# Given Information - The random variable \( X \) has the probability density function (pdf): \[ f(x) = kx^2(1-x), \quad < x < 1 \] # What Have to Find - (a) The constant \( k \) - (b) The expected value \( E[X] \) - (c) The probability \( P(X > .5) \) # Definition or Concept Used - The total probability for a continuous random variable must be 1: \(\int_{-\infty}^{\infty} f(x)\,dx = 1\). - Expected value: \(E[X] = \int_{-\infty}^{\infty} x f(x)\,dx\) - Probability over an interval: \(P(X > a) = \int_{a}^{\infty} f(x)\,dx\) # Step-by-Step Solution **(a) Finding the constant \( k \):** Set the total probability to 1: \[ \int_^1 kx^2(1-x)\,dx = 1 \] \[ k \int_^1 (x^2 - x^3)\,dx = 1 \] \[ k \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_^1 = 1 \] \[ k \left( \frac{1}{3} - \frac{1}{4} \right) = 1 \] \[ k \left( \frac{4-3}{12} \right) = 1 \] \[ k \cdot \frac{1}{12} = 1 \] \[ k = 12 \] **(b) Finding the expected value \( E[X] \):** \[ E[X] = \int_^1 x f(x)\,dx = \int_^1 x \cdot 12x^2(1-x)\,dx \] \[ = 12 \int_^1 x^3(1-x)\,dx \] \[ = 12 \int_^1 (x^3 - x^4)\,dx \] \[ = 12 \left[ \frac{x^4}{4} - \frac{x^5}{5} \right]_^1 \] \[ = 12 \left( \frac{1}{4} - \frac{1}{5} \right) \] \[ = 12 \cdot \frac{5-4}{20} \] \[ = 12 \cdot \frac{1}{20} \] \[ = \frac{12}{20} = \frac{3}{5} \] **(c) Finding \( P(X > .5) \):** \[ P(X > .5) = \int_{.5}^1 12x^2(1-x)\,dx \] \[ = 12 \int_{.5}^1 (x^2 - x^3)\,dx \] \[ = 12 \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{.5}^1 \] \[ = 12 \left( \left[ \frac{1}{3} - \frac{1}{4} \right] - \left[ \frac{(.5)^3}{3} - \frac{(.5)^4}{4} \right] \right) \] \[ = 12 \left( \frac{1}{12} - \left[ \frac{.125}{3} - \frac{.0625}{4} \right] \right) \] \[ = 12 \left( \frac{1}{12} - \left[ .04167 - .015625 \right] \right) \] \[ = 12 \left( \frac{1}{12} - .026045 \right) \] \[ = 12 \left( .08333 - .026045 \right) \] \[ = 12 \times .057285 \] \[ = .68742 \] # Summary - \( k = 12 \) - \( E[X] = \frac{3}{5} \) - \( P(X > .5) \approx .687 \)

# Given Information - The random variable \( X \) has the probability density function (pdf): \[ f(x) = kx^2(1-x), \quad 0 < x < 1 \] # What We Need to Find - (a) The constant \( k \) - (b) The expected value \( E[X] \) - (c) The probability \( P(X > 0.5) \) # Definitions or Concepts Used - **Total Probability**: The total probability for a continuous random variable must equal 1: \[ \int_{-\infty}^{\infty} f(x)\,dx = 1 \] - **Expected Value**: The expected value is calculated as: \[ E[X] = \int_{-\infty}^{\infty} x f(x)\,dx \] - **Probability Over an Interval**: The probability that \( X \) exceeds a certain value \( a \) is given by: \[ P(X > a) = \int_{a}^{\infty} f(x)\,dx \] # Step-by-Step Solution **Finding the Constant \( k \)** Set the total probability to 1: \[ \int_0^1 kx^2(1-x)\,dx = 1 \] This simplifies to: \[ k \int_0^1 (x^2 - x^3)\,dx = 1 \] Calculating the integral: \[ k \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = 1 \] \[ k \left( \frac{1}{3} - \frac{1}{4} \right) = 1 \] \[ k \cdot \frac{1}{12} = 1 \quad \Rightarrow \quad k = 12 \] **Finding the Expected Value \( E[X] \)** Calculate \( E[X] \): \[ E[X] = \int_0^1 x f(x)\,dx = \int_0^1 x \cdot 12x^2(1-x)\,dx \] This becomes: \[ = 12 \int_0^1 (x^3 - x^4)\,dx \] Calculating the integral: \[ = 12 \left[ \frac{x^4}{4} - \frac{x^5}{5} \right]_0^1 \] \[ = 12 \left( \frac{1}{4} - \frac{1}{5} \right) = 12 \cdot \frac{1}{20} = \frac{12}{20} = \frac{3}{5} \] **Finding \( P(X > 0.5) \)** Calculate the probability: \[ P(X > 0.5) = \int_{0.5}^1 12x^2(1-x)\,dx \] This simplifies to: \[ = 12 \int_{0.5}^1 (x^2 - x^3)\,dx \] Calculating the integral: \[ = 12 \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0.5}^1 \] Evaluating: \[ = 12 \left( \left[ \frac{1}{3} - \frac{1}{4} \right] - \left[ \frac{(0.5)^3}{3} - \frac{(0.5)^4}{4} \right] \right) \] Calculating each term: \[ = 12 \left( \frac{1}{12} - \left[ \frac{0.125}{3} - \frac{0.0625}{4} \right] \right) \] \[ = 12 \left( \frac{1}{12} - \left[ 0.04167 - 0.015625 \right] \right) \] Continuing the calculation: \[ = 12 \left( 0.08333 - 0.026045 \right) = 12 \cdot 0.057285 \approx 0.68742 \] # Summary - Constant \( k = 12 \) - Expected value \( E[X] = \frac{3}{5} \) - Probability \( P(X > 0.5) \approx 0.687 \)

# Given Information - A standardized intelligence test has scores that are normally distributed with: - Mean \( \mu = 100 \) - Standard deviation \( \sigma = 15 \) # What We Need to Find - (a) The probability that a randomly selected individual scores below 85. - (b) The probability that a randomly selected individual scores above 130. - (c) The probability that a randomly selected individual scores between 85 and 130. # Definitions or Concepts Used - **Z-score**: The Z-score for a value \( x \) is calculated as: \[ Z = \frac{x - \mu}{\sigma} \] - **Normal Distribution**: The probabilities can be found using the cumulative distribution function (CDF) of the normal distribution. # Step-by-Step Solution **(a) Finding the Probability for Scores Below 85** Calculate the Z-score for \( x = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Using the standard normal distribution table or a calculator: \[ P(X < 85) = P(Z < -1) \approx 0.1587 \] **(b) Finding the Probability for Scores Above 130** Calculate the Z-score for \( x = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Using the standard normal distribution table or a calculator: \[ P(X > 130) = P(Z > 2) \approx 1 - P(Z < 2) \approx 1 - 0.9772 = 0.0228 \] **(c) Finding the Probability for Scores Between 85 and 130** We already calculated \( P(X < 85) \) and \( P(X > 130) \). Now, we find \( P(X < 130) \): \[ P(X < 130) = P(Z < 2) \approx 0.9772 \] Now, calculate: \[ P(85 < X < 130) = P(X < 130) - P(X < 85) \approx 0.9772 - 0.1587 = 0.8185 \] # Summary - Probability of scoring below 85: \( P(X < 85) \approx 0.1587 \) - Probability of scoring above 130: \( P(X > 130) \approx 0.0228 \) - Probability of scoring between 85 and 130: \( P(85 < X < 130) \approx 0.8185 \)

# Given Information - A standardized intelligence test has scores that are normally distributed with: - Mean \( \mu = 100 \) - Standard deviation \( \sigma = 15 \) # What We Need to Find - (a) The probability that a randomly selected individual scores below 85. - (b) The probability that a randomly selected individual scores above 130. - (c) The probability that a randomly selected individual scores between 85 and 130. # Definitions or Concepts Used - **Z-score**: The Z-score for a value \( x \) is calculated as: \[ Z = \frac{x - \mu}{\sigma} \] - **Cumulative Distribution Function (CDF)**: The CDF of the standard normal distribution provides the probability that a standard normal random variable is less than or equal to a given value. # Step-by-Step Solution **(a) Finding the Probability for Scores Below 85** To find the Z-score for \( x = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Using the standard normal distribution table or calculator: \[ P(X < 85) = P(Z < -1) \approx 0.1587 \] **(b) Finding the Probability for Scores Above 130** Calculate the Z-score for \( x = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Using the standard normal distribution table or calculator: \[ P(X > 130) = P(Z > 2) \approx 1 - P(Z < 2) \approx 1 - 0.9772 = 0.0228 \] **(c) Finding the Probability for Scores Between 85 and 130** We already have \( P(X < 85) \) and need to find \( P(X < 130) \): \[ P(X < 130) = P(Z < 2) \approx 0.9772 \] Now calculate the probability between 85 and 130: \[ P(85 < X < 130) = P(X < 130) - P(X < 85) \approx 0.9772 - 0.1587 = 0.8185 \] # Summary - Probability of scoring below 85: \[ P(X < 85) \approx 0.1587 \] - Probability of scoring above 130: \[ P(X > 130) \approx 0.0228 \] - Probability of scoring between 85 and 130: \[ P(85 < X < 130) \approx 0.8185 \]

# Given Information - A standardized intelligence test has scores that follow a normal distribution with: - Mean \( \mu = 100 \) - Standard deviation \( \sigma = 15 \) # What We Need to Find - (a) The probability that a randomly selected individual scores below 85. - (b) The probability that a randomly selected individual scores above 130. - (c) The probability that a randomly selected individual scores between 85 and 130. # Definitions or Concepts Used - **Z-score**: The Z-score for a value \( x \) is derived from the formula: \[ Z = \frac{x - \mu}{\sigma} \] - **Cumulative Distribution Function (CDF)**: The CDF of the normal distribution gives the probability that a normal random variable is less than or equal to a certain value. # Step-by-Step Solution **(a) Finding the Probability for Scores Below 85** Calculate the Z-score for \( x = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Using the standard normal distribution table or a calculator, we find: \[ P(X < 85) = P(Z < -1) \approx 0.1587 \] **(b) Finding the Probability for Scores Above 130** Calculate the Z-score for \( x = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Using the standard normal distribution table or a calculator: \[ P(X > 130) = P(Z > 2) \approx 1 - P(Z < 2) \approx 1 - 0.9772 = 0.0228 \] **(c) Finding the Probability for Scores Between 85 and 130** We have already calculated \( P(X < 85) \) and need to find \( P(X < 130) \): \[ P(X < 130) = P(Z < 2) \approx 0.9772 \] Now, calculate the probability for scores between 85 and 130: \[ P(85 < X < 130) = P(X < 130) - P(X < 85) \approx 0.9772 - 0.1587 = 0.8185 \] # Summary - Probability of scoring below 85: \[ P(X < 85) \approx 0.1587 \] - Probability of scoring above 130: \[ P(X > 130) \approx 0.0228 \] - Probability of scoring between 85 and 130: \[ P(85 < X < 130) \approx 0.8185 \]

# Given Information - A standardized intelligence test has scores that are normally distributed with: - Mean \( \mu = 100 \) - Standard deviation \( \sigma = 15 \) # What We Need to Find - (a) The probability that a randomly selected individual scores below 85. - (b) The probability that a randomly selected individual scores above 130. - (c) The probability that a randomly selected individual scores between 85 and 130. # Definitions or Concepts Used - **Z-score**: The Z-score for a value \( x \) is calculated using the formula: \[ Z = \frac{x - \mu}{\sigma} \] - **Cumulative Distribution Function (CDF)**: The CDF of the standard normal distribution provides the probability that a standard normal variable is less than or equal to a given value. # Step-by-Step Solution **(a) Finding the Probability for Scores Below 85** First, calculate the Z-score for \( x = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Using a standard normal distribution table or calculator: \[ P(X < 85) = P(Z < -1) \approx 0.1587 \] **(b) Finding the Probability for Scores Above 130** Next, calculate the Z-score for \( x = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Using a standard normal distribution table or calculator: \[ P(X > 130) = P(Z > 2) \approx 1 - P(Z < 2) \approx 1 - 0.9772 = 0.0228 \] **(c) Finding the Probability for Scores Between 85 and 130** Now, we need to find \( P(X < 130) \): \[ P(X < 130) = P(Z < 2) \approx 0.9772 \] To find the probability between 85 and 130: \[ P(85 < X < 130) = P(X < 130) - P(X < 85) \approx 0.9772 - 0.1587 = 0.8185 \] # Summary - Probability of scoring below 85: \[ P(X < 85) \approx 0.1587 \] - Probability of scoring above 130: \[ P(X > 130) \approx 0.0228 \] - Probability of scoring between 85 and 130: \[ P(85 < X < 130) \approx 0.8185 \]

# Given Information - A standardized intelligence test has scores that are normally distributed with: - Mean \( \mu = 100 \) - Standard deviation \( \sigma = 15 \) # What We Need to Find - (a) The probability that a randomly selected individual scores below 85. - (b) The probability that a randomly selected individual scores above 130. - (c) The probability that a randomly selected individual scores between 85 and 130. # Definitions or Concepts Used - **Z-score**: The Z-score for a specific score \( x \) is calculated as follows: \[ Z = \frac{x - \mu}{\sigma} \] - **Cumulative Distribution Function (CDF)**: The CDF of the normal distribution gives the probability that a variable takes on a value less than or equal to a given score. # Step-by-Step Solution **(a) Finding the Probability for Scores Below 85** First, compute the Z-score for \( x = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Using a Z-table or standard normal distribution calculator: \[ P(X < 85) = P(Z < -1) \approx 0.1587 \] **(b) Finding the Probability for Scores Above 130** Next, calculate the Z-score for \( x = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Using the Z-table or calculator: \[ P(X > 130) = P(Z > 2) \approx 1 - P(Z < 2) \approx 1 - 0.9772 = 0.0228 \] **(c) Finding the Probability for Scores Between 85 and 130** To find this probability, we also need \( P(X < 130) \): \[ P(X < 130) = P(Z < 2) \approx 0.9772 \] Now, calculate the probability of scores between 85 and 130: \[ P(85 < X < 130) = P(X < 130) - P(X < 85) \approx 0.9772 - 0.1587 = 0.8185 \] # Summary - Probability of scoring below 85: \[ P(X < 85) \approx 0.1587 \] - Probability of scoring above 130: \[ P(X > 130) \approx 0.0228 \] - Probability of scoring between 85 and 130: \[ P(85 < X < 130) \approx 0.8185 \]

# Given Information - A standardized intelligence test has scores that are normally distributed with: - Mean \( \mu = 100 \) - Standard deviation \( \sigma = 15 \) # What We Need to Find - (a) The probability that a randomly selected individual scores below 85. - (b) The probability that a randomly selected individual scores above 130. - (c) The probability that a randomly selected individual scores between 85 and 130. # Definitions or Concepts Used - **Z-score**: The Z-score for a given score \( x \) is calculated using the formula: \[ Z = \frac{x - \mu}{\sigma} \] - **Cumulative Distribution Function (CDF)**: The CDF for the standard normal distribution provides the probability that a variable takes on a value less than or equal to a specific score. # Step-by-Step Solution **(a) Finding the Probability for Scores Below 85** Calculate the Z-score for \( x = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Using a Z-table or a calculator, we find: \[ P(X < 85) = P(Z < -1) \approx 0.1587 \] **(b) Finding the Probability for Scores Above 130** Next, compute the Z-score for \( x = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Using the Z-table or calculator: \[ P(X > 130) = P(Z > 2) \approx 1 - P(Z < 2) \approx 1 - 0.9772 = 0.0228 \] **(c) Finding the Probability for Scores Between 85 and 130** To find this probability, first determine \( P(X < 130) \): \[ P(X < 130) = P(Z < 2) \approx 0.9772 \] Now, calculate the probability of scores between 85 and 130: \[ P(85 < X < 130) = P(X < 130) - P(X < 85) \approx 0.9772 - 0.1587 = 0.8185 \] # Summary - Probability of scoring below 85: \[ P(X < 85) \approx 0.1587 \] - Probability of scoring above 130: \[ P(X > 130) \approx 0.0228 \] - Probability of scoring between 85 and 130: \[ P(85 < X < 130) \approx 0.8185 \]

# Given Information - A standardized intelligence test has scores that are normally distributed with: - Mean \( \mu = 100 \) - Standard deviation \( \sigma = 15 \) # What We Need to Find - (a) The probability that a randomly selected individual scores below 85. - (b) The probability that a randomly selected individual scores above 130. - (c) The probability that a randomly selected individual scores between 85 and 130. # Definitions or Concepts Used - **Z-score**: The Z-score for a specific score \( x \) is calculated as: \[ Z = \frac{x - \mu}{\sigma} \] - **Cumulative Distribution Function (CDF)**: The CDF of the normal distribution gives the probability that a random variable is less than or equal to a specified value. # Step-by-Step Solution **(a) Finding the Probability for Scores Below 85** Calculate the Z-score for \( x = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Using a Z-table or a normal distribution calculator: \[ P(X < 85) = P(Z < -1) \approx 0.1587 \] **(b) Finding the Probability for Scores Above 130** Calculate the Z-score for \( x = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Using the Z-table or a calculator: \[ P(X > 130) = P(Z > 2) \approx 1 - P(Z < 2) \approx 1 - 0.9772 = 0.0228 \] **(c) Finding the Probability for Scores Between 85 and 130** To find the probability, calculate \( P(X < 130) \): \[ P(X < 130) = P(Z < 2) \approx 0.9772 \] Now, compute the probability of scoring between 85 and 130: \[ P(85 < X < 130) = P(X < 130) - P(X < 85) \approx 0.9772 - 0.1587 = 0.8185 \] # Summary - Probability of scoring below 85: \[ P(X < 85) \approx 0.1587 \] - Probability of scoring above 130: \[ P(X > 130) \approx 0.0228 \] - Probability of scoring between 85 and 130: \[ P(85 < X < 130) \approx 0.8185 \]

# Given Information - A standardized intelligence test has scores that are normally distributed with: - Mean \( \mu = 100 \) - Standard deviation \( \sigma = 15 \) # What We Need to Find - (a) The probability that a randomly selected individual scores below 85. - (b) The probability that a randomly selected individual scores above 130. - (c) The probability that a randomly selected individual scores between 85 and 130. # Definitions or Concepts Used - **Z-score**: The Z-score for a given score \( x \) is calculated using: \[ Z = \frac{x - \mu}{\sigma} \] - **Cumulative Distribution Function (CDF)**: The CDF for the normal distribution gives the probability that a variable takes on a value less than or equal to a specified score. - **Excel Function**: The probability can also be calculated using the Excel function: \[ =NORM.DIST(z, 0, 1, TRUE) \] # Step-by-Step Solution **(a) Finding the Probability for Scores Below 85** Calculate the Z-score for \( x = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Using the Z-score in Excel: \[ P(X < 85) = NORM.DIST(-1, 0, 1, TRUE) \approx 0.1587 \] **(b) Finding the Probability for Scores Above 130** Calculate the Z-score for \( x = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Using the Z-score in Excel: \[ P(X > 130) = 1 - NORM.DIST(2, 0, 1, TRUE) \approx 1 - 0.9772 = 0.0228 \] **(c) Finding the Probability for Scores Between 85 and 130** First, calculate \( P(X < 130) \): \[ P(X < 130) = NORM.DIST(2, 0, 1, TRUE) \approx 0.9772 \] Now, compute the probability for scores between 85 and 130: \[ P(85 < X < 130) = P(X < 130) - P(X < 85) \approx 0.9772 - 0.1587 = 0.8185 \] # Summary - Probability of scoring below 85: \[ P(X < 85) \approx 0.1587 \] - Probability of scoring above 130: \[ P(X > 130) \approx 0.0228 \] - Probability of scoring between 85 and 130: \[ P(85 < X < 130) \approx 0.8185 \]

# Given Information - A standardized intelligence test has scores that are normally distributed with: - Mean \( \mu = 100 \) - Standard deviation \( \sigma = 15 \) # What We Need to Find - (a) The probability that a randomly selected individual scores below 85. - (b) The probability that a randomly selected individual scores above 130. - (c) The probability that a randomly selected individual scores between 85 and 130. # Definitions or Concepts Used - **Z-score**: The Z-score for a specific score \( x \) can be calculated using: \[ Z = \frac{x - \mu}{\sigma} \] - **Cumulative Distribution Function (CDF)**: The CDF of the normal distribution gives the probability that a random variable is less than or equal to a specified value. - **Excel Function**: The probabilities can be calculated using the Excel function: \[ =NORM.DIST(z, 0, 1, TRUE) \] # Step-by-Step Solution **(a) Finding the Probability for Scores Below 85** First, compute the Z-score for \( x = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Using the Excel function to find the probability: \[ P(X < 85) = NORM.DIST(-1, 0, 1, TRUE) \approx 0.1587 \] **(b) Finding the Probability for Scores Above 130** Calculate the Z-score for \( x = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Using the Excel function to find the probability: \[ P(X > 130) = 1 - NORM.DIST(2, 0, 1, TRUE) \approx 1 - 0.9772 = 0.0228 \] **(c) Finding the Probability for Scores Between 85 and 130** First, determine \( P(X < 130) \): \[ P(X < 130) = NORM.DIST(2, 0, 1, TRUE) \approx 0.9772 \] Now, compute the probability for scores between 85 and 130: \[ P(85 < X < 130) = P(X < 130) - P(X < 85) \approx 0.9772 - 0.1587 = 0.8185 \] # Summary - Probability of scoring below 85: \[ P(X < 85) \approx 0.1587 \] - Probability of scoring above 130: \[ P(X > 130) \approx 0.0228 \] - Probability of scoring between 85 and 130: \[ P(85 < X < 130) \approx 0.8185 \]

# Given Information - A standardized intelligence test has scores that follow a normal distribution characterized by: - Mean \( \mu = 100 \) - Standard deviation \( \sigma = 15 \) # What We Need to Find - (a) The probability that a randomly selected individual scores below 85. - (b) The probability that a randomly selected individual scores above 130. - (c) The probability that a randomly selected individual scores between 85 and 130. # Definitions or Concepts Used - **Z-score**: The Z-score for a specific score \( x \) is calculated using the equation: \[ Z = \frac{x - \mu}{\sigma} \] - **Cumulative Distribution Function (CDF)**: The CDF provides the probability that a normally distributed random variable is less than or equal to a particular value. - **Excel Function**: The probabilities can also be assessed using the Excel function: \[ =NORM.DIST(z, 0, 1, TRUE) \] # Step-by-Step Solution **(a) Finding the Probability for Scores Below 85** First, we calculate the Z-score for \( x = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Using the Excel function to find the probability: \[ P(X < 85) = NORM.DIST(-1, 0, 1, TRUE) \approx 0.1587 \] **(b) Finding the Probability for Scores Above 130** Next, we calculate the Z-score for \( x = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Using the Excel function to find the probability: \[ P(X > 130) = 1 - NORM.DIST(2, 0, 1, TRUE) \approx 1 - 0.9772 = 0.0228 \] **(c) Finding the Probability for Scores Between 85 and 130** To find this probability, we first compute \( P(X < 130) \): \[ P(X < 130) = NORM.DIST(2, 0, 1, TRUE) \approx 0.9772 \] Now, we determine the probability for scores between 85 and 130: \[ P(85 < X < 130) = P(X < 130) - P(X < 85) \approx 0.9772 - 0.1587 = 0.8185 \] # Summary - Probability of scoring below 85: \[ P(X < 85) \approx 0.1587 \] - Probability of scoring above 130: \[ P(X > 130) \approx 0.0228 \] - Probability of scoring between 85 and 130: \[ P(85 < X < 130) \approx 0.8185 \]

# Given Regression Equation The regression equation provided is: \[ \hat{y} = 12 + 3x \] where: - \( \hat{y} \) is the predicted value of the dependent variable. - \( x \) is the independent variable. The coefficient of determination, \( R^2 = 0.81 \). # Interpretation of the Slope The slope of the regression line is represented by the coefficient of \( x \), which is \( 3 \) in this case. Here’s how to interpret it: - **Change in Predicted Value**: For every one-unit increase in \( x \), the predicted value of \( y \) increases by \( 3 \) units. This means there is a positive relationship between \( x \) and \( y \): as \( x \) increases, \( y \) also increases. - **Contextual Application**: If \( x \) represents a specific variable (e.g., hours studied), this interpretation might mean that for each additional hour studied, the predicted score (or outcome) increases by \( 3 \) points. # Understanding \( R^2 \) The \( R^2 \) value of \( 0.81 \) indicates that approximately \( 81\% \) of the variability in the dependent variable \( y \) can be explained by the independent variable \( x \) using this regression model. This suggests a strong relationship between \( x \) and \( y \) and a good fit for the model. # Summary - The slope of \( 3 \) indicates that for every one-unit increase in \( x \), \( y \) increases by \( 3 \) units. - The \( R^2 \) value of \( 0.81 \) shows that \( 81\% \) of the variance in \( y \) is accounted for by \( x \), indicating a strong relationship.

# Given Regression Equation The regression equation is given as: \[ \hat{y} = 12 + 3x \] where: - \( \hat{y} \) is the estimated value of the dependent variable. - \( x \) is the independent variable. The coefficient of determination is \( R^2 = 0.81 \). # Interpretation of the Slope The slope of the regression line is the coefficient of \( x \), which is \( 3 \). Here’s a breakdown of its interpretation: - **Rate of Change**: The slope of \( 3 \) indicates that for each one-unit increase in \( x \), the predicted value of \( y \) increases by \( 3 \) units. This signifies a direct positive correlation between \( x \) and \( y \). - **Practical Example**: If \( x \) represents the number of hours worked, then for each additional hour worked, the predicted output (or any relevant measure represented by \( y \)) increases by \( 3 \) units. This shows how sensitive the predicted outcome is to changes in \( x \). # Understanding \( R^2 \) The \( R^2 \) value of \( 0.81 \) implies that approximately \( 81\% \) of the variability in the dependent variable \( y \) can be explained by the independent variable \( x \) in this regression model. This suggests that the model provides a good fit for the data, indicating that \( x \) is a strong predictor of \( y \). # Summary - The slope of \( 3 \) indicates that for every one-unit increase in \( x \), the predicted value of \( y \) increases by \( 3 \) units. - An \( R^2 \) value of \( 0.81 \) suggests that \( 81\% \) of the variance in \( y \) is explained by the independent variable \( x \), indicating a strong relationship and fit of the model.

# Given Information - The regression equation is: \[ \hat{y} = 12 + 3x \] - The coefficient of determination is: \[ R^2 = 0.81 \] # What We Need to Find - Interpret the slope of the regression equation. - Calculate the correlation coefficient. - Determine how the predicted \( y \) changes if \( x \) increases by 2 units. # Definitions or Concepts Used - **Slope**: The slope of the regression line represents the change in the predicted value of \( y \) for each one-unit increase in \( x \). - **Correlation Coefficient**: The correlation coefficient \( r \) can be derived from the \( R^2 \) value using the formula: \[ r = \sqrt{R^2} \] - **Change in Predicted Value**: The change in the predicted value of \( y \) can be calculated by multiplying the slope by the change in \( x \). # Step-by-Step Solution **Interpreting the Slope** The slope of the regression line is \( 3 \). This indicates that for every one-unit increase in \( x \), the predicted value of \( y \) increases by \( 3 \) units. **Calculating the Correlation Coefficient** To find the correlation coefficient \( r \): \[ r = \sqrt{R^2} = \sqrt{0.81} = 0.9 \] This suggests a strong positive linear relationship between \( x \) and \( y \). **Determining the Change in Predicted \( y \)** If \( x \) increases by \( 2 \) units, the change in the predicted \( y \) can be calculated as follows: \[ \text{Change in } y = \text{slope} \times \text{change in } x = 3 \times 2 = 6 \] Thus, the predicted \( y \) will increase by \( 6 \) units when \( x \) increases by \( 2 \) units. # Summary - The slope of \( 3 \) indicates that for each one-unit increase in \( x \), the predicted value of \( y \) increases by \( 3 \) units. - The correlation coefficient is \( r = 0.9 \). - If \( x \) increases by \( 2 \) units, the predicted \( y \) increases by \( 6 \) units.

# Given Information - The regression equation is: \[ \hat{y} = 12 + 3x \] - The coefficient of determination is: \[ R^2 = 0.81 \] # What We Need to Find - Provide an interpretation of the slope in the regression equation. - Calculate the correlation coefficient. - Explain how the predicted \( y \) changes if \( x \) is increased by 2 units. # Definitions or Concepts Used - **Slope**: The slope of the regression line quantifies the expected change in the predicted value of \( y \) for each one-unit increase in \( x \). - **Correlation Coefficient**: The correlation coefficient \( r \) is derived from the \( R^2 \) value using the equation: \[ r = \sqrt{R^2} \] - **Change in Predicted Value**: The change in the predicted \( y \) can be computed by multiplying the slope by the change in \( x \). # Step-by-Step Solution **Interpreting the Slope** The slope of the regression equation is \( 3 \). This means that for every single unit increase in \( x \), the predicted value of \( y \) increases by \( 3 \) units. This indicates a direct positive relationship between the variables. **Calculating the Correlation Coefficient** To determine the correlation coefficient \( r \): \[ r = \sqrt{R^2} = \sqrt{0.81} = 0.9 \] This value signifies a strong positive correlation between \( x \) and \( y \). **Determining the Change in Predicted \( y \)** When \( x \) increases by \( 2 \) units, the change in the predicted \( y \) is calculated as follows: \[ \text{Change in } y = \text{slope} \times \text{change in } x = 3 \times 2 = 6 \] Thus, an increase of \( 2 \) units in \( x \) results in the predicted value of \( y \) rising by \( 6 \) units. # Summary - The slope of \( 3 \) indicates that for each unit increase in \( x \), the predicted value of \( y \) increases by \( 3 \) units. - The correlation coefficient is \( r = 0.9 \). - If \( x \) is raised by \( 2 \) units, the predicted \( y \) will increase by \( 6 \) units.

# Given Information - The regression equation is: \[ \hat{y} = 12 + 3x \] - The coefficient of determination is: \[ R^2 = 0.81 \] # What We Need to Find - Interpret the slope of the regression equation. - Calculate the correlation coefficient. - Explain the change in predicted \( y \) when \( x \) increases by 2 units. # Definitions or Concepts Used - **Slope**: The slope in the regression equation indicates the amount by which the predicted value of \( y \) changes for each unit increase in \( x \). - **Correlation Coefficient**: The correlation coefficient \( r \) is calculated from the \( R^2 \) value using the formula: \[ r = \sqrt{R^2} \] - **Change in Predicted Value**: The change in the predicted value of \( y \) can be derived from multiplying the slope by the change in \( x \). # Step-by-Step Solution **Interpreting the Slope** The slope of the regression line is \( 3 \). This indicates that for each unit increase in \( x \), the predicted value of \( y \) increases by \( 3 \) units. Thus, there is a positive association between \( x \) and \( y \). **Calculating the Correlation Coefficient** To find the correlation coefficient \( r \): \[ r = \sqrt{R^2} = \sqrt{0.81} = 0.9 \] This value suggests a strong positive linear relationship between \( x \) and \( y \). **Determining the Change in Predicted \( y \)** When \( x \) increases by \( 2 \) units, the change in predicted \( y \) can be computed as: \[ \text{Change in } y = \text{slope} \times \text{change in } x = 3 \times 2 = 6 \] Therefore, if \( x \) increases by \( 2 \) units, the predicted \( y \) will increase by \( 6 \) units. # Summary - The slope of \( 3 \) indicates that for every unit increase in \( x \), the predicted value of \( y \) increases by \( 3 \) units. - The correlation coefficient is \( r = 0.9 \). - If \( x \) increases by \( 2 \) units, the predicted \( y \) increases by \( 6 \) units.

# Given Information - Prevalence of the disease in the population: \( P(D) = 0.02 \) (2%) - Sensitivity of the test: \( P(T^+ | D) = 0.95 \) (95%) - Specificity of the test: \( P(T^- | D^-) = 0.90 \) (90%) # What We Need to Find - The probability that a person has the disease given that they tested positive: \( P(D | T^+) \). # Definitions or Concepts Used - **Bayes' Theorem**: This theorem is used to find the probability of an event based on prior knowledge of conditions related to the event. It can be expressed as: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] - **Total Probability of Testing Positive**: The total probability of testing positive, \( P(T^+) \), can be calculated as: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^-) \cdot P(D^-) \] Where: - \( P(D^-) = 1 - P(D) = 0.98 \) - \( P(T^+ | D^-) = 1 - P(T^- | D^-) = 1 - 0.90 = 0.10 \) # Step-by-Step Solution **Calculate Total Probability of Testing Positive \( P(T^+) \)** 1. Calculate \( P(T^+ | D^-) \): \[ P(T^+ | D^-) = 0.10 \] 2. Calculate \( P(D^-) \): \[ P(D^-) = 1 - P(D) = 1 - 0.02 = 0.98 \] 3. Use the total probability formula: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^-) \cdot P(D^-) \] \[ P(T^+) = (0.95 \cdot 0.02) + (0.10 \cdot 0.98) \] \[ P(T^+) = 0.019 + 0.098 = 0.117 \] **Calculate \( P(D | T^+) \) Using Bayes' Theorem** \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] \[ P(D | T^+) = \frac{0.95 \cdot 0.02}{0.117} \] \[ P(D | T^+) = \frac{0.019}{0.117} \approx 0.1624 \] # Summary - The probability that a person actually has the disease given that they tested positive is approximately \( P(D | T^+) \approx 0.1624 \) or \( 16.24\% \).

# Given Information - Prevalence of the disease in the population: \( P(D) = 0.02 \) (2%) - Sensitivity of the test: \( P(T^+ | D) = 0.95 \) (95%) - Specificity of the test: \( P(T^- | D^-) = 0.90 \) (90%) # What We Need to Find - The probability that a person has the disease given that they tested positive: \( P(D | T^+) \). # Definitions or Concepts Used - **Bayes' Theorem**: This theorem allows us to find the probability of a condition based on prior knowledge. It can be expressed as: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] - **Total Probability of Testing Positive**: The overall probability of testing positive, \( P(T^+) \), can be derived from the following formula: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^-) \cdot P(D^-) \] Where: - \( P(D^-) = 1 - P(D) = 0.98 \) - \( P(T^+ | D^-) = 1 - P(T^- | D^-) = 1 - 0.90 = 0.10 \) # Step-by-Step Solution **Calculate Total Probability of Testing Positive \( P(T^+) \)** 1. Determine \( P(D^-) \): \[ P(D^-) = 1 - P(D) = 1 - 0.02 = 0.98 \] 2. Calculate \( P(T^+ | D^-) \): \[ P(T^+ | D^-) = 0.10 \] 3. Apply the total probability formula: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^-) \cdot P(D^-) \] \[ P(T^+) = (0.95 \cdot 0.02) + (0.10 \cdot 0.98) \] \[ P(T^+) = 0.019 + 0.098 = 0.117 \] **Calculate \( P(D | T^+) \) Using Bayes' Theorem** \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] \[ P(D | T^+) = \frac{0.95 \cdot 0.02}{0.117} \] \[ P(D | T^+) = \frac{0.019}{0.117} \approx 0.1624 \] # Summary - The probability that a person actually has the disease given that they tested positive is approximately \( P(D | T^+) \approx 0.1624 \) or \( 16.24\% \).

# Given Information - The prevalence of the disease in the population: \( P(D) = 0.02 \) (2%) - The sensitivity of the test: \( P(T^+ | D) = 0.95 \) (95%) - The specificity of the test: \( P(T^- | D^-) = 0.90 \) (90%) # What We Need to Find - The probability that a person has the disease given that they tested positive: \( P(D | T^+) \). # Definitions or Concepts Used - **Bayes' Theorem**: This theorem is used to find the probability of an event based on prior knowledge of conditions related to the event: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] - **Total Probability of Testing Positive**: The overall probability of testing positive, \( P(T^+) \), can be calculated as: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^-) \cdot P(D^-) \] Where: - \( P(D^-) = 1 - P(D) = 0.98 \) - \( P(T^+ | D^-) = 1 - P(T^- | D^-) = 1 - 0.90 = 0.10 \) # Step-by-Step Solution **Calculate Total Probability of Testing Positive \( P(T^+) \)** 1. Calculate \( P(D^-) \): \[ P(D^-) = 1 - P(D) = 1 - 0.02 = 0.98 \] 2. Determine \( P(T^+ | D^-) \): \[ P(T^+ | D^-) = 0.10 \] 3. Utilize the total probability formula: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^-) \cdot P(D^-) \] \[ P(T^+) = (0.95 \cdot 0.02) + (0.10 \cdot 0.98) \] \[ P(T^+) = 0.019 + 0.098 = 0.117 \] **Calculate \( P(D | T^+) \) Using Bayes' Theorem** \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] \[ P(D | T^+) = \frac{0.95 \cdot 0.02}{0.117} \] \[ P(D | T^+) = \frac{0.019}{0.117} \approx 0.1624 \] # Summary - The probability that a person actually has the disease given that they tested positive is approximately \( P(D | T^+) \approx 0.1624 \) or \( 16.24\% \).

# Given Information - The random variables \( x_1, x_2, \ldots, x_n \) are independent and identically distributed (i.i.d.) from an exponential distribution with the probability density function (pdf): \[ f(x | \lambda) = \lambda e^{-\lambda x}, \quad x > 0 \] # What We Need to Find - Derive the likelihood function for \( \lambda \). - Find the maximum likelihood estimate (MLE) of \( \lambda \). # Definitions or Concepts Used - **Likelihood Function**: The likelihood function \( L(\lambda) \) for a set of observations \( x_1, x_2, \ldots, x_n \) is the joint probability of the observed data given the parameter \( \lambda \): \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) \] - **Maximum Likelihood Estimate (MLE)**: The MLE is the value of \( \lambda \) that maximizes the likelihood function. # Step-by-Step Solution **Derive the Likelihood Function** The likelihood function for the exponential distribution can be expressed as: \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) = \prod_{i=1}^{n} \left( \lambda e^{-\lambda x_i} \right) \] This expands to: \[ L(\lambda) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] **Finding the MLE of \( \lambda \)** To find the MLE, we need to maximize the likelihood function \( L(\lambda) \). It is often more convenient to maximize the log-likelihood function \( \ell(\lambda) \): \[ \ell(\lambda) = \ln(L(\lambda)) = \ln \left( \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \right) \] This simplifies to: \[ \ell(\lambda) = n \ln(\lambda) - \lambda \sum_{i=1}^{n} x_i \] Next, differentiate \( \ell(\lambda) \) with respect to \( \lambda \) and set the derivative to zero: \[ \frac{d\ell(\lambda)}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0 \] Solving for \( \lambda \) gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} x_i \] \[ \lambda = \frac{n}{\sum_{i=1}^{n} x_i} \] Therefore, the maximum likelihood estimate of \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] # Summary - The likelihood function is given by: \[ L(\lambda) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] - The maximum likelihood estimate of \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \]

# Given Information - The random variables \( x_1, x_2, \ldots, x_n \) are independent and identically distributed (i.i.d.) from an exponential distribution with the probability density function (pdf): \[ f(x | \lambda) = \lambda e^{-\lambda x}, \quad x > 0 \] # What We Need to Find - Derive the likelihood function for \( \lambda \). - Determine the maximum likelihood estimate (MLE) of \( \lambda \). # Definitions or Concepts Used - **Likelihood Function**: The likelihood function \( L(\lambda) \) is defined as the product of the individual probability densities for each observed data point: \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) \] - **Maximum Likelihood Estimate (MLE)**: The MLE is the parameter value that maximizes the likelihood function. # Step-by-Step Solution **Derive the Likelihood Function** The likelihood function for the i.i.d. exponential distribution is given by: \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) = \prod_{i=1}^{n} \left( \lambda e^{-\lambda x_i} \right) \] This can be expanded to: \[ L(\lambda) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] **Finding the MLE of \( \lambda \)** To find the MLE, we can maximize the log-likelihood function \( \ell(\lambda) \): \[ \ell(\lambda) = \ln(L(\lambda)) = \ln \left( \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \right) \] This simplifies to: \[ \ell(\lambda) = n \ln(\lambda) - \lambda \sum_{i=1}^{n} x_i \] Next, we differentiate \( \ell(\lambda) \) with respect to \( \lambda \) and set the derivative to zero: \[ \frac{d\ell(\lambda)}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0 \] Rearranging this gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} x_i \] Solving for \( \lambda \) yields: \[ \lambda = \frac{n}{\sum_{i=1}^{n} x_i} \] Thus, the maximum likelihood estimate of \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] # Summary - The likelihood function is: \[ L(\lambda) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] - The maximum likelihood estimate of \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \]

# Given Information - The random variables \( x_1, x_2, \ldots, x_n \) are independent and identically distributed (i.i.d.) from an exponential distribution with the probability density function (pdf): \[ f(x | \lambda) = \lambda e^{-\lambda x}, \quad x > 0 \] # What We Need to Find - Derive the likelihood function for \( \lambda \). - Calculate the maximum likelihood estimate (MLE) of \( \lambda \). # Definitions or Concepts Used - **Likelihood Function**: The likelihood function \( L(\lambda) \) represents the probability of observing the given data as a function of the parameter \( \lambda \): \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) \] - **Maximum Likelihood Estimate (MLE)**: The MLE is the value of \( \lambda \) that maximizes the likelihood function. # Step-by-Step Solution **Deriving the Likelihood Function** Starting from the definition of the likelihood function, we can express it as: \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) = \prod_{i=1}^{n} \left( \lambda e^{-\lambda x_i} \right) \] This product can be simplified to: \[ L(\lambda) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] **Finding the MLE of \( \lambda \)** To find the MLE, we maximize the log-likelihood function \( \ell(\lambda) \): \[ \ell(\lambda) = \ln(L(\lambda)) = \ln \left( \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \right) \] This expands to: \[ \ell(\lambda) = n \ln(\lambda) - \lambda \sum_{i=1}^{n} x_i \] Next, we differentiate \( \ell(\lambda) \) with respect to \( \lambda \) and set the derivative to zero: \[ \frac{d\ell(\lambda)}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0 \] Rearranging this equation gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} x_i \] Solving for \( \lambda \) results in: \[ \lambda = \frac{n}{\sum_{i=1}^{n} x_i} \] Thus, the maximum likelihood estimate of \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] # Summary - The likelihood function is expressed as: \[ L(\lambda) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] - The maximum likelihood estimate for \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \]

# Given Information - The random variables \( x_1, x_2, \ldots, x_n \) are independent and identically distributed (i.i.d.) from an exponential distribution characterized by the probability density function (pdf): \[ f(x | \lambda) = \lambda e^{-\lambda x}, \quad x > 0 \] # What We Need to Find - Derive the likelihood function for the parameter \( \lambda \). - Determine the maximum likelihood estimate (MLE) of \( \lambda \). # Definitions or Concepts Used - **Likelihood Function**: The likelihood function \( L(\lambda) \) quantifies how likely a given set of observations is, given a specific value of the parameter \( \lambda \). It is defined as: \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) \] - **Maximum Likelihood Estimate (MLE)**: The MLE is the value of \( \lambda \) that maximizes the likelihood function. This is often achieved by maximizing the log-likelihood function instead. # Step-by-Step Solution **Deriving the Likelihood Function** The likelihood function for the observed data \( x_1, x_2, \ldots, x_n \) can be expressed as: \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) = \prod_{i=1}^{n} \left( \lambda e^{-\lambda x_i} \right) \] This product can be rewritten as: \[ L(\lambda) = \lambda^n \cdot e^{-\lambda \sum_{i=1}^{n} x_i} \] **Finding the MLE of \( \lambda \)** To find the MLE, we will maximize the log-likelihood function \( \ell(\lambda) \): \[ \ell(\lambda) = \ln(L(\lambda)) = \ln \left( \lambda^n \cdot e^{-\lambda \sum_{i=1}^{n} x_i} \right) \] This simplifies to: \[ \ell(\lambda) = n \ln(\lambda) - \lambda \sum_{i=1}^{n} x_i \] Next, we differentiate the log-likelihood function with respect to \( \lambda \) and set the derivative equal to zero: \[ \frac{d\ell(\lambda)}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0 \] Rearranging this equation gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} x_i \] Solving for \( \lambda \) leads to: \[ \lambda = \frac{n}{\sum_{i=1}^{n} x_i} \] Thus, the maximum likelihood estimate of \( \lambda \) can be expressed as: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] # Summary - The likelihood function is given by: \[ L(\lambda) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] - The maximum likelihood estimate for \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \]

# Given Information - The random variables \( x_1, x_2, \ldots, x_n \) are independent and identically distributed (i.i.d.) from an exponential distribution, defined by the probability density function (pdf): \[ f(x | \lambda) = \lambda e^{-\lambda x}, \quad x > 0 \] # What We Need to Find - Derive the likelihood function for the parameter \( \lambda \). - Find the maximum likelihood estimate (MLE) of \( \lambda \). # Definitions or Concepts Used - **Likelihood Function**: The likelihood function \( L(\lambda) \) expresses the probability of observing the given data as a function of the parameter \( \lambda \): \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) \] - **Maximum Likelihood Estimate (MLE)**: The MLE is the parameter value that maximizes the likelihood function. In practice, we often maximize the log-likelihood function. # Step-by-Step Solution **Deriving the Likelihood Function** The likelihood function for the observed values \( x_1, x_2, \ldots, x_n \) can be formulated as: \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) = \prod_{i=1}^{n} \left( \lambda e^{-\lambda x_i} \right) \] This can be simplified to: \[ L(\lambda) = \lambda^n \cdot e^{-\lambda \sum_{i=1}^{n} x_i} \] **Finding the MLE of \( \lambda \)** To find the MLE, we will maximize the log-likelihood function \( \ell(\lambda) \): \[ \ell(\lambda) = \ln(L(\lambda)) = \ln \left( \lambda^n \cdot e^{-\lambda \sum_{i=1}^{n} x_i} \right) \] This simplifies to: \[ \ell(\lambda) = n \ln(\lambda) - \lambda \sum_{i=1}^{n} x_i \] Next, we differentiate the log-likelihood function with respect to \( \lambda \) and set the derivative equal to zero: \[ \frac{d\ell(\lambda)}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0 \] Rearranging gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} x_i \] From which we solve for \( \lambda \): \[ \lambda = \frac{n}{\sum_{i=1}^{n} x_i} \] Thus, the maximum likelihood estimate of \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] # Summary - The likelihood function is: \[ L(\lambda) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] - The maximum likelihood estimate for \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \]

# Given Information - We have a set of random variables \( x_1, x_2, \ldots, x_n \) that are independent and identically distributed (i.i.d.) according to an exponential distribution, which is characterized by the probability density function (pdf): \[ f(x | \lambda) = \lambda e^{-\lambda x}, \quad x > 0 \] # What We Need to Find - We need to derive the likelihood function for the parameter \( \lambda \). - We must also calculate the maximum likelihood estimate (MLE) of \( \lambda \). # Definitions or Concepts Used - **Likelihood Function**: The likelihood function \( L(\lambda) \) quantifies how probable the observed data is for different values of the parameter \( \lambda \): \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) \] - **Maximum Likelihood Estimate (MLE)**: The MLE is the value of \( \lambda \) that maximizes the likelihood function. To make this process easier, we often work with the log-likelihood function. # Step-by-Step Solution **Deriving the Likelihood Function** We begin by expressing the likelihood function for the observed data points \( x_1, x_2, \ldots, x_n \): \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) = \prod_{i=1}^{n} \left( \lambda e^{-\lambda x_i} \right) \] This can be simplified into a more manageable form: \[ L(\lambda) = \lambda^n \cdot e^{-\lambda \sum_{i=1}^{n} x_i} \] **Finding the MLE of \( \lambda \)** To find the MLE, we will maximize the log-likelihood function \( \ell(\lambda) \), which is the natural logarithm of the likelihood function: \[ \ell(\lambda) = \ln(L(\lambda)) = \ln \left( \lambda^n \cdot e^{-\lambda \sum_{i=1}^{n} x_i} \right) \] This simplifies to: \[ \ell(\lambda) = n \ln(\lambda) - \lambda \sum_{i=1}^{n} x_i \] Next, we differentiate the log-likelihood function with respect to \( \lambda \) and set the result to zero to find the critical points: \[ \frac{d\ell(\lambda)}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0 \] Reorganizing this equation yields: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} x_i \] From here, we can solve for \( \lambda \): \[ \lambda = \frac{n}{\sum_{i=1}^{n} x_i} \] Thus, the maximum likelihood estimate of \( \lambda \) is given by: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] # Summary - The derived likelihood function is expressed as: \[ L(\lambda) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] - The maximum likelihood estimate for the parameter \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] This approach effectively captures the relationship between the observed data and the parameter \( \lambda \), providing a clear pathway to estimate \( \lambda \) based on the data at hand.

# Given Information - We have a collection of random variables \( x_1, x_2, \ldots, x_n \) that are independently and identically distributed (i.i.d.) following an exponential distribution, represented by the probability density function (pdf): \[ f(x | \lambda) = \lambda e^{-\lambda x}, \quad x > 0 \] # What We Need to Find - Our goal is to derive the likelihood function for the parameter \( \lambda \). - Additionally, we need to determine the maximum likelihood estimate (MLE) of \( \lambda \). # Definitions or Concepts Used - **Likelihood Function**: The likelihood function \( L(\lambda) \) represents the probability of obtaining the observed data given a specific value of the parameter \( \lambda \): \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) \] - **Maximum Likelihood Estimate (MLE)**: The MLE is the value of \( \lambda \) that maximizes the likelihood function. This is often achieved by maximizing the log-likelihood function, which simplifies the calculations. # Step-by-Step Solution **Deriving the Likelihood Function** We start by composing the likelihood function for the observed data points \( x_1, x_2, \ldots, x_n \): \[ L(\lambda) = \prod_{i=1}^{n} f(x_i | \lambda) = \prod_{i=1}^{n} \left( \lambda e^{-\lambda x_i} \right) \] This product can be condensed into the following form: \[ L(\lambda) = \lambda^n \cdot e^{-\lambda \sum_{i=1}^{n} x_i} \] **Finding the MLE of \( \lambda \)** To find the MLE, we will maximize the log-likelihood function \( \ell(\lambda) \), which is the logarithm of the likelihood function: \[ \ell(\lambda) = \ln(L(\lambda)) = \ln \left( \lambda^n \cdot e^{-\lambda \sum_{i=1}^{n} x_i} \right) \] This further simplifies to: \[ \ell(\lambda) = n \ln(\lambda) - \lambda \sum_{i=1}^{n} x_i \] Next, we differentiate the log-likelihood function with respect to \( \lambda \) and set the derivative equal to zero: \[ \frac{d\ell(\lambda)}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0 \] Rearranging this equation gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} x_i \] From here, we solve for \( \lambda \): \[ \lambda = \frac{n}{\sum_{i=1}^{n} x_i} \] Thus, the maximum likelihood estimate of \( \lambda \) is expressed as: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] # Summary - The derived likelihood function is: \[ L(\lambda) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] - The maximum likelihood estimate for \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] This formulation effectively captures the relationship between the observed data and the parameter \( \lambda \), providing a clear and efficient method to estimate \( \lambda \) based on the observed sample.