type="NAT" question="A manufacturing process produces 5% defective items. If a random sample of 20 items is selected, what is the expected number of defective items in the sample?" answer="1.0" hint="Recall the formula for the expected value of a Binomial distribution." solution="Let $X$ be the number of defective items in a sample of 20. This follows a Binomial distribution.

Step 1: Identify the parameters.
> Number of trials, $n = 20$.
> Probability of success (item being defective), $p = 0.05$.

Step 2: Apply the formula for the expected value of a Binomial distribution.
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Answer: The expected number of defective items is $1.0$."
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3. Geometric Distribution

The Geometric distribution models the number of Bernoulli trials required to achieve the first success. This includes the success itself.

Worked Example:

A quality control process involves testing items until a non-defective item is found. The probability of an item being non-defective is $0.8$. We want to find the probability that the first non-defective item is found on the 3rd trial.

Step 1: Define the random variable and parameters.

> Let $X$ be the number of trials until the first non-defective item is found. This follows a Geometric distribution.
> Probability of success (non-defective item), $p = 0.8$.
> Number of desired trials, $k = 3$.

Step 2: Apply the Geometric PMF.

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Answer: The probability that the first non-defective item is found on the 3rd trial is $0.032$.

:::question type="MCQ" question="A basketball player makes a free throw with a probability of $0.7$. What is the probability that the player's first successful free throw occurs on their 4th attempt?" options=["0.7","0.021","0.0063","0.0189"] answer="0.0189" hint="The first success occurs on the $k$-th trial means $k-1$ failures followed by 1 success." solution="Let $X$ be the number of attempts until the first successful free throw. This follows a Geometric distribution.

Step 1: Identify the parameters.
> Probability of success, $p = 0.7$.
> Number of desired trials, $k = 4$.

Step 2: Apply the Geometric PMF.
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Answer: The probability that the player's first successful free throw occurs on their 4th attempt is $0.0189$."
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4. Negative Binomial Distribution

The Negative Binomial distribution generalizes the Geometric distribution. It models the number of Bernoulli trials required to achieve a specified number of successes, $r$.

Worked Example:

A software company is hiring developers, and each candidate has a $0.2$ probability of passing the coding interview, independently. We want to find the probability that the 5th successful hire occurs on the 10th interview.

Step 1: Define the random variable and parameters.

> Let $X$ be the number of interviews until the 5th successful hire. This follows a Negative Binomial distribution.
> Number of desired successes, $r = 5$.
> Probability of success (passing interview), $p = 0.2$.
> Total number of trials, $k = 10$.

Step 2: Apply the Negative Binomial PMF.

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Answer: The probability that the 5th successful hire occurs on the 10th interview is approximately $0.0132$.

:::question type="MCQ" question="A machine produces items, and the probability of an item being non-defective is $0.9$. We are interested in finding the 3rd non-defective item. What is the expected number of items we need to examine until we find the 3rd non-defective item?" options=["3.33","2.7","3","10"] answer="3.33" hint="Use the expected value formula for the Negative Binomial distribution." solution="Let $X$ be the number of items examined until the 3rd non-defective item is found. This follows a Negative Binomial distribution.

Step 1: Identify the parameters.
> Number of desired successes, $r = 3$.
> Probability of success (non-defective item), $p = 0.9$.

Step 2: Apply the formula for the expected value of a Negative Binomial distribution.
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Answer: The expected number of items we need to examine is approximately $3.33$."
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5. Poisson Distribution

The Poisson distribution models the number of events occurring in a fixed interval of time or space, given that these events occur with a known constant mean rate and independently of the time since the last event. It is often used for rare events.

Worked Example:

A call center receives an average of $4$ calls per hour. We want to find the probability that the call center receives exactly $2$ calls in a specific hour.

Step 1: Define the random variable and parameters.

> Let $X$ be the number of calls received in an hour. This follows a Poisson distribution.
> The average rate of calls, $\lambda = 4$.
> Number of desired calls, $k = 2$.

Step 2: Apply the Poisson PMF.

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Answer: The probability that the call center receives exactly 2 calls in an hour is approximately $0.1465$.

:::question type="NAT" question="The number of typos on a page of a certain book follows a Poisson distribution with an average of $0.5$ typos per page. What is the probability that a randomly selected page has no typos? (Round to 4 decimal places)" answer="0.6065" hint="Use the Poisson PMF with $k=0$." solution="Let $X$ be the number of typos on a page. This follows a Poisson distribution.

Step 1: Identify the parameters.
> Average rate of typos, $\lambda = 0.5$.
> Number of desired typos, $k = 0$.

Step 2: Apply the Poisson PMF.
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Answer: The probability that a randomly selected page has no typos is approximately $0.6065$."
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6. Hypergeometric Distribution

The Hypergeometric distribution models the number of successes in a sample drawn without replacement from a finite population containing a known number of successes and failures.

Worked Example:

A box contains $10$ items, of which $3$ are defective. If a sample of $4$ items is drawn randomly without replacement, we want to find the probability that exactly $1$ of the sampled items is defective.

Step 1: Define the random variable and parameters.

> Let $X$ be the number of defective items in the sample. This follows a Hypergeometric distribution.
> Total population size, $N = 10$.
> Total number of defective items in population, $K = 3$.
> Sample size, $n = 4$.
> Number of desired defective items in sample, $k = 1$.

Step 2: Apply the Hypergeometric PMF.

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Answer: The probability that exactly 1 of the sampled items is defective is $0.5$.

:::question type="MCQ" question="A deck of 52 cards contains 4 aces. If 5 cards are drawn randomly without replacement, what is the probability that exactly 2 of them are aces?" options=["$\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}$","$\frac{\binom{4}{2}\binom{48}{5}}{\binom{52}{5}}$","$\frac{\binom{4}{2}\binom{52}{3}}{\binom{52}{5}}$","$\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{3}}$"] answer="$\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}$" hint="Identify the total population, number of successes in population, sample size, and number of successes in sample." solution="Let $X$ be the number of aces in the sample of 5 cards. This follows a Hypergeometric distribution.

Step 1: Identify the parameters.
> Total population size, $N = 52$ (total cards).
> Total number of 'success' items (aces) in population, $K = 4$.
> Sample size, $n = 5$ (cards drawn).
> Number of desired 'success' items (aces) in sample, $k = 2$.

Step 2: Apply the Hypergeometric PMF.
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Answer: The correct expression for the probability is $\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}$."
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7. Discrete Uniform Distribution

The Discrete Uniform distribution assigns equal probability to each outcome in a finite set of possible values.

Worked Example:

A fair six-sided die is rolled. We want to find the probability of rolling a 4.

Step 1: Define the random variable and parameters.

> Let $X$ be the outcome of the die roll. This follows a Discrete Uniform distribution.
> The possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.
> Number of possible outcomes, $n = 6$.
> Desired outcome, $x = 4$.

Step 2: Apply the Discrete Uniform PMF.

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Answer: The probability of rolling a 4 is $\frac{1}{6}$.

:::question type="NAT" question="A random number generator produces integers from 1 to 100 (inclusive), with each integer having an equal probability of being generated. What is the expected value of the generated number?" answer="50.5" hint="Identify the range of outcomes and use the expected value formula for a Discrete Uniform distribution." solution="Let $X$ be the generated integer. This follows a Discrete Uniform distribution.

Step 1: Identify the parameters.
> The range of outcomes is from $a=1$ to $b=100$.
> Total number of outcomes, $n = b-a+1 = 100-1+1 = 100$.

Step 2: Apply the formula for the expected value of a Discrete Uniform distribution.
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Answer: The expected value of the generated number is $50.5$."
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Advanced Applications

Worked Example:

An online server experiences failures at an average rate of $0.5$ failures per day.

What is the probability of having exactly 2 failures in a 3-day period?

If the server is critical, what is the probability that it runs for 5 days without any failures?

Step 1: Identify the distribution and parameters for part 1.

> The number of failures in a fixed period follows a Poisson distribution.
> The average rate of failures per day is $0.5$.
> For a 3-day period, the new average rate $\lambda$ is $0.5 \text{ failures/day} \times 3 \text{ days} = 1.5$.
> We want $k=2$ failures.

Step 2: Calculate the probability for part 1.

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Step 3: Identify the distribution and parameters for part 2.

> For the server to run for 5 days without any failures, this means 0 failures in a 5-day period. This is still a Poisson distribution.
> The average rate of failures per day is $0.5$.
> For a 5-day period, the new average rate $\lambda$ is $0.5 \text{ failures/day} \times 5 \text{ days} = 2.5$.
> We want $k=0$ failures.

Step 4: Calculate the probability for part 2.

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Answer:

The probability of exactly 2 failures in a 3-day period is approximately $0.2510$.

The probability that the server runs for 5 days without any failures is approximately $0.0821$.

:::question type="MSQ" question="A company sends out marketing emails, and historically, 20% of the recipients open the email. If 10 emails are sent to different recipients, which of the following statements are correct? (Select all that apply)" options=["The probability that exactly 3 emails are opened is $\binom{10}{3}(0.2)^3(0.8)^7$.","The expected number of opened emails is 2.","The probability that the first opened email is the 5th one sent is $(0.8)^4(0.2)$.","The variance of the number of opened emails is 2."] answer="The probability that exactly 3 emails are opened is \binom{10}{3}(0.2)^3(0.8)^7}.,The expected number of opened emails is 2.,The probability that the first opened email is the 5th one sent is $(0.8)^4(0.2)$." hint="Analyze each statement based on Binomial and Geometric distributions." solution="Let $X$ be the number of opened emails out of 10. This is a Binomial distribution with $n=10$ and $p=0.2$.

Statement 1: The probability that exactly 3 emails are opened is $\binom{10}{3}(0.2)^3(0.8)^7$.
> This is directly from the Binomial PMF: $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$.
> For $k=3$, $P(X=3) = \binom{10}{3}(0.2)^3(0.8)^{10-3} = \binom{10}{3}(0.2)^3(0.8)^7$.
> This statement is Correct.

Statement 2: The expected number of opened emails is 2.
> For a Binomial distribution, $E[X] = np$.
> $E[X] = 10 \times 0.2 = 2$.
> This statement is Correct.

Statement 3: The probability that the first opened email is the 5th one sent is $(0.8)^4(0.2)$.
> Let $Y$ be the number of emails sent until the first one is opened. This is a Geometric distribution with $p=0.2$.
> $P(Y=k) = (1-p)^{k-1}p$.
> For $k=5$, $P(Y=5) = (1-0.2)^{5-1}(0.2) = (0.8)^4(0.2)$.
> This statement is Correct.

Statement 4: The variance of the number of opened emails is 2.
> For a Binomial distribution, $\operatorname{Var}(X) = np(1-p)$.
> $\operatorname{Var}(X) = 10 \times 0.2 \times (1-0.2) = 10 \times 0.2 \times 0.8 = 1.6$.
> The statement says the variance is 2, which is incorrect.
> This statement is Incorrect.

Answer: The correct options are 'The probability that exactly 3 emails are opened is $\binom{10}{3}(0.2)^3(0.8)^7$.', 'The expected number of opened emails is 2.', 'The probability that the first opened email is the 5th one sent is $(0.8)^4(0.2)$.' "
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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="NAT" question="A particular type of integrated circuit has a failure rate of 1 in 100 during the first 1000 hours of operation. If a batch of 50 such circuits is tested, what is the variance of the number of circuits that fail within the first 1000 hours?" answer="0.495" hint="Identify the distribution and its parameters. Recall the variance formula." solution="Let $X$ be the number of circuits that fail in the batch of 50. This follows a Binomial distribution.

Step 1: Identify the parameters.
> Number of trials (circuits), $n = 50$.
> Probability of success (a circuit failing), $p = 1/100 = 0.01$.

Step 2: Apply the formula for the variance of a Binomial distribution.
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Answer: The variance of the number of failing circuits is $0.495$."
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:::question type="MCQ" question="A biased coin has a probability of landing heads as $0.6$. What is the probability that the 3rd head occurs on the 5th toss?" options=["$\binom{5}{3}(0.6)^3(0.4)^2$","$\binom{4}{2}(0.6)^3(0.4)^2$","$(0.4)^2(0.6)$","$\binom{5}{2}(0.6)^3(0.4)^2$"] answer="$\binom{4}{2}(0.6)^3(0.4)^2$" hint="This involves a specific number of successes on a specific trial number." solution="Let $X$ be the number of tosses until the 3rd head. This follows a Negative Binomial distribution.

Step 1: Identify the parameters.
> Number of desired successes, $r = 3$.
> Probability of success (heads), $p = 0.6$.
> Total number of trials, $k = 5$.

Step 2: Apply the Negative Binomial PMF.
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Answer: The correct option is '$\binom{4}{2}(0.6)^3(0.4)^2$'."
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:::question type="MSQ" question="Which of the following scenarios can be appropriately modeled by a Poisson distribution? (Select all that apply)" options=["The number of defective items in a sample of 1000 taken from a large production line with a known defect rate.","The number of cars passing a specific point on a highway in a 5-minute interval, given the average traffic flow.","The number of customers arriving at a store between 9 AM and 10 AM, given an average arrival rate.","The number of heads obtained when flipping a coin 20 times."] answer="The number of cars passing a specific point on a highway in a 5-minute interval, given the average traffic flow.,The number of customers arriving at a store between 9 AM and 10 AM, given an average arrival rate." hint="Poisson models events in an interval with an average rate." solution="Option 1: The number of defective items in a sample of 1000 taken from a large production line with a known defect rate.
> This is a Binomial distribution ($n=1000$, $p=$ defect rate). It can be approximated by Poisson if $n$ is large and $p$ is small, but it's fundamentally Binomial. So, not appropriately modeled as its primary distribution.

Option 2: The number of cars passing a specific point on a highway in a 5-minute interval, given the average traffic flow.
> This is a classic Poisson scenario: counting events (cars) in a fixed interval (5 minutes) with an average rate.
> This statement is Correct.

Option 3: The number of customers arriving at a store between 9 AM and 10 AM, given an average arrival rate.
> Another classic Poisson scenario: counting events (customer arrivals) in a fixed time interval (1 hour) with an average rate.
> This statement is Correct.

Option 4: The number of heads obtained when flipping a coin 20 times.
> This is a Binomial distribution ($n=20$, $p=0.5$).
> This statement is Incorrect.

Answer: The correct options are 'The number of cars passing a specific point on a highway in a 5-minute interval, given the average traffic flow.', 'The number of customers arriving at a store between 9 AM and 10 AM, given an average arrival rate.' "
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:::question type="NAT" question="A box contains 8 red balls and 4 blue balls. If 3 balls are drawn randomly without replacement, what is the probability that all 3 balls are red? (Round to 4 decimal places)" answer="0.2545" hint="This is sampling without replacement from a finite population." solution="Let $X$ be the number of red balls drawn. This follows a Hypergeometric distribution.

Step 1: Identify the parameters.
> Total population size, $N = 8 \text{ (red)} + 4 \text{ (blue)} = 12$.
> Total number of 'success' items (red balls) in population, $K = 8$.
> Sample size, $n = 3$.
> Number of desired 'success' items (red balls) in sample, $k = 3$.

Step 2: Apply the Hypergeometric PMF.
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Answer: The probability that all 3 balls are red is approximately $0.2545$."
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:::question type="MCQ" question="A computer program generates a random integer $X$ between 1 and 10 (inclusive), such that each integer has an equal probability of being chosen. What is the variance of $X$?" options=["2.5","8.25","10","9.1667"] answer="8.25" hint="This is a Discrete Uniform distribution. Use the variance formula for integers $a$ to $b$." solution="Let $X$ be the random integer generated. This follows a Discrete Uniform distribution.

Step 1: Identify the parameters.
> The range of outcomes is from $a=1$ to $b=10$.
> Total number of outcomes, $n = b-a+1 = 10-1+1 = 10$.

Step 2: Apply the formula for the variance of a Discrete Uniform distribution.
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Answer: The variance of $X$ is $8.25$."
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Summary

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What's Next?

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Part 2: Continuous Distributions

Continuous distributions model random variables that can take any value within a given range, providing a framework for analyzing probabilities of events over continuous scales. We apply these distributions to solve problems involving quantities such as time, distance, or measurement errors.

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Core Concepts

1. Probability Density Function (PDF) and Cumulative Distribution Function (CDF)

The Probability Density Function (PDF), $f_X(x)$, describes the relative likelihood for a continuous random variable $X$ to take on a given value $x$. The Cumulative Distribution Function (CDF), $F_X(x)$, gives the probability that $X$ will take a value less than or equal to $x$.

Worked Example:
Consider a random variable $X$ with CDF given by:

Find the PDF, $f_X(x)$.

Step 1: Differentiate $F_X(x)$ with respect to $x$ for the interval where it is non-constant.

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Step 2: Combine with the piecewise definition.

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Answer: The PDF is $f_X(x) = 2x$ for $0 \leq x < 1$ and $0$ otherwise.

:::question type="MCQ" question="A continuous random variable $X$ has a PDF $f_X(x) = \frac{1}{2}e^{-x/2}$ for $x \geq 0$ and $0$ otherwise. What is $P(X > 2)$?" options=["$e^{-1}$","$e^{-2}$","$1 - e^{-1}$","$1 - e^{-2}$"] answer="$e^{-1}$" hint="Use the CDF or direct integration of the PDF." solution="Step 1: Calculate the probability by integrating the PDF from $2$ to $\infty$.

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Step 2: Perform the integration.

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2. Expectation and Variance for Continuous Random Variables

The expectation (mean) $\operatorname{E}[X]$ represents the average value of a continuous random variable $X$. The variance $\operatorname{Var}(X)$ measures the spread or dispersion of the values of $X$ around its mean.

Worked Example:
A continuous random variable $X$ has the PDF $f_X(x) = 3x^2$ for $0 \leq x \leq 1$ and $0$ otherwise. Find $\operatorname{E}[X]$ and $\operatorname{Var}(X)$.

Step 1: Calculate $\operatorname{E}[X]$.

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Step 2: Calculate $\operatorname{E}[X^2]$.

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Step 3: Calculate $\operatorname{Var}(X)$.

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Answer: $\operatorname{E}[X] = \frac{3}{4}$ and $\operatorname{Var}(X) = \frac{3}{80}$.

:::question type="NAT" question="Let $X$ be a continuous random variable with PDF $f_X(x) = c(1-x^2)$ for $0 \leq x \leq 1$ and $0$ otherwise. Find $\operatorname{E}[X]$. Round your answer to two decimal places." answer="0.38" hint="First, find the constant $c$ by integrating the PDF over its domain and setting it to 1. Then calculate $\operatorname{E}[X]$ using the formula." solution="Step 1: Find the constant $c$. The total probability must be 1.

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Step 2: Calculate $\operatorname{E}[X]$ using the determined PDF $f_X(x) = \frac{3}{2}(1-x^2)$.

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Step 3: Convert to two decimal places.

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3. Uniform Distribution

A continuous random variable $X$ has a Uniform distribution over the interval $[a, b]$ if its PDF is constant within this interval and zero elsewhere. This implies that all values within the interval are equally likely.

Worked Example:
A bus arrives at a stop at a random time between 10:00 AM and 10:15 AM. Let $X$ be the arrival time in minutes past 10:00 AM. Find the probability that the bus arrives between 10:05 AM and 10:10 AM.

Step 1: Define the distribution parameters.
The arrival time $X$ is uniformly distributed over the interval $[0, 15]$ minutes. So, $a=0$ and $b=15$.

Step 2: Write the PDF.

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Step 3: Calculate the probability $P(5 < X < 10)$.

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Answer: The probability that the bus arrives between 10:05 AM and 10:10 AM is $\frac{1}{3}$.

:::question type="MCQ" question="The lifespan of a certain electronic component, in hours, is uniformly distributed between 500 and 1500. What is the expected lifespan of the component?" options=["750 hours","1000 hours","1250 hours","1500 hours"] answer="1000 hours" hint="For a uniform distribution $U(a, b)$, the expected value is simply the midpoint of the interval." solution="Step 1: Identify the parameters of the uniform distribution.
The lifespan $X$ is uniformly distributed over $[500, 1500]$. So, $a=500$ and $b=1500$.

Step 2: Apply the formula for the mean of a uniform distribution.

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Answer: The expected lifespan of the component is 1000 hours."
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4. Exponential Distribution

The Exponential distribution models the time until an event occurs in a Poisson process, where events occur continuously and independently at a constant average rate. It exhibits the memoryless property.

Worked Example:
The time (in minutes) a customer spends waiting for service at a bank follows an exponential distribution with an average waiting time of 5 minutes. What is the probability that a customer waits more than 10 minutes?

Step 1: Determine the rate parameter $\lambda$.
The average waiting time is $\operatorname{E}[X] = 5$ minutes. For an exponential distribution, $\operatorname{E}[X] = \frac{1}{\lambda}$.

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Step 2: Use the CDF or direct integration to find $P(X > 10)$.

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Answer: The probability that a customer waits more than 10 minutes is $e^{-2}$.

:::question type="MCQ" question="The lifetime of a light bulb follows an exponential distribution with a mean of 800 hours. Given that a light bulb has already lasted 600 hours, what is the probability that it will last for at least another 200 hours?" options=["$e^{-1/4}$","$e^{-1/2}$","$e^{-1}$","$1 - e^{-1/4}$"] answer="$e^{-1/4}$" hint="This question tests the memoryless property of the exponential distribution." solution="Step 1: Determine the rate parameter $\lambda$.
The mean lifetime is $\operatorname{E}[X] = 800$ hours. So, $\frac{1}{\lambda} = 800 \implies \lambda = \frac{1}{800}$.

Step 2: Apply the memoryless property.
The memoryless property states that $P(X > s+t \mid X > s) = P(X > t)$.
Here, $s = 600$ hours (already lasted) and $t = 200$ hours (another 200 hours).
We need to find $P(X > 600+200 \mid X > 600) = P(X > 200)$.

Step 3: Calculate $P(X > 200)$.

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Answer: The probability that it will last for at least another 200 hours is $e^{-1/4}$.
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5. Normal (Gaussian) Distribution

The Normal distribution is a symmetric, bell-shaped distribution characterized by its mean $\mu$ and variance $\sigma^2$. It is fundamental in statistics due to the Central Limit Theorem.

Worked Example:
Suppose the scores on a standardized test are normally distributed with a mean of 70 and a standard deviation of 10. What is the probability that a randomly selected student scores between 60 and 85? (Use $\Phi(1.5) \approx 0.9332$ and $\Phi(-1) \approx 0.1587$).

Step 1: Define the distribution parameters.
$X \sim N(70, 10^2)$, so $\mu = 70$ and $\sigma = 10$.

Step 2: Standardize the values $x_1 = 60$ and $x_2 = 85$ to $Z$-scores.

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Step 3: Calculate the probability using the standard normal CDF.

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Answer: The probability that a student scores between 60 and 85 is approximately $0.7745$.

:::question type="MCQ" question="The weights of adult males in a city are normally distributed with a mean of 75 kg and a standard deviation of 5 kg. What percentage of adult males weigh between 70 kg and 80 kg? (Use the empirical rule or $\Phi(1) \approx 0.8413$)" options=["68.26%","95.45%","99.73%","50%"] answer="68.26%" hint="Recognize that the interval $[70, 80]$ is within one standard deviation of the mean. Recall the empirical rule for normal distributions." solution="Step 1: Identify the distribution parameters.
$X \sim N(75, 5^2)$, so $\mu = 75$ and $\sigma = 5$.

Step 2: Observe the given interval.
The interval is $[70, 80]$.
$70 = 75 - 5 = \mu - \sigma$
$80 = 75 + 5 = \mu + \sigma$
We are looking for the probability $P(\mu - \sigma < X < \mu + \sigma)$.

Step 3: Apply the empirical rule.
For a normal distribution, approximately 68.26% of data falls within one standard deviation of the mean.

Alternatively, using Z-scores:
$Z_1 = \frac{70-75}{5} = -1$
$Z_2 = \frac{80-75}{5} = 1$
$P(70 < X < 80) = P(-1 < Z < 1) = \Phi(1) - \Phi(-1)$
Using symmetry, $\Phi(-1) = 1 - \Phi(1)$.
$P(-1 < Z < 1) = \Phi(1) - (1 - \Phi(1)) = 2\Phi(1) - 1$
Given $\Phi(1) \approx 0.8413$:
$P(-1 < Z < 1) \approx 2(0.8413) - 1 = 1.6826 - 1 = 0.6826$.

Answer: Approximately 68.26% of adult males weigh between 70 kg and 80 kg."
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6. Gamma Distribution

The Gamma distribution is a versatile distribution used to model waiting times or the sum of independent exponentially distributed random variables. It is characterized by two parameters: shape ($\alpha$) and rate ($\beta$ or scale $\theta = 1/\beta$).

Worked Example:
Suppose the waiting time for the first customer at a store is exponentially distributed with a mean of 2 minutes. What is the mean waiting time for the 3rd customer, assuming waiting times are independent?

Step 1: Identify the parameters for a single exponential waiting time.
The mean waiting time for the first customer (an exponential random variable) is 2 minutes. So for $\operatorname{Exp}(\beta)$, $\operatorname{E}[X] = \frac{1}{\beta} = 2 \implies \beta = \frac{1}{2}$.

Step 2: Relate to the Gamma distribution.
The waiting time for the 3rd customer is the sum of 3 independent exponential random variables, each with rate $\beta = 1/2$. This sum follows a Gamma distribution with shape parameter $\alpha = 3$ and rate parameter $\beta = 1/2$.
So, $X \sim \operatorname{Gamma}(3, 1/2)$.

Step 3: Calculate the mean of the Gamma distribution.

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Answer: The mean waiting time for the 3rd customer is 6 minutes.

:::question type="MCQ" question="A system has three identical components, each with a lifetime that follows an exponential distribution with a mean of 100 hours. If the system fails when all three components fail (they operate in parallel), what is the expected time until system failure? Assume component failures are independent." options=["100 hours","200 hours","300 hours","Not enough information"] answer="300 hours" hint="The sum of independent exponential random variables follows a Gamma distribution. Identify the parameters for the Gamma distribution." solution="Step 1: Determine the rate parameter $\beta$ for each component's exponential lifetime.
Mean lifetime $\operatorname{E}[X_i] = 100$ hours. For $\operatorname{Exp}(\beta)$, $\operatorname{E}[X_i] = \frac{1}{\beta}$.
So, $\frac{1}{\beta} = 100 \implies \beta = \frac{1}{100}$.

Step 2: Recognize that the system failure time is the sum of three independent exponential random variables.
Let $X_1, X_2, X_3$ be the lifetimes of the three components. The system fails when all three fail, meaning the system lifetime is $Y = X_1 + X_2 + X_3$.
Since $X_i \sim \operatorname{Exp}(1/100)$ independently, their sum $Y$ follows a Gamma distribution with shape parameter $\alpha = 3$ and rate parameter $\beta = 1/100$.
So, $Y \sim \operatorname{Gamma}(3, 1/100)$.

Step 3: Calculate the expected value of the Gamma distribution.

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Answer: The expected time until system failure is 300 hours."
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7. Beta Distribution

The Beta distribution is defined on the interval $[0, 1]$ and is widely used to model probabilities or proportions. It is characterized by two positive shape parameters, $\alpha$ and $\beta$.

Worked Example:
A random variable $X$ representing the proportion of time a machine is operational follows a Beta distribution with parameters $\alpha=2$ and $\beta=3$. What is the expected proportion of time the machine is operational?

Step 1: Identify the parameters of the Beta distribution.
$X \sim \operatorname{Beta}(2, 3)$, so $\alpha = 2$ and $\beta = 3$.

Step 2: Apply the formula for the mean of a Beta distribution.

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Answer: The expected proportion of time the machine is operational is $\frac{2}{5}$ or $0.4$.

:::question type="MCQ" question="The proportion of defective items produced by a manufacturing process is modeled by a Beta distribution with parameters $\alpha=1$ and $\beta=1$. What is the PDF of this proportion?" options=["$f(x) = 1$ for $0 \leq x \leq 1$","$f(x) = x$ for $0 \leq x \leq 1$","$f(x) = 1-x$ for $0 \leq x \leq 1$","$f(x) = 2x(1-x)$ for $0 \leq x \leq 1$"] answer="$f(x) = 1$ for $0 \leq x \leq 1$" hint="Recall the definition of the Beta function $B(\alpha, \beta)$ and simplify the PDF for the given parameters. What distribution is this equivalent to?" solution="Step 1: Identify the parameters of the Beta distribution.
$\alpha = 1$ and $\beta = 1$.

Step 2: Write down the PDF formula for $\operatorname{Beta}(\alpha, \beta)$.

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Step 3: Substitute $\alpha=1$ and $\beta=1$ into the PDF.

>

Step 4: Calculate the Beta function $B(1, 1)$.
$B(1, 1) = \frac{\Gamma(1)\Gamma(1)}{\Gamma(1+1)} = \frac{(1-1)!(1-1)!}{(2-1)!} = \frac{0! \cdot 0!}{1!} = \frac{1 \cdot 1}{1} = 1$.

Step 5: Substitute $B(1, 1)$ back into the PDF.

>

This is the PDF of a Uniform distribution $U(0,1)$.

Answer: The PDF is $f(x) = 1$ for $0 \leq x \leq 1$."
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8. Lognormal Distribution

A random variable $X$ is Lognormally distributed if its logarithm, $\ln(X)$, is normally distributed. This distribution is suitable for modeling variables that are positively skewed and bounded below by zero, such as financial asset prices or income.

Worked Example:
The price of a stock, $X$, is lognormally distributed. If $\ln(X)$ is normally distributed with mean $\mu=0.05$ and standard deviation $\sigma=0.10$, what is the expected price of the stock?

Step 1: Identify the parameters of the underlying normal distribution.
For $\ln(X) \sim N(\mu, \sigma^2)$, we have $\mu = 0.05$ and $\sigma = 0.10$.
Thus, $\sigma^2 = (0.10)^2 = 0.01$.

Step 2: Apply the formula for the mean of a Lognormal distribution.

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>
>
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Answer: The expected price of the stock is $e^{0.055}$. (Numerical value is approx. $1.0565$).

:::question type="MCQ" question="A random variable $X$ follows a Lognormal distribution such that $\ln(X) \sim N(2, 0.25)$. Which of the following statements is true regarding $X$?" options=["$\operatorname{E}[X] = e^2$","The median of $X$ is $e^2$","The variance of $X$ is $0.25$","$X$ can take negative values"] answer="The median of $X$ is $e^2$" hint="Recall the properties of the Lognormal distribution, particularly how its median relates to the mean of the underlying normal distribution." solution="Step 1: Identify the parameters of the underlying normal distribution.
For $\ln(X) \sim N(\mu, \sigma^2)$, we have $\mu = 2$ and $\sigma^2 = 0.25$.

Step 2: Evaluate each option.
* $\operatorname{E}[X] = e^2$: The mean of a Lognormal distribution is $e^{\mu + \sigma^2/2}$. Here, $\operatorname{E}[X] = e^{2 + 0.25/2} = e^{2.125}$. So, this statement is false.
* The median of $X$ is $e^2$: For a Lognormal distribution, the median is $e^\mu$. Since $\mu=2$, the median is $e^2$. This statement is true.
* The variance of $X$ is $0.25$: The variance of $X$ is $(e^{\sigma^2} - 1)e^{2\mu + \sigma^2}$. This is not $0.25$. The variance of $\ln(X)$ is $0.25$. So, this statement is false.
* $X$ can take negative values: The Lognormal distribution is defined for $x > 0$. So, this statement is false.

Answer: The median of $X$ is $e^2$"
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9. Weibull Distribution

The Weibull distribution is a flexible distribution commonly used in reliability engineering to model lifetimes of components. It can model decreasing, constant, or increasing failure rates depending on its shape parameter.

Worked Example:
The lifetime of a certain type of bearing (in years) follows a Weibull distribution with shape parameter $k=1$ and scale parameter $\lambda=5$. What is the mean lifetime of these bearings?

Step 1: Identify the parameters of the Weibull distribution.
$k=1$ and $\lambda=5$.

Step 2: Apply the formula for the mean of a Weibull distribution.

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>
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Step 3: Evaluate the Gamma function.
Recall that for integer $n$, $\Gamma(n) = (n-1)!$. So, $\Gamma(2) = (2-1)! = 1! = 1$.

>

Answer: The mean lifetime of the bearings is 5 years. (Note: Since $k=1$, this is an exponential distribution with mean $\lambda=5$).

:::question type="MCQ" question="A device's time to failure (in months) is modeled by a Weibull distribution with parameters $k=2$ and $\lambda=10$. What is the initial failure rate trend of this device?" options=["Constant failure rate","Increasing failure rate","Decreasing failure rate","Cannot be determined without more information"] answer="Increasing failure rate" hint="The shape parameter $k$ determines the failure rate trend for a Weibull distribution. Consider the cases $k<1$, $k=1$, and $k>1$." solution="Step 1: Identify the shape parameter $k$.
Here, $k=2$.

Step 2: Relate the shape parameter to the failure rate trend.
* If $k < 1$, the failure rate is decreasing over time. This indicates 'infant mortality' or devices that improve with age.
* If $k = 1$, the failure rate is constant (equivalent to an exponential distribution). This implies random failures, independent of age.
* If $k > 1$, the failure rate is increasing over time. This indicates 'wear-out' or devices that are more likely to fail as they age.

Step 3: Conclude based on $k=2$.
Since $k=2 > 1$, the device exhibits an increasing failure rate.

Answer: Increasing failure rate"
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10. Cauchy Distribution

The Cauchy distribution is a peculiar continuous distribution known for its heavy tails and the fact that its mean and variance are undefined. It is a classic example of a distribution for which the Law of Large Numbers does not apply.

Worked Example:
A random variable $X$ follows a Cauchy distribution with location parameter $x_0=0$ and scale parameter $\gamma=1$. What is the median of this distribution?

Step 1: Identify the location parameter.
For a Cauchy distribution, the location parameter $x_0$ is also its median. Here, $x_0=0$.

Step 2: State the median.
The median of the Cauchy distribution is $x_0$.

>

Answer: The median of the distribution is 0.

:::question type="MCQ" question="Which of the following statements is true about a random variable $X$ following a Cauchy distribution?" options=["Its mean is always zero.","Its variance is finite.","The Central Limit Theorem applies to sums of Cauchy random variables.","Its PDF is symmetric around its location parameter."] answer="Its PDF is symmetric around its location parameter." hint="Recall the unique properties of the Cauchy distribution, especially regarding its moments and symmetry." solution="Step 1: Evaluate each statement based on the properties of the Cauchy distribution.
* Its mean is always zero. False. The mean of a Cauchy distribution is undefined, not necessarily zero.
* Its variance is finite. False. The variance of a Cauchy distribution is undefined.
* The Central Limit Theorem applies to sums of Cauchy random variables. False. The Central Limit Theorem requires finite variance for the sum to converge to a normal distribution. For Cauchy random variables, the average of $n$ i.i.d. Cauchy variables is itself a Cauchy variable, not a normal one.
* Its PDF is symmetric around its location parameter. True. The PDF $f_X(x) = \frac{1}{\pi \gamma \left[ 1 + \left(\frac{x - x_0}{\gamma}\right)^2 \right]}$ is clearly symmetric around $x_0$, as replacing $(x-x_0)$ with $-(x-x_0)$ does not change the value of the PDF.

Answer: Its PDF is symmetric around its location parameter."
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Advanced Applications

We apply the concepts of continuous distributions to solve problems involving transformations of random variables and finding probabilities in more complex scenarios.

Worked Example:
Let $X$ be a continuous random variable with PDF $f_X(x) = 2x$ for $0 \leq x \leq 1$ and $0$ otherwise. Find the PDF of $Y = X^2$.

Step 1: Find the CDF of $X$.
For $0 \leq x \leq 1$:

>

So, $F_X(x) = x^2$ for $0 \leq x \leq 1$.

Step 2: Find the CDF of $Y$.
Since $Y = X^2$, for $0 \leq y \leq 1$, we have $X = \sqrt{Y}$.
The CDF of $Y$, $F_Y(y)$, is $P(Y \leq y)$.

>

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And $F_Y(y) = 0$ for $y < 0$, $F_Y(y) = 1$ for $y \geq 1$.

Step 3: Find the PDF of $Y$ by differentiating $F_Y(y)$.

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Answer: The PDF of $Y = X^2$ is $f_Y(y) = 1$ for $0 \leq y \leq 1$ and $0$ otherwise, which is a Uniform distribution $U(0,1)$.

:::question type="NAT" question="Let $X$ be an exponential random variable with parameter $\lambda = 1$. Find the median of $X$. Round your answer to two decimal places." answer="0.69" hint="The median $m$ is the value such that $P(X \leq m) = 0.5$. Use the CDF of the exponential distribution." solution="Step 1: Write the CDF for an exponential distribution with $\lambda=1$.

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Step 2: Set $F_X(m) = 0.5$ and solve for $m$.

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Step 3: Calculate the numerical value and round to two decimal places.

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Answer: 0.69"
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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="The amount of time (in hours) a student spends studying for an exam is a continuous random variable $X$ with PDF $f_X(x) = kx(2-x)$ for $0 \leq x \leq 2$ and $0$ otherwise. What is the value of $k$?" options=["$1/2$","$3/4$","$3/8$","$1/4$"] answer="$3/4$" hint="The total probability over the domain of the PDF must integrate to 1." solution="Step 1: Set the integral of the PDF over its domain equal to 1.

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Step 2: Evaluate the integral.

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Step 3: Solve for $k$.

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"
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:::question type="NAT" question="The lifetime of a product (in years) is exponentially distributed with a mean of 4 years. What is the probability that the product lasts exactly 5 years? Round your answer to two decimal places." answer="0.00" hint="For a continuous random variable, the probability of it taking an exact value is zero." solution="Step 1: Understand the nature of continuous random variables.
For any continuous random variable $X$ and any specific value $x$, the probability $P(X=x)$ is always 0. This is because the probability is defined as the area under the PDF curve, and the area under a single point is zero.

Step 2: Apply this principle to the question.
The question asks for the probability that the product lasts exactly 5 years. Since lifetime is a continuous variable, this probability is 0.

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Answer: 0.00"
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:::question type="MSQ" question="Let $X$ be a random variable with PDF $f_X(x) = \frac{1}{x^2}$ for $x \geq 1$ and $0$ otherwise. Select ALL correct statements." options=["$F_X(x) = 1 - 1/x$ for $x \geq 1$","$\operatorname{E}[X]$ is finite.","$\operatorname{Var}(X)$ is finite.","$P(X > 2) = 1/2$"] answer="$F_X(x) = 1 - 1/x$ for $x \geq 1$,$P(X > 2) = 1/2$" hint="Calculate the CDF, expectation, and variance. Remember to integrate carefully over the domain." solution="Step 1: Calculate the CDF $F_X(x)$.
For $x \geq 1$:

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So, $F_X(x) = 1 - 1/x$ for $x \geq 1$. This statement is correct.

Step 2: Calculate $\operatorname{E}[X]$.

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Since $\operatorname{E}[X]$ is infinite, it is not finite. This statement is incorrect.

Step 3: Calculate $\operatorname{Var}(X)$.
Since $\operatorname{E}[X]$ is infinite, $\operatorname{Var}(X)$ must also be infinite (or undefined).
$\operatorname{Var}(X) = \operatorname{E}[X^2] - (\operatorname{E}[X])^2$. If $\operatorname{E}[X]$ is infinite, $\operatorname{E}[X^2]$ will also be infinite.
>

So, $\operatorname{Var}(X)$ is infinite. This statement is incorrect.

Step 4: Calculate $P(X > 2)$.
Using the CDF:

>

Alternatively, by integrating the PDF:

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This statement is correct.

Answer: $F_X(x) = 1 - 1/x$ for $x \geq 1$,$P(X > 2) = 1/2$"
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:::question type="MCQ" question="If $X \sim N(5, 4)$, what is the distribution of $Y = 2X + 3$?" options=["$N(13, 16)$","$N(13, 8)$","$N(13, 4)$","$N(10, 16)$"] answer="$N(13, 16)$" hint="Recall the properties of linear transformations of normal random variables: $\operatorname{E}[aX+b] = a\operatorname{E}[X]+b$ and $\operatorname{Var}(aX+b) = a^2\operatorname{Var}(X)$." solution="Step 1: Identify the mean and variance of $X$.
$X \sim N(5, 4)$, so $\operatorname{E}[X] = 5$ and $\operatorname{Var}(X) = 4$.

Step 2: Calculate the mean of $Y$.

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Step 3: Calculate the variance of $Y$.

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Step 4: State the distribution of $Y$.
A linear transformation of a normal random variable is also a normal random variable.
So, $Y \sim N(13, 16)$.

Answer: $N(13, 16)$"
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:::question type="NAT" question="A random variable $X$ has a PDF $f_X(x) = \frac{1}{2}e^{-|x|}$ for $-\infty < x < \infty$. What is $\operatorname{E}[|X|]$? Round your answer to two decimal places." answer="1.00" hint="Split the integral into two parts due to the absolute value in the PDF and the expectation formula." solution="Step 1: Write the formula for $\operatorname{E}[|X|]$.

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Step 2: Split the integral due to the absolute value.
Since $f_X(x)$ is symmetric around 0, we can simplify the integral.

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Step 3: Evaluate the integral using integration by parts, or recognize it as the mean of an exponential distribution.
The integral $\int_{0}^{\infty} x e^{-x} \, dx$ is the mean of an exponential distribution with $\lambda = 1$. The mean of $\operatorname{Exp}(\lambda)$ is $1/\lambda$.
Here, $\lambda=1$, so the integral value is $1/1 = 1$.

Alternatively, using integration by parts:
Let $u=x$, $dv=e^{-x}dx$. Then $du=dx$, $v=-e^{-x}$.

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Answer: 1.00"
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Summary

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What's Next?

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="A call center receives calls at an average rate of 5 calls per hour. Assuming the number of calls follows a Poisson distribution, what is the probability of receiving exactly 3 calls in a 30-minute period?" options=["0.1404","0.2138","0.0821","0.2565"] answer="0.2138" hint="Adjust the Poisson rate parameter ($\lambda$) to match the specified time period before applying the PMF." solution="For a 30-minute period, the average rate of calls is $\lambda = 5 \text{ calls/hour} \times 0.5 \text{ hours} = 2.5$.
The probability of receiving exactly $k$ calls in a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
For $k=3$ and $\lambda=2.5$:

"
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:::question type="NAT" question="The lifetime of a certain electronic component (in hours) is exponentially distributed with a mean of 200 hours. If a component has already functioned for 150 hours, what is the probability that it will last for at least another 100 hours? (Provide answer to 4 decimal places.)" answer="0.6065" hint="Recall the memoryless property of the exponential distribution." solution="For an exponentially distributed random variable $X$ with mean $\mu$, the rate parameter is $\lambda = 1/\mu$. Here, $\mu=200$, so $\lambda = 1/200$.
The memoryless property states that $P(X > t+s | X > t) = P(X > s)$.
In this case, $t=150$ hours (time already functioned) and $s=100$ hours (additional time needed).
So, $P(X > 150+100 | X > 150) = P(X > 100)$.
The CDF of an exponential distribution is $F(x) = 1 - e^{-\lambda x}$, so $P(X > x) = e^{-\lambda x}$.
$P(X > 100) = e^{-(1/200) \times 100} = e^{-0.5} \approx 0.60653$.
Rounding to 4 decimal places, the answer is 0.6065."
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:::question type="MCQ" question="Which of the following distributions describes the number of trials required to achieve the first success in a sequence of independent Bernoulli trials, where the probability of success remains constant for each trial?" options=["Binomial distribution","Poisson distribution","Geometric distribution","Hypergeometric distribution"] answer="Geometric distribution" hint="Consider the stopping condition for each distribution type." solution="The Geometric distribution models the number of independent Bernoulli trials needed to get the first success. The Binomial distribution models the number of successes in a fixed number of trials. The Poisson distribution models the number of events in a fixed interval of time or space. The Hypergeometric distribution models the number of successes in draws without replacement."
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:::question type="NAT" question="A discrete random variable $X$ has the following probability mass function: $P(X=1) = 0.3$, $P(X=2) = 0.5$, $P(X=3) = 0.2$. What is the variance of $X$?" answer="0.49" hint="The variance can be calculated as $\operatorname{Var}(X) = E[X^2] - (E[X])^2$." solution="First, calculate the expected value $E[X]$:
$E[X] = (1 \times 0.3) + (2 \times 0.5) + (3 \times 0.2) = 0.3 + 1.0 + 0.6 = 1.9$.
Next, calculate $E[X^2]$:
$E[X^2] = (1^2 \times 0.3) + (2^2 \times 0.5) + (3^2 \times 0.2) = (1 \times 0.3) + (4 \times 0.5) + (9 \times 0.2) = 0.3 + 2.0 + 1.8 = 4.1$.
Finally, calculate the variance:
$\operatorname{Var}(X) = E[X^2] - (E[X])^2 = 4.1 - (1.9)^2 = 4.1 - 3.61 = 0.49$."
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What's Next?