# Given Information - Class intervals and frequencies are: - –10: 6 - 10–20: ? - 20–30: 14 - 30–40: 10 - Total frequency = 40 # What Have to Find - The missing frequency for the interval 10–20. - Draw the histogram for the data. # Definition or Concept Used - **Frequency**: Number of observations in each class interval. - **Histogram**: A bar graph showing frequency distribution of a continuous dataset. - The sum of all frequencies equals the total given frequency. # Step by Step Solution Let the missing frequency for 10–20 be \( x \). According to the total frequency, \[ 6 + x + 14 + 10 = 40 \] \[ x + 30 = 40 \] \[ x = 40 - 30 \] \[ x = 10 \] The frequencies are now: - –10: 6 - 10–20: 10 - 20–30: 14 - 30–40: 10 To draw the histogram: - On the x-axis, mark the class intervals: –10, 10–20, 20–30, 30–40. - On the y-axis, mark the frequencies: –15. - Draw bars for each class interval up to their respective frequencies. ## Histogram Below is a simple representation of the histogram: ``` Frequency ^ 15| 14| ████ 13| ████ 12| ████ 11| ████ 10| ████ ████ 9| ████ ████ 8| ████ ████ 7| ████ ████ 6| ████ ████ ████ 5| ████ ████ ████ 4| ████ ████ ████ 3| ████ ████ ████ 2| ████ ████ ████ 1| ████ ████ ████ +------------------------> –10 10–20 20–30 30–40 ``` # Summary The missing frequency is **10**. The histogram is drawn using the frequencies: 6, 10, 14, 10 for the intervals –10, 10–20, 20–30, 30–40 respectively.

# Given Information - Class intervals and frequencies: - 0–10: 6 - 10–20: ? - 20–30: 14 - 30–40: 10 - Total frequency = 40 # What Have to Find - The missing frequency for the interval 10-20. - Draw the histogram for the data. # Definition or Concept Used - **Frequency**: The number of observations in each class interval. - **Histogram**: A graphical representation of the frequency distribution of continuous data. - The sum of all frequencies equals the total frequency provided. # Step by Step Solution Let the missing frequency for the interval 10–20 be \( x \). According to the total frequency, \[ 6 + x + 14 + 10 = 40 \] \[ x + 30 = 40 \] \[ x = 40 - 30 \] \[ x = 10 \] The updated frequencies are: - 0–10: 6 - 10–20: 10 - 20–30: 14 - 30–40: 10 To draw the histogram: - On the x-axis, label the class intervals: 0–10, 10–20, 20–30, 30–40. - On the y-axis, label the frequencies up to 15. - Draw bars for each class interval corresponding to their frequencies. ## Histogram Below is a simple text representation of the histogram: ``` Frequency ^ 15| 14| ████ 13| ████ 12| ████ 11| ████ 10| ████ ████ 9| ████ ████ 8| ████ ████ 7| ████ ████ 6| ████ ████ ████ 5| ████ ████ ████ 4| ████ ████ ████ 3| ████ ████ ████ 2| ████ ████ ████ 1| ████ ████ ████ +------------------------> 0–10 10–20 20–30 30–40 ``` # Summary The missing frequency is **10**. The histogram is drawn using the frequencies: 6, 10, 14, 10 for the intervals 0–10, 10–20, 20–30, 30–40 respectively.

# Given Information - Two fair dice are rolled. - Condition: The sum of the two dice is greater than 8. # What Have to Find - The probability that at least one die shows a 6, given that the sum is greater than 8. # Definition or Concept Used - **Sample Space**: The total possible outcomes when rolling two dice. - **Conditional Probability**: The probability of an event occurring given that another event has already occurred, defined as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] where \( A \) is the event of interest and \( B \) is the condition. # Step by Step Solution 1. **Calculate the total outcomes when rolling two dice**: - Each die has 6 faces, thus the total outcomes is \( 6 \times 6 = 36 \). 2. **Identify outcomes where the sum is greater than 8**: - The combinations are: - Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes - Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes - Sum = 11: (5,6), (6,5) → 2 outcomes - Sum = 12: (6,6) → 1 outcome - Total outcomes where the sum is greater than 8: \( 4 + 3 + 2 + 1 = 10 \). 3. **Identify outcomes where at least one die shows a 6 and the sum is greater than 8**: - Possible outcomes: - (6,3) → Sum = 9 - (6,4) → Sum = 10 - (6,5) → Sum = 11 - (6,6) → Sum = 12 - (3,6) → Sum = 9 - (4,6) → Sum = 10 - (5,6) → Sum = 11 - Total outcomes with at least one 6 and sum > 8: 6 outcomes. 4. **Calculate the conditional probability**: \[ P(\text{at least one 6} | \text{sum} > 8) = \frac{P(\text{at least one 6} \cap \text{sum} > 8)}{P(\text{sum} > 8)} = \frac{6}{10} = 0.6 \] # Summary The probability that at least one die shows a 6, given that the sum is greater than 8, is **0.6**.

# Given Information - Two fair dice are rolled. - Condition: The sum of the two dice is greater than 8. # What Have to Find - The probability that at least one die shows a 6, given that the sum is greater than 8. # Definition or Concept Used - **Sample Space**: The total possible outcomes when rolling two dice, which is \( 6 \times 6 = 36 \). - **Conditional Probability**: The probability of an event occurring given that another event has already occurred, defined as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] where \( A \) is the event of interest and \( B \) is the condition. # Step by Step Solution 1. **Calculate the total outcomes when rolling two dice**: - Total outcomes = \( 36 \). 2. **Identify outcomes where the sum is greater than 8**: - Possible combinations: - Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes - Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes - Sum = 11: (5,6), (6,5) → 2 outcomes - Sum = 12: (6,6) → 1 outcome - Total outcomes where the sum is greater than 8: \( 4 + 3 + 2 + 1 = 10 \). 3. **Identify outcomes where at least one die shows a 6 and the sum is greater than 8**: - Valid outcomes: - (6,3) → Sum = 9 - (6,4) → Sum = 10 - (6,5) → Sum = 11 - (6,6) → Sum = 12 - (3,6) → Sum = 9 - (4,6) → Sum = 10 - (5,6) → Sum = 11 - Total outcomes with at least one 6 and sum > 8: **7 outcomes**. 4. **Calculate the conditional probability**: \[ P(\text{at least one 6} | \text{sum} > 8) = \frac{P(\text{at least one 6} \cap \text{sum} > 8)}{P(\text{sum} > 8)} = \frac{7}{10} \] # Summary The probability that at least one die shows a 6, given that the sum is greater than 8, is **\(\frac{7}{10}\)**.

# Given Information - A box contains: - Gold coins: 3 - Silver coins: 5 - Fake coins: 2 - Total coins: \(3 + 5 + 2 = 10\) - Condition: At least one coin picked is gold. # What Have to Find - The probability that both coins picked are gold, given that at least one is gold. # Definition or Concept Used - **Conditional Probability**: The probability of an event occurring given that another event has already occurred, defined as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] where \( A \) is the event of both coins being gold and \( B \) is the event of at least one coin being gold. # Step by Step Solution 1. **Calculate the total ways to pick 2 coins from 10**: \[ \text{Total ways} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \] 2. **Calculate the total ways to pick 2 coins such that at least one is gold**: - Total ways to pick 2 coins with no gold: - Pick from silver and fake coins only (7 coins total): \[ \text{Ways without gold} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \] - Thus, ways to pick at least one gold: \[ \text{Ways with at least one gold} = \text{Total ways} - \text{Ways without gold} = 45 - 21 = 24 \] 3. **Calculate the total ways to pick 2 gold coins**: \[ \text{Ways to pick 2 gold} = \binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3 \] 4. **Calculate the conditional probability**: \[ P(\text{both gold} | \text{at least one gold}) = \frac{P(\text{both gold} \cap \text{at least one gold})}{P(\text{at least one gold})} = \frac{3}{24} = \frac{1}{8} \] # Summary The probability that both coins picked are gold, given that at least one is gold, is **\(\frac{1}{8}\)**.

# Given Information - Prevalence of the disease: \( P(D) = 0.01 \) (1%) - Accuracy of the test when the disease is present: \( P(T^+ | D) = 0.98 \) - False positive rate: \( P(T^+ | D^c) = 0.03 \) - Probability of not having the disease: \( P(D^c) = 1 - P(D) = 0.99 \) # What Have to Find - The probability that a person actually has the disease given they tested positive twice independently: \( P(D | T^+_1 \cap T^+_2) \) # Definition or Concept Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} \] - **Total Probability**: \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D^c) P(D^c) \] # Step by Step Solution 1. **Calculate the probability of testing positive once**: \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D^c) P(D^c) \] \[ P(T^+) = (0.98)(0.01) + (0.03)(0.99) \] \[ P(T^+) = 0.0098 + 0.0297 = 0.0395 \] 2. **Calculate the conditional probability of having the disease given one positive test**: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} = \frac{(0.98)(0.01)}{0.0395} \] \[ P(D | T^+) = \frac{0.0098}{0.0395} \approx 0.2481 \] 3. **Since the tests are independent, calculate the probability after two positive tests**: \[ P(D | T^+_1 \cap T^+_2) = \frac{P(T^+_2 | D) P(D | T^+_1)}{P(T^+)} \] - Use \( P(D | T^+_1) \) from the previous result: \[ P(D | T^+_1 \cap T^+_2) = \frac{(0.98)(0.2481)}{0.0395} \] \[ P(D | T^+_1 \cap T^+_2) = \frac{0.2434}{0.0395} \approx 0.6166 \] # Summary The probability that a person actually has the disease given they tested positive twice is approximately **0.6166** or **61.66%**.

# Given Information - Prevalence of the disease: \( P(D) = 0.01 \) (1%) - Accuracy of the test when the disease is present: \( P(T^+ | D) = 0.98 \) - False positive rate: \( P(T^+ | D^c) = 0.03 \) - Probability of not having the disease: \( P(D^c) = 1 - P(D) = 0.99 \) # What Have to Find - The probability that a person actually has the disease given they tested positive twice independently: \( P(D | T^+_1 \cap T^+_2) \) # Definition or Concept Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} \] - **Total Probability**: \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D^c) P(D^c) \] # Step by Step Solution 1. **Calculate the probability of testing positive once**: \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D^c) P(D^c) \] \[ P(T^+) = (0.98)(0.01) + (0.03)(0.99) \] \[ P(T^+) = 0.0098 + 0.0297 = 0.0395 \] 2. **Calculate the conditional probability of having the disease given one positive test**: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} = \frac{(0.98)(0.01)}{0.0395} \] \[ P(D | T^+) = \frac{0.0098}{0.0395} \approx 0.2481 \] 3. **Calculate the probability of a second positive test**: - Given that the first test was positive, the new probability of having the disease remains \( P(D | T^+) \). Thus: \[ P(T^+ | D) = 0.98 \] \[ P(T^+ | D^c) = 0.03 \] 4. **Calculate the total probability of testing positive twice**: \[ P(T^+_1 \cap T^+_2) = P(T^+ | D) P(D | T^+) + P(T^+ | D^c) P(D^c | T^+) \] - The probability of not having the disease given a positive test is: \[ P(D^c | T^+) = 1 - P(D | T^+) \approx 1 - 0.2481 = 0.7519 \] - Therefore: \[ P(T^+_1 \cap T^+_2) = (0.98)(0.2481) + (0.03)(0.7519) \] \[ P(T^+_1 \cap T^+_2) = 0.2434 + 0.0226 = 0.266 \] 5. **Calculate the conditional probability given two positive tests**: \[ P(D | T^+_1 \cap T^+_2) = \frac{P(T^+_1 \cap T^+_2 | D) P(D)}{P(T^+_1 \cap T^+_2)} \] \[ P(D | T^+_1 \cap T^+_2) = \frac{(0.98)(0.98)(0.01)}{0.266} \] \[ P(D | T^+_1 \cap T^+_2) = \frac{0.009604}{0.266} \approx 0.0361 \] # Summary The probability that a person actually has the disease given they tested positive twice is approximately **0.0361** or **3.61%**.

# Given Information - Prevalence of the disease: \( P(D) = 0.01 \) (1% have the disease) - Probability of not having the disease: \( P(D^c) = 0.99 \) (99% do not have it) - Test characteristics: - Sensitivity (true positive rate): \( P(T^+ | D) = 0.98 \) (98% probability of a positive test when the disease is present) - False positive rate: \( P(T^+ | D^c) = 0.03 \) (3% probability of a positive test when the disease is not present) # What Have to Find - The probability that a person actually has the disease given they tested positive twice independently: \( P(D | T^+_1 \cap T^+_2) \) # Definition or Concept Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} \] - **Total Probability**: \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D^c) P(D^c) \] # Step by Step Solution 1. **Calculate the probability of testing positive twice given the disease**: \[ P(T^+_1 \cap T^+_2 | D) = (0.98)^2 = 0.9604 \] 2. **Calculate the probability of testing positive twice given no disease**: \[ P(T^+_1 \cap T^+_2 | D^c) = (0.03)^2 = 0.0009 \] 3. **Calculate the overall probability of testing positive twice**: \[ P(T^+_1 \cap T^+_2) = P(T^+_1 \cap T^+_2 | D) P(D) + P(T^+_1 \cap T^+_2 | D^c) P(D^c) \] Substituting the values: \[ P(T^+_1 \cap T^+_2) = (0.9604)(0.01) + (0.0009)(0.99) \] \[ = 0.009604 + 0.000891 = 0.010495 \] 4. **Apply Bayes' Theorem**: \[ P(D | T^+_1 \cap T^+_2) = \frac{P(T^+_1 \cap T^+_2 | D) P(D)}{P(T^+_1 \cap T^+_2)} \] Substituting the values: \[ P(D | T^+_1 \cap T^+_2) = \frac{(0.9604)(0.01)}{0.010495} \] \[ = \frac{0.009604}{0.010495} \approx 0.9151 \] # Summary The probability that a person actually has the disease given they tested positive twice is approximately **0.9151** or **91.51%**.

# Given Information - Prevalence of the disease: \( P(D) = 0.01 \) (1% have the disease) - Probability of not having the disease: \( P(D^c) = 0.99 \) (99% do not have it) - Test characteristics: - Sensitivity (true positive rate): \( P(T^+ | D) = 0.98 \) (98% probability of a positive test when the disease is present) - False positive rate: \( P(T^+ | D^c) = 0.03 \) (3% probability of a positive test when the disease is not present) # What Have to Find - The probability that a person actually has the disease given they tested positive twice independently: \( P(D | T^+_1 \cap T^+_2) \) # Definition or Concept Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} \] - **Total Probability**: \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D^c) P(D^c) \] # Step by Step Solution 1. **Calculate the probability of testing positive twice given the disease**: \[ P(T^+_1 \cap T^+_2 | D) = (0.98)^2 = 0.9604 \] 2. **Calculate the probability of testing positive twice given no disease**: \[ P(T^+_1 \cap T^+_2 | D^c) = (0.03)^2 = 0.0009 \] 3. **Calculate the overall probability of testing positive twice**: \[ P(T^+_1 \cap T^+_2) = P(T^+_1 \cap T^+_2 | D) P(D) + P(T^+_1 \cap T^+_2 | D^c) P(D^c) \] Substituting the values: \[ P(T^+_1 \cap T^+_2) = (0.9604)(0.01) + (0.0009)(0.99) \] \[ = 0.009604 + 0.000891 = 0.010495 \] 4. **Apply Bayes' Theorem**: \[ P(D | T^+_1 \cap T^+_2) = \frac{P(T^+_1 \cap T^+_2 | D) P(D)}{P(T^+_1 \cap T^+_2)} \] Substituting the values: \[ P(D | T^+_1 \cap T^+_2) = \frac{(0.9604)(0.01)}{0.010495} \] \[ = \frac{0.009604}{0.010495} \approx 0.916 \] # Summary The probability that a person actually has the disease given they tested positive twice is approximately **0.916** or **91.6%**.

# Given Information - Prevalence of the disease: \( P(D) = 0.01 \) (1% of the population has the disease) - Probability of not having the disease: \( P(D^c) = 0.99 \) (99% do not have the disease) - Test characteristics: - Sensitivity (true positive rate): \( P(T^+ | D) = 0.98 \) (The test correctly identifies the disease 98% of the time) - False positive rate: \( P(T^+ | D^c) = 0.03 \) (The test incorrectly identifies the disease 3% of the time when it is not present) # What Have to Find - The probability that a person actually has the disease given they tested positive twice independently: \( P(D | T^+_1 \cap T^+_2) \) # Definition or Concept Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} \] - **Total Probability**: \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D^c) P(D^c) \] # Step by Step Solution 1. **Calculate the probability of testing positive twice given the disease**: \[ P(T^+_1 \cap T^+_2 | D) = (0.98)^2 = 0.9604 \] 2. **Calculate the probability of testing positive twice given no disease**: \[ P(T^+_1 \cap T^+_2 | D^c) = (0.03)^2 = 0.0009 \] 3. **Calculate the overall probability of testing positive twice**: \[ P(T^+_1 \cap T^+_2) = P(T^+_1 \cap T^+_2 | D) P(D) + P(T^+_1 \cap T^+_2 | D^c) P(D^c) \] Substituting the values: \[ P(T^+_1 \cap T^+_2) = (0.9604)(0.01) + (0.0009)(0.99) \] \[ = 0.009604 + 0.000891 = 0.010495 \] 4. **Apply Bayes' Theorem**: \[ P(D | T^+_1 \cap T^+_2) = \frac{P(T^+_1 \cap T^+_2 | D) P(D)}{P(T^+_1 \cap T^+_2)} \] Substituting the values: \[ P(D | T^+_1 \cap T^+_2) = \frac{(0.9604)(0.01)}{0.010495} \] \[ = \frac{0.009604}{0.010495} \approx 0.916 \] # Summary The probability that a person actually has the disease given they tested positive twice is approximately **0.916** or **91.6%**.

# Given Information - A box contains: - Gold coins: 3 - Silver coins: 5 - Fake coins: 2 - Total coins: \(3 + 5 + 2 = 10\) - Condition: At least one coin picked is gold. # What Have to Find - The probability that both coins picked are gold given that at least one is gold: \( P(\text{both gold} | \text{at least one gold}) \) # Definition or Concept Used - **Conditional Probability**: The probability of an event occurring given that another event has already occurred, defined as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] where \( A \) is the event of interest and \( B \) is the condition. # Step by Step Solution 1. **Calculate the total ways to pick 2 coins from 10**: \[ \text{Total ways} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \] 2. **Calculate the total ways to pick 2 coins such that at least one is gold**: - Calculate the total ways to pick 2 coins without any gold: - Pick from silver and fake coins only (7 coins total): \[ \text{Ways without gold} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \] - Thus, ways to pick at least one gold: \[ \text{Ways with at least one gold} = \text{Total ways} - \text{Ways without gold} = 45 - 21 = 24 \] 3. **Calculate the total ways to pick 2 gold coins**: \[ \text{Ways to pick 2 gold} = \binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3 \] 4. **Calculate the conditional probability**: \[ P(\text{both gold} | \text{at least one gold}) = \frac{P(\text{both gold} \cap \text{at least one gold})}{P(\text{at least one gold})} = \frac{3}{24} = \frac{1}{8} \] # Summary The probability that both coins picked are gold given that at least one is gold is **\(\frac{1}{8}\)**.

# Given Information - A box contains: - Gold coins: 3 - Silver coins: 5 - Fake coins: 2 - Total number of coins: \(3 + 5 + 2 = 10\) - Condition: At least one coin picked is gold. # What Have to Find - The probability that both coins picked are gold given that at least one coin is gold: \( P(\text{both gold} | \text{at least one gold}) \) # Definition or Concept Used - **Conditional Probability**: The probability of an event occurring given another event has occurred, expressed as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] where \( A \) is the event of interest and \( B \) is the condition. # Step by Step Solution 1. **Total ways to pick 2 coins from 10**: \[ \text{Total combinations} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \] 2. **Calculate combinations for at least one gold**: - Count combinations without any gold coins: - Select from silver and fake coins (7 total): \[ \text{Combinations without gold} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \] - Therefore, combinations with at least one gold coin: \[ \text{Combinations with at least one gold} = \text{Total combinations} - \text{Combinations without gold} = 45 - 21 = 24 \] 3. **Calculate combinations for picking 2 gold coins**: \[ \text{Combinations for 2 gold} = \binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3 \] 4. **Calculate the conditional probability**: \[ P(\text{both gold} | \text{at least one gold}) = \frac{P(\text{both gold} \cap \text{at least one gold})}{P(\text{at least one gold})} = \frac{3}{24} = \frac{1}{8} \] # Summary The probability that both coins picked are gold given that at least one is gold is **\(\frac{1}{8}\)**.

# Given Information - A box contains: - Gold coins: 3 - Silver coins: 5 - Fake coins: 2 - Total coins: \(3 + 5 + 2 = 10\) - Condition: At least one coin picked is gold. # What Have to Find - The probability that both coins picked are gold, given that at least one of them is gold: \( P(\text{both gold} | \text{at least one gold}) \) # Definition or Concept Used - **Conditional Probability**: The probability of an event occurring given another event has occurred, defined as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] where \( A \) is the event of interest and \( B \) is the condition. # Step by Step Solution 1. **Calculate the total number of ways to choose 2 coins from 10**: \[ \text{Total combinations} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \] 2. **Determine combinations without any gold coins**: - Total coins excluding gold = Silver + Fake = \(5 + 2 = 7\) \[ \text{Ways without gold} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \] 3. **Calculate combinations with at least one gold coin**: \[ \text{Ways with at least one gold} = \text{Total ways} - \text{Ways without gold} = 45 - 21 = 24 \] 4. **Calculate the ways to pick 2 gold coins**: \[ \text{Ways to pick 2 gold} = \binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3 \] 5. **Calculate the conditional probability**: \[ P(\text{both gold} | \text{at least one gold}) = \frac{P(\text{both gold} \cap \text{at least one gold})}{P(\text{at least one gold})} = \frac{3}{24} = \frac{1}{8} \] # Summary The probability that both coins picked are gold, given that at least one coin is gold, is **\(\frac{1}{8}\)**.

# Given Information - A box contains: - Gold coins: 3 - Silver coins: 5 - Fake coins: 2 - Total number of coins: \(3 + 5 + 2 = 10\) - Condition: At least one coin picked is gold. # What Have to Find - The probability that both coins picked are gold, given that at least one of them is gold: \( P(\text{both gold} | \text{at least one gold}) \) # Definition or Concept Used - **Conditional Probability**: The probability of an event occurring given that another event has occurred, defined as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] where \( A \) is the event of interest and \( B \) is the condition. # Step by Step Solution 1. **Calculate the total ways to choose 2 coins from 10**: \[ \text{Total combinations} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \] 2. **Determine the combinations that do not include any gold coins**: - Total non-gold coins: Silver + Fake = \(5 + 2 = 7\) \[ \text{Ways without gold} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \] 3. **Calculate the combinations that have at least one gold coin**: \[ \text{Ways with at least one gold} = \text{Total ways} - \text{Ways without gold} = 45 - 21 = 24 \] 4. **Calculate the combinations for picking 2 gold coins**: \[ \text{Ways to pick 2 gold} = \binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3 \] 5. **Calculate the conditional probability**: \[ P(\text{both gold} | \text{at least one gold}) = \frac{P(\text{both gold} \cap \text{at least one gold})}{P(\text{at least one gold})} = \frac{3}{24} = \frac{1}{8} \] # Summary The probability that both coins picked are gold, given that at least one coin is gold, is **\(\frac{1}{8}\)**.

# Given Information - A box contains: - Gold coins: 3 - Silver coins: 5 - Fake coins: 2 - Total coins: \(3 + 5 + 2 = 10\) - Condition: At least one coin picked is gold. # What Have to Find - The probability that both coins picked are gold, given that at least one of them is gold: \( P(\text{both gold} | \text{at least one gold}) \) # Definition or Concept Used - **Conditional Probability**: The probability of an event occurring given that another event has occurred, defined as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] where \( A \) is the event of interest (both coins being gold) and \( B \) is the condition (at least one coin being gold). # Step by Step Solution 1. **Calculate the total combinations of selecting 2 coins from 10**: \[ \text{Total combinations} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \] 2. **Determine combinations that do not include any gold coins**: - Total non-gold coins: Silver + Fake = \(5 + 2 = 7\) \[ \text{Ways without gold} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \] 3. **Calculate the combinations that include at least one gold coin**: \[ \text{Ways with at least one gold} = \text{Total ways} - \text{Ways without gold} = 45 - 21 = 24 \] 4. **Calculate the combinations for picking 2 gold coins**: \[ \text{Ways to pick 2 gold} = \binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3 \] 5. **Calculate the conditional probability**: \[ P(\text{both gold} | \text{at least one gold}) = \frac{P(\text{both gold and at least one gold})}{P(\text{at least one gold})} = \frac{3}{24} = \frac{1}{8} \] # Summary The probability that both coins picked are gold, given that at least one coin is gold, is **\(\frac{1}{8}\)**.

# Given Information - A box contains: - Gold coins: 3 - Silver coins: 5 - Fake coins: 2 - Total coins: \(3 + 5 + 2 = 10\) - Condition: At least one of the coins picked is gold. # What Have to Find - The probability that both coins picked are gold, given that at least one of them is gold: \( P(\text{both gold} | \text{at least one gold}) \) # Definition or Concept Used - **Conditional Probability**: The probability of an event occurring given that another event has occurred, defined as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] where \( A \) is the event of interest (both coins being gold) and \( B \) is the condition (at least one coin being gold). # Step by Step Solution 1. **Calculate the total ways to choose 2 coins from 10**: \[ \text{Total combinations} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \] 2. **Determine combinations that do not include any gold coins**: - Total non-gold coins: Silver + Fake = \(5 + 2 = 7\) \[ \text{Ways without gold} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \] 3. **Calculate the combinations that include at least one gold coin**: \[ \text{Ways with at least one gold} = \text{Total ways} - \text{Ways without gold} = 45 - 21 = 24 \] 4. **Calculate the combinations for picking 2 gold coins**: \[ \text{Ways to pick 2 gold} = \binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3 \] 5. **Calculate the conditional probability**: \[ P(\text{both gold} | \text{at least one gold}) = \frac{P(\text{both gold and at least one gold})}{P(\text{at least one gold})} = \frac{3}{24} = \frac{1}{8} \] # Summary The probability that both coins picked are gold, given that at least one of them is gold, is **\(\frac{1}{8}\)**.

# Given Information - Prevalence of the disease: \( P(D) = 0.01 \) (1% of the population has the disease) - Probability of not having the disease: \( P(D^c) = 0.99 \) (99% do not have the disease) - Test characteristics: - Sensitivity (true positive rate): \( P(T^+ | D) = 0.98 \) (98% probability of a positive test when the disease is present) - False positive rate: \( P(T^+ | D^c) = 0.03 \) (3% probability of a positive test when the disease is not present) # What Have to Find - The probability that a person actually has the disease given they tested positive twice independently: \( P(D | T^+_1 \cap T^+_2) \) # Definition or Concept Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} \] - **Total Probability**: \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D^c) P(D^c) \] # Step by Step Solution 1. **Calculate the probability of testing positive once**: \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D^c) P(D^c) \] Substituting the values: \[ P(T^+) = (0.98)(0.01) + (0.03)(0.99) \] \[ P(T^+) = 0.0098 + 0.0297 = 0.0395 \] 2. **Calculate the conditional probability of having the disease after one positive test**: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} \] Substituting the values: \[ P(D | T^+) = \frac{(0.98)(0.01)}{0.0395} \approx 0.2481 \] 3. **Calculate the probability of testing positive twice given the disease**: - Since tests are independent: \[ P(T^+_1 \cap T^+_2 | D) = (0.98)^2 = 0.9604 \] 4. **Calculate the probability of testing positive twice given no disease**: \[ P(T^+_1 \cap T^+_2 | D^c) = (0.03)^2 = 0.0009 \] 5. **Calculate the overall probability of testing positive twice**: \[ P(T^+_1 \cap T^+_2) = P(T^+_1 \cap T^+_2 | D) P(D) + P(T^+_1 \cap T^+_2 | D^c) P(D^c) \] Substituting the values: \[ P(T^+_1 \cap T^+_2) = (0.9604)(0.01) + (0.0009)(0.99) \] \[ = 0.009604 + 0.000891 = 0.010495 \] 6. **Apply Bayes' Theorem for two positive tests**: \[ P(D | T^+_1 \cap T^+_2) = \frac{P(T^+_1 \cap T^+_2 | D) P(D)}{P(T^+_1 \cap T^+_2)} \] Substituting the values: \[ P(D | T^+_1 \cap T^+_2) = \frac{(0.9604)(0.01)}{0.010495} \] \[ = \frac{0.009604}{0.010495} \approx 0.916 \] # Summary The probability that a person actually has the disease given they tested positive twice is approximately **0.916** or **91.6%**.

# Given Information - A disease affects **1%** of the population: \( P(D) = 0.01 \) - The probability of not having the disease: \( P(D^c) = 0.99 \) - Test characteristics: - Sensitivity (true positive rate): \( P(T^+ | D) = 0.98 \) (the test detects the disease correctly 98% of the time) - False positive rate: \( P(T^+ | D^c) = 0.03 \) (the test incorrectly detects the disease 3% of the time when it is not present) # What Have to Find - The probability that a person actually has the disease given that they tested positive twice independently: \( P(D | T^+_1 \cap T^+_2) \) # Definition or Concept Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] - **Total Probability**: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^c) \cdot P(D^c) \] # Step by Step Solution 1. **Calculate the probability of testing positive once**: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^c) \cdot P(D^c) \] Substituting the known values: \[ P(T^+) = (0.98 \times 0.01) + (0.03 \times 0.99) \] \[ P(T^+) = 0.0098 + 0.0297 = 0.0395 \] 2. **Calculate the conditional probability of having the disease after one positive test**: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] Plugging in the values: \[ P(D | T^+) = \frac{(0.98 \times 0.01)}{0.0395} \approx 0.2481 \] 3. **Calculate the probability of testing positive twice given the disease**: \[ P(T^+_1 \cap T^+_2 | D) = (0.98)^2 = 0.9604 \] 4. **Calculate the probability of testing positive twice given no disease**: \[ P(T^+_1 \cap T^+_2 | D^c) = (0.03)^2 = 0.0009 \] 5. **Calculate the overall probability of testing positive twice**: \[ P(T^+_1 \cap T^+_2) = P(T^+_1 \cap T^+_2 | D) \cdot P(D) + P(T^+_1 \cap T^+_2 | D^c) \cdot P(D^c) \] Substituting the known values: \[ P(T^+_1 \cap T^+_2) = (0.9604 \times 0.01) + (0.0009 \times 0.99) \] \[ P(T^+_1 \cap T^+_2) = 0.009604 + 0.000891 = 0.010495 \] 6. **Apply Bayes' Theorem for two positive tests**: \[ P(D | T^+_1 \cap T^+_2) = \frac{P(T^+_1 \cap T^+_2 | D) \cdot P(D)}{P(T^+_1 \cap T^+_2)} \] Substituting the values: \[ P(D | T^+_1 \cap T^+_2) = \frac{(0.9604 \times 0.01)}{0.010495} \] \[ P(D | T^+_1 \cap T^+_2) = \frac{0.009604}{0.010495} \approx 0.916 \] # Summary The probability that a person actually has the disease given that they tested positive twice is approximately **0.916** or **91.6%**.

# Given Information - **Disease Prevalence**: \( P(D) = 0.01 \) (1% of the population is affected) - **Probability of Not Having the Disease**: \( P(D^c) = 0.99 \) (99% of the population is unaffected) - **Test Characteristics**: - **Sensitivity** (True Positive Rate): \( P(T^+ | D) = 0.98 \) (The test correctly identifies the disease 98% of the time) - **False Positive Rate**: \( P(T^+ | D^c) = 0.03 \) (The test incorrectly identifies the disease in 3% of cases when it is not present) # What We Need to Find - The probability that an individual has the disease given that they tested positive twice independently: \( P(D | T^+_1 \cap T^+_2) \) # Definitions and Concepts Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] - **Law of Total Probability**: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^c) \cdot P(D^c) \] # Step-by-Step Solution 1. **Calculate the Probability of Testing Positive Once**: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^c) \cdot P(D^c) \] Substituting the values: \[ P(T^+) = (0.98 \cdot 0.01) + (0.03 \cdot 0.99) \] \[ P(T^+) = 0.0098 + 0.0297 = 0.0395 \] 2. **Calculate the Conditional Probability of Having the Disease After One Positive Test**: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] Plugging in the values: \[ P(D | T^+) = \frac{(0.98 \cdot 0.01)}{0.0395} \approx 0.2481 \] 3. **Probability of Testing Positive Twice Given the Disease**: \[ P(T^+_1 \cap T^+_2 | D) = (0.98)^2 = 0.9604 \] 4. **Probability of Testing Positive Twice Given No Disease**: \[ P(T^+_1 \cap T^+_2 | D^c) = (0.03)^2 = 0.0009 \] 5. **Overall Probability of Testing Positive Twice**: \[ P(T^+_1 \cap T^+_2) = P(T^+_1 \cap T^+_2 | D) \cdot P(D) + P(T^+_1 \cap T^+_2 | D^c) \cdot P(D^c) \] Substituting values: \[ P(T^+_1 \cap T^+_2) = (0.9604 \cdot 0.01) + (0.0009 \cdot 0.99) \] \[ P(T^+_1 \cap T^+_2) = 0.009604 + 0.000891 = 0.010495 \] 6. **Apply Bayes' Theorem for Two Positive Tests**: \[ P(D | T^+_1 \cap T^+_2) = \frac{P(T^+_1 \cap T^+_2 | D) \cdot P(D)}{P(T^+_1 \cap T^+_2)} \] Plugging in the values: \[ P(D | T^+_1 \cap T^+_2) = \frac{(0.9604 \cdot 0.01)}{0.010495} \] \[ P(D | T^+_1 \cap T^+_2) = \frac{0.009604}{0.010495} \approx 0.916 \] # Summary The probability that an individual actually has the disease given that they tested positive twice is approximately **0.916** or **91.6%**.

# Given Information - **Prevalence of the Disease**: \( P(D) = 0.01 \) (1% of individuals in the population are affected) - **Probability of Not Having the Disease**: \( P(D^c) = 0.99 \) (99% of individuals are unaffected) - **Test Characteristics**: - **True Positive Rate** (Sensitivity): \( P(T^+ | D) = 0.98 \) (The test is accurate 98% of the time when the disease is present) - **False Positive Rate**: \( P(T^+ | D^c) = 0.03 \) (The test incorrectly indicates the disease in 3% of cases when it is absent) # Objective - To find the probability that an individual has the disease given they tested positive twice: \( P(D | T^+_1 \cap T^+_2) \) # Definitions and Concepts Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] - **Law of Total Probability**: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^c) \cdot P(D^c) \] # Step-by-Step Calculation 1. **Probability of Testing Positive Once**: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^c) \cdot P(D^c) \] Substituting the values: \[ P(T^+) = (0.98 \cdot 0.01) + (0.03 \cdot 0.99) \] \[ P(T^+) = 0.0098 + 0.0297 = 0.0395 \] 2. **Calculating the Probability of Having the Disease After One Positive Test**: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] Plugging in the values: \[ P(D | T^+) = \frac{(0.98 \cdot 0.01)}{0.0395} \approx 0.2481 \] 3. **Probability of Two Positive Tests Given the Disease**: \[ P(T^+_1 \cap T^+_2 | D) = (0.98)^2 = 0.9604 \] 4. **Probability of Two Positive Tests Given No Disease**: \[ P(T^+_1 \cap T^+_2 | D^c) = (0.03)^2 = 0.0009 \] 5. **Overall Probability of Testing Positive Twice**: \[ P(T^+_1 \cap T^+_2) = P(T^+_1 \cap T^+_2 | D) \cdot P(D) + P(T^+_1 \cap T^+_2 | D^c) \cdot P(D^c) \] Substituting values: \[ P(T^+_1 \cap T^+_2) = (0.9604 \cdot 0.01) + (0.0009 \cdot 0.99) \] \[ P(T^+_1 \cap T^+_2) = 0.009604 + 0.000891 = 0.010495 \] 6. **Final Calculation Using Bayes' Theorem**: \[ P(D | T^+_1 \cap T^+_2) = \frac{P(T^+_1 \cap T^+_2 | D) \cdot P(D)}{P(T^+_1 \cap T^+_2)} \] Substituting the values: \[ P(D | T^+_1 \cap T^+_2) = \frac{(0.9604 \cdot 0.01)}{0.010495} \] \[ P(D | T^+_1 \cap T^+_2) = \frac{0.009604}{0.010495} \approx 0.916 \] # Summary The probability that an individual actually has the disease, given that they tested positive twice, is approximately **0.916** or **91.6%**.

# Given Information - **Prevalence of the Disease**: \( P(D) = 0.01 \) (1% of the population is affected) - **Probability of Not Having the Disease**: \( P(D^c) = 0.99 \) (99% of the population is unaffected) - **Test Characteristics**: - **Sensitivity** (True Positive Rate): \( P(T^+ | D) = 0.95 \) (The test correctly identifies the disease 95% of the time) - **Specificity** (True Negative Rate): \( P(T^- | D^c) = 0.90 \) (The test correctly identifies non-disease cases 90% of the time) # What We Need to Find - The probability that an individual has the disease given they tested positive twice: \( P(D | T^+_1 \cap T^+_2) \) # Definitions and Concepts Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] - **Total Probability**: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^c) \cdot P(D^c) \] where \( P(T^+ | D^c) = 1 - P(T^- | D^c) = 1 - 0.90 = 0.10 \). # Step-by-Step Calculation 1. **Calculate the Probability of Testing Positive Once**: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^c) \cdot P(D^c) \] Substituting the values: \[ P(T^+) = (0.95 \cdot 0.01) + (0.10 \cdot 0.99) \] \[ P(T^+) = 0.0095 + 0.099 = 0.1085 \] 2. **Calculate the Conditional Probability of Having the Disease After One Positive Test**: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] Plugging in the values: \[ P(D | T^+) = \frac{(0.95 \cdot 0.01)}{0.1085} \approx 0.0876 \] 3. **Probability of Testing Positive Twice Given the Disease**: \[ P(T^+_1 \cap T^+_2 | D) = (0.95)^2 = 0.9025 \] 4. **Probability of Testing Positive Twice Given No Disease**: \[ P(T^+_1 \cap T^+_2 | D^c) = (0.10)^2 = 0.01 \] 5. **Overall Probability of Testing Positive Twice**: \[ P(T^+_1 \cap T^+_2) = P(T^+_1 \cap T^+_2 | D) \cdot P(D) + P(T^+_1 \cap T^+_2 | D^c) \cdot P(D^c) \] Substituting values: \[ P(T^+_1 \cap T^+_2) = (0.9025 \cdot 0.01) + (0.01 \cdot 0.99) \] \[ P(T^+_1 \cap T^+_2) = 0.009025 + 0.0099 = 0.018925 \] 6. **Final Calculation Using Bayes' Theorem**: \[ P(D | T^+_1 \cap T^+_2) = \frac{P(T^+_1 \cap T^+_2 | D) \cdot P(D)}{P(T^+_1 \cap T^+_2)} \] Substituting the values: \[ P(D | T^+_1 \cap T^+_2) = \frac{(0.9025 \cdot 0.01)}{0.018925} \] \[ P(D | T^+_1 \cap T^+_2) = \frac{0.009025}{0.018925} \approx 0.476 \] # Summary The probability that a person actually has the disease given that they tested positive twice is approximately **0.476** or **47.6%**.

# Given Information - **Prevalence of the Disease**: \( P(D) = 0.01 \) (1% of the population is affected) - **Probability of Not Having the Disease**: \( P(D^c) = 0.99 \) (99% are unaffected) - **Test Characteristics**: - **Sensitivity** (True Positive Rate): \( P(T^+ | D) = 0.95 \) (The test correctly detects the disease 95% of the time) - **Specificity** (True Negative Rate): \( P(T^- | D^c) = 0.90 \) (The test accurately identifies non-disease cases 90% of the time) # Objective - To determine the probability that an individual has the disease given that they tested positive twice: \( P(D | T^+_1 \cap T^+_2) \) # Definitions and Concepts Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] - **Total Probability**: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^c) \cdot P(D^c) \] where \( P(T^+ | D^c) = 1 - P(T^- | D^c) = 0.10 \). # Step-by-Step Calculation 1. **Calculate the Probability of Testing Positive Once**: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^c) \cdot P(D^c) \] Plugging in the values: \[ P(T^+) = (0.95 \cdot 0.01) + (0.10 \cdot 0.99) \] \[ P(T^+) = 0.0095 + 0.099 = 0.1085 \] 2. **Calculate the Probability of Having the Disease After One Positive Test**: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] Substituting the values: \[ P(D | T^+) = \frac{(0.95 \cdot 0.01)}{0.1085} \approx 0.0876 \] 3. **Probability of Testing Positive Twice Given the Disease**: \[ P(T^+_1 \cap T^+_2 | D) = (0.95)^2 = 0.9025 \] 4. **Probability of Testing Positive Twice Given No Disease**: \[ P(T^+_1 \cap T^+_2 | D^c) = (0.10)^2 = 0.01 \] 5. **Overall Probability of Testing Positive Twice**: \[ P(T^+_1 \cap T^+_2) = P(T^+_1 \cap T^+_2 | D) \cdot P(D) + P(T^+_1 \cap T^+_2 | D^c) \cdot P(D^c) \] Substituting values: \[ P(T^+_1 \cap T^+_2) = (0.9025 \cdot 0.01) + (0.01 \cdot 0.99) \] \[ P(T^+_1 \cap T^+_2) = 0.009025 + 0.0099 = 0.018925 \] 6. **Final Calculation Using Bayes' Theorem**: \[ P(D | T^+_1 \cap T^+_2) = \frac{P(T^+_1 \cap T^+_2 | D) \cdot P(D)}{P(T^+_1 \cap T^+_2)} \] Substituting the values: \[ P(D | T^+_1 \cap T^+_2) = \frac{(0.9025 \cdot 0.01)}{0.018925} \] \[ P(D | T^+_1 \cap T^+_2) = \frac{0.009025}{0.018925} \approx 0.476 \] # Summary The probability that an individual actually has the disease given that they tested positive twice is approximately **0.476** or **47.6%**.

# Given Information - Mean of the distribution: \( \mu = 75 \) - Standard deviation: \( \sigma = 12 \) # What Have to Find - The coefficient of variation (CV). # Definition or Concept Used - **Coefficient of Variation (CV)**: A measure of relative variability, calculated as: \[ CV = \frac{\sigma}{\mu} \times 100\% \] where \( \sigma \) is the standard deviation and \( \mu \) is the mean. # Step by Step Calculation 1. **Substitute the values into the formula**: \[ CV = \frac{12}{75} \times 100\% \] 2. **Calculate the fraction**: \[ CV = 0.16 \times 100\% \] 3. **Calculate the percentage**: \[ CV = 16\% \] # Summary The coefficient of variation is **16%**.

# Given Information The dataset is: - \(8, 12, 15, 18, 20, 22, 25, 28, 30\) # What Have to Find - The first quartile \(Q_1\) and the third quartile \(Q_3\). # Definitions or Concepts Used - **Quartiles**: - The first quartile \(Q_1\) is the median of the first half of the data. - The third quartile \(Q_3\) is the median of the second half of the data. # Step by Step Calculation 1. **Sort the data** (already sorted): - \(8, 12, 15, 18, 20, 22, 25, 28, 30\) 2. **Calculate the position of \(Q_1\)**: - The formula for \(Q_1\) is given by: \[ Q_1 = \frac{n + 1}{4} \] where \(n\) is the number of data points. - Here, \(n = 9\): \[ Q_1 = \frac{9 + 1}{4} = 2.5 \] - This means \(Q_1\) is the average of the 2nd and 3rd values. 3. **Find \(Q_1\)**: - The 2nd value is \(12\) and the 3rd value is \(15\). \[ Q_1 = \frac{12 + 15}{2} = \frac{27}{2} = 13.5 \] 4. **Calculate the position of \(Q_3\)**: - The formula for \(Q_3\) is given by: \[ Q_3 = \frac{3(n + 1)}{4} \] - Using \(n = 9\): \[ Q_3 = \frac{3(9 + 1)}{4} = \frac{30}{4} = 7.5 \] - This means \(Q_3\) is the average of the 7th and 8th values. 5. **Find \(Q_3\)**: - The 7th value is \(25\) and the 8th value is \(28\). \[ Q_3 = \frac{25 + 28}{2} = \frac{53}{2} = 26.5 \] # Summary The first quartile \(Q_1\) is **13.5** and the third quartile \(Q_3\) is **26.5**.