# Given Data - Two dice are thrown repeatedly. - We are interested in the **first occurrence** of a sum of 8. - We want the probability that the **first 8 appears on the 4th throw**. # Definition and Concept Used **Geometric Distribution:** The experiment is repeated until the first success (sum 8) occurs. The probability that the first success occurs on the \( n \)th trial is given by: \[ P(\text{first success on } n\text{th trial}) = (1 - p)^{n-1}p \] where \( p \) is the probability of success on a single trial. **Probability of Getting a Sum of 8 in One Throw:** There are 36 possible outcomes when throwing two dice. The favorable pairs for sum 8 are: \((2,6), (3,5), (4,4), (5,3), (6,2)\) So, \( p = \frac{5}{36} \). # Step-by-Step Solution The probability of **not** getting a sum of 8 in a single throw is \( 1 - p = 1 - \frac{5}{36} = \frac{31}{36} \). We want the first 3 throws to **not** result in a sum of 8, and the 4th throw to result in a sum of 8. Applying the geometric distribution formula: \[ P(\text{first 8 on 4th throw}) = (1-p)^{4-1} \times p = (1-p)^3 \times p \] \[ = \left(\frac{31}{36}\right)^3 \times \frac{5}{36} \] \[ = \frac{31^3}{36^3} \times \frac{5}{36} \] \[ = \frac{29791}{46656} \times \frac{5}{36} \] \[ = \frac{29791 \times 5}{46656 \times 36} \] \[ = \frac{148955}{1679616} \] Thus, the required probability is \[ \boxed{\frac{148955}{1679616}} \]

# Given Data - Two dice are thrown repeatedly. - We are interested in the **first occurrence** of a sum of 8. - We want the probability that the **first 8 appears on the 4th throw**. # Definition and Concept Used **Geometric Distribution:** The experiment is repeated until the first success (sum 8) occurs. The probability that the first success occurs on the \( n \)th trial is given by: \[ P(\text{first success on } n\text{th trial}) = (1 - p)^{n-1}p \] where \( p \) is the probability of success on a single trial. **Probability of Getting a Sum of 8 in One Throw:** There are 36 possible outcomes when throwing two dice. The favorable pairs for sum 8 are: \((2,6), (3,5), (4,4), (5,3), (6,2)\) Thus, \( p = \frac{5}{36} \). # Step-by-Step Solution The probability of **not** getting a sum of 8 in a single throw is calculated as: \[ 1 - p = 1 - \frac{5}{36} = \frac{31}{36} \] We need the first 3 throws to **not** yield a sum of 8, and the 4th throw to yield a sum of 8. Using the geometric distribution formula: \[ P(\text{first 8 on 4th throw}) = (1-p)^{3} \times p \] Substituting the values: \[ = \left(\frac{31}{36}\right)^{3} \times \frac{5}{36} \] Calculating each component: \[ = \frac{31^3}{36^3} \times \frac{5}{36} \] \[ = \frac{29791}{46656} \times \frac{5}{36} \] Now, compute the final multiplication: \[ = \frac{29791 \times 5}{46656 \times 36} \] \[ = \frac{148955}{1679616} \] Thus, the required probability is \[ \boxed{\frac{148955}{1679616}} \] # Decimal Representation Calculating the decimal value: \[ \frac{148955}{1679616} \approx 0.0886 \]

# Given Data - Two dice are thrown repeatedly. - We are interested in the **first occurrence** of a sum of 8. - We want the probability that the **first 8 appears on the 4th throw**. # Definition and Concept Used **Geometric Distribution:** The experiment is repeated until the first success (sum 8) occurs. The probability that the first success occurs on the \( n \)th trial is given by: \[ P(\text{first success on } n\text{th trial}) = (1 - p)^{n-1}p \] where \( p \) is the probability of success on a single trial. **Probability of Getting a Sum of 8 in One Throw:** There are 36 possible outcomes when throwing two dice. The favorable pairs for sum 8 are: \((2,6), (3,5), (4,4), (5,3), (6,2)\) Thus, \( p = \frac{5}{36} \). # Step-by-Step Solution The probability of **not** getting a sum of 8 in a single throw is calculated as: \[ 1 - p = 1 - \frac{5}{36} = \frac{31}{36} \] We need the first 3 throws to **not** yield a sum of 8, and the 4th throw to yield a sum of 8. Using the geometric distribution formula: \[ P(\text{first 8 on 4th throw}) = (1-p)^{3} \times p \] Substituting the values: \[ = \left(\frac{31}{36}\right)^{3} \times \frac{5}{36} \] Calculating each component: \[ = \frac{31^3}{36^3} \times \frac{5}{36} \] Calculating \( 31^3 \) and \( 36^3 \): \[ = \frac{29791}{46656} \times \frac{5}{36} \] Now, compute the final multiplication: \[ = \frac{29791 \times 5}{46656 \times 36} \] \[ = \frac{148955}{1679616} \] Thus, the required probability is \[ \boxed{\frac{148955}{1679616}} \] # Decimal Representation To find the decimal representation of \( p \): \[ p = \frac{5}{36} \approx 0.1389 \] This indicates that the probability of rolling a sum of 8 in one throw of two dice is approximately 0.1389.

# Given Data - Two dice are thrown repeatedly. - We are interested in the **first occurrence** of a sum of 8. - We want the probability that the **first 8 appears on the 4th throw**. # Definition and Concept Used **Geometric Distribution:** The experiment is repeated until the first success (sum 8) occurs. The probability that the first success occurs on the \( n \)th trial is given by: \[ P(\text{first success on } n\text{th trial}) = (1 - p)^{n-1}p \] where \( p \) is the probability of success on a single trial. **Probability of Getting a Sum of 8 in One Throw:** There are 36 possible outcomes when throwing two dice. The favorable pairs for sum 8 are: \((2,6), (3,5), (4,4), (5,3), (6,2)\) Thus, \( p = \frac{5}{36} \approx 0.1389 \). # Step-by-Step Solution The probability of **not** getting a sum of 8 in a single throw is: \[ 1 - p = 1 - \frac{5}{36} = \frac{31}{36} \approx 0.8611 \] We need the first 3 throws to **not** yield a sum of 8, and the 4th throw to yield a sum of 8. Using the geometric distribution formula: \[ P(\text{first 8 on 4th throw}) = (1-p)^{3} \times p \] Substituting the values: \[ = \left(\frac{31}{36}\right)^{3} \times \frac{5}{36} \] Calculating \( \left(\frac{31}{36}\right)^{3} \): \[ \left(\frac{31}{36}\right)^{3} \approx 0.6360 \] Now, use this value to find: \[ P(\text{first 8 on 4th throw}) \approx 0.6360 \times 0.1389 \approx 0.0886 \] Thus, the required probability that the first 8 appears on the 4th throw is approximately: \[ \boxed{0.0886} \] # Summary of Values - Probability of getting a sum of 8 in one throw \( p \approx 0.1389 \) - Probability of not getting a sum of 8 in one throw \( 1 - p \approx 0.8611 \) - Probability that the first 8 appears on the 4th throw \( \approx 0.0886 \)

# Given Data - Two dice are thrown repeatedly. - We are interested in the **first occurrence** of a sum of 8. - We want the probability that the **first 8 appears on the 4th throw**. # Definition and Terms Used **Geometric Distribution:** This distribution models the number of trials until the first success occurs. The probability that the first success occurs on the \( n \)th trial is expressed as: \[ P(\text{first success on } n\text{th trial}) = (1 - p)^{n-1}p \] where \( p \) is the probability of success on a single trial. **Probability of Getting a Sum of 8 in One Throw:** When two dice are rolled, there are 36 possible outcomes. The favorable pairs for achieving a sum of 8 are: - \((2,6)\) - \((3,5)\) - \((4,4)\) - \((5,3)\) - \((6,2)\) Thus, the probability \( p \) of rolling a sum of 8 is: \[ p = \frac{5}{36} \approx 0.1389 \] # Step-by-Step Solution 1. **Calculate the Probability of Not Getting a Sum of 8:** \[ 1 - p = 1 - \frac{5}{36} = \frac{31}{36} \approx 0.8611 \] 2. **Determine the Probability for the First 3 Throws Not Yielding a Sum of 8:** We need to find the probability that the first three throws do not yield a sum of 8, followed by a successful throw on the 4th: \[ P(\text{first 8 on 4th throw}) = (1-p)^{3} \times p \] 3. **Substituting the Values:** \[ = \left(\frac{31}{36}\right)^{3} \times \frac{5}{36} \] 4. **Calculating \( \left(\frac{31}{36}\right)^{3} \):** \[ \left(\frac{31}{36}\right)^{3} \approx 0.6360 \] 5. **Final Calculation:** \[ P(\text{first 8 on 4th throw}) \approx 0.6360 \times 0.1389 \approx 0.0886 \] Thus, the required probability that the first 8 appears on the 4th throw is approximately: \[ \boxed{0.0886} \] # Summary of Values - Probability of getting a sum of 8 in one throw \( p \approx 0.1389 \) - Probability of not getting a sum of 8 in one throw \( 1 - p \approx 0.8611 \) - Probability that the first 8 appears on the 4th throw \( \approx 0.0886 \)

# Given Data - Two dice are thrown repeatedly. - We want to determine the probability that a sum of 8 appears for the **first time on the 4th throw**. # Definition and Terms Used **Geometric Distribution:** This distribution describes the number of trials needed to achieve the first success. The probability that the first success occurs on the \( n \)th trial is given by: \[ P(\text{first success on } n\text{th trial}) = (1 - p)^{n-1}p \] where \( p \) is the probability of success in a single trial. **Probability of Getting a Sum of 8 in One Throw:** When rolling two dice, there are 36 total outcomes. The pairs that yield a sum of 8 are: - \((2,6)\) - \((3,5)\) - \((4,4)\) - \((5,3)\) - \((6,2)\) Thus, the probability \( p \) of rolling a sum of 8 is: \[ p = \frac{5}{36} \approx 0.1389 \] # Step-by-Step Solution - **Calculate the Probability of Not Getting a Sum of 8:** \[ 1 - p = 1 - \frac{5}{36} = \frac{31}{36} \approx 0.8611 \] - **Calculate the Probability for the First Three Throws Not Yielding a Sum of 8:** The first three throws must not yield a sum of 8, and the 4th must be a success: \[ P(\text{first 8 on 4th throw}) = (1-p)^{3} \times p \] - **Substituting the Values:** \[ = \left(\frac{31}{36}\right)^{3} \times \frac{5}{36} \] - **Calculating \( \left(\frac{31}{36}\right)^{3} \):** \[ \left(\frac{31}{36}\right)^{3} \approx 0.6360 \] - **Final Calculation:** \[ P(\text{first 8 on 4th throw}) \approx 0.6360 \times 0.1389 \approx 0.0886 \] Thus, the required probability that the first 8 appears on the 4th throw is approximately: \[ \boxed{0.0886} \] # Summary of Values - Probability of getting a sum of 8 in one throw \( p \approx 0.1389 \) - Probability of not getting a sum of 8 in one throw \( 1 - p \approx 0.8611 \) - Probability that the first 8 appears on the 4th throw \( \approx 0.0886 \)

# Given Data - Two dice are thrown repeatedly. - We need to find the probability that a sum of 8 appears for the **first time on the 4th throw**. # Definition and Terms Used **Geometric Distribution:** This distribution describes the number of trials until the first success occurs. The probability that the first success occurs on the \( n \)th trial is given by: \[ P(\text{first success on } n\text{th trial}) = (1 - p)^{n-1}p \] where \( p \) is the probability of success on a single trial. **Probability of Getting a Sum of 8 in One Throw:** The total outcomes when rolling two dice is 36. The combinations that yield a sum of 8 are: - \((2,6)\) - \((3,5)\) - \((4,4)\) - \((5,3)\) - \((6,2)\) Thus, the probability \( p \) of rolling a sum of 8 is: \[ p = \frac{5}{36} \approx 0.1389 \] The probability of not getting a sum of 8 in a single throw is: \[ 1 - p = 1 - \frac{5}{36} = \frac{31}{36} \approx 0.8611 \] To find the probability that the first 8 occurs on the 4th throw, we substitute into the geometric distribution formula: \[ P(\text{first 8 on 4th throw}) = (1-p)^{3} \times p \] Substituting the values: \[ = \left(\frac{31}{36}\right)^{3} \times \frac{5}{36} \] Calculating \( \left(\frac{31}{36}\right)^{3} \): \[ \approx 0.6360 \] Now, multiply this result by \( p \): \[ P(\text{first 8 on 4th throw}) \approx 0.6360 \times 0.1389 \approx 0.0886 \] Thus, the required probability that the first 8 appears on the 4th throw is: \[ \boxed{0.0886} \] # Summary of Values - Probability of getting a sum of 8 in one throw \( p \approx 0.1389 \) - Probability of not getting a sum of 8 in one throw \( 1 - p \approx 0.8611 \) - Probability that the first 8 appears on the 4th throw \( \approx 0.0886 \)

# Given Data - Two dice are thrown repeatedly. - We want to find the probability that a sum of 8 appears for the **first time on the 4th throw**. # Definition and Terms Used **Geometric Distribution:** This distribution applies when we count the number of trials until the first success occurs. The probability that the first success occurs on the \( n \)th trial is expressed as: \[ P(\text{first success on } n\text{th trial}) = (1 - p)^{n-1}p \] where \( p \) represents the probability of success in a single trial. **Probability of Getting a Sum of 8 in One Throw:** When rolling two dice, there are 36 total outcomes. The combinations that result in a sum of 8 are: - \((2,6)\) - \((3,5)\) - \((4,4)\) - \((5,3)\) - \((6,2)\) Thus, the probability \( p \) of rolling a sum of 8 is: \[ p = \frac{5}{36} \approx 0.1389 \] The probability of not rolling a sum of 8 in a single throw is: \[ 1 - p = 1 - \frac{5}{36} = \frac{31}{36} \approx 0.8611 \] To find the probability that the first 8 occurs on the 4th throw, we use the geometric distribution formula: \[ P(\text{first 8 on 4th throw}) = (1-p)^{3} \times p \] Substituting the values: \[ = \left(\frac{31}{36}\right)^{3} \times \frac{5}{36} \] Calculating \( \left(\frac{31}{36}\right)^{3} \): \[ \approx 0.6360 \] Now, multiply this by \( p \): \[ P(\text{first 8 on 4th throw}) \approx 0.6360 \times 0.1389 \approx 0.0886 \] Thus, the required probability that the first 8 appears on the 4th throw is: \[ \boxed{0.0886} \]

# Given Data - Two dice are thrown repeatedly. - The task is to find the probability that a sum of 8 appears for the **first time on the 4th throw**. # Definition and Terms Used **Geometric Distribution:** This distribution is used to model the number of trials until the first success occurs. The probability that the first success happens on the \( n \)th trial is given by: \[ P(\text{first success on } n\text{th trial}) = (1 - p)^{n-1}p \] where \( p \) is the probability of success on a single trial. **Probability of Getting a Sum of 8 in One Throw:** When rolling two dice, there are 36 possible outcomes. The combinations that yield a sum of 8 are: - \((2,6)\) - \((3,5)\) - \((4,4)\) - \((5,3)\) - \((6,2)\) Thus, the probability \( p \) of rolling a sum of 8 is: \[ p = \frac{5}{36} \approx 0.1389 \] The probability of **not** rolling a sum of 8 in a single throw is: \[ 1 - p = 1 - \frac{5}{36} = \frac{31}{36} \approx 0.8611 \] # Solution Using the geometric distribution formula, we calculate the probability that the first occurrence of a sum of 8 happens on the 4th throw: \[ P(\text{first 8 on 4th throw}) = (1-p)^{3} \times p \] Substituting the values into the equation: \[ = \left(\frac{31}{36}\right)^{3} \times \frac{5}{36} \] Calculating \( \left(\frac{31}{36}\right)^{3} \): \[ \approx 0.6360 \] Now, using this result to find the final probability: \[ P(\text{first 8 on 4th throw}) \approx 0.6360 \times 0.1389 \approx 0.0886 \] Thus, the required probability that the first 8 appears on the 4th throw is: \[ \boxed{0.0886} \]

# Given Data - Person A speaks the truth \( p_A = 0.75 \). - Person B speaks the truth \( p_B = 0.60 \). - Both A and B claim that a certain event occurred. # Definition and Terms Used **Total Probability Theorem:** To find the probability that the event actually occurred given that both A and B claim it did, we can use the law of total probability. Let \( E \) be the event that the event actually occurred, \( T_A \) be the event that A speaks the truth, and \( T_B \) be the event that B speaks the truth. The probability of the event actually occurring given both A and B speak is given by: \[ P(E | T_A \cap T_B) = \frac{P(T_A \cap T_B | E) P(E)}{P(T_A \cap T_B)} \] Assuming the event could either occur or not occur, we denote \( P(E) \) as the prior probability of the event occurring. # Solution Assuming \( P(E) = p \) (the probability we want to determine) and \( P(\neg E) = 1 - p \). 1. **Calculate \( P(T_A \cap T_B | E) \)**: If the event occurred, both A and B tell the truth: \[ P(T_A \cap T_B | E) = P(T_A | E) \cdot P(T_B | E) = p_A \cdot p_B = 0.75 \times 0.60 = 0.45 \] 2. **Calculate \( P(T_A \cap T_B | \neg E) \)**: If the event did not occur, both A and B would lie: \[ P(T_A \cap T_B | \neg E) = P(\neg T_A | \neg E) \cdot P(\neg T_B | \neg E) = (1 - p_A) \cdot (1 - p_B) = 0.25 \times 0.40 = 0.10 \] 3. **Total Probability \( P(T_A \cap T_B) \)**: Using the law of total probability: \[ P(T_A \cap T_B) = P(T_A \cap T_B | E) \cdot P(E) + P(T_A \cap T_B | \neg E) \cdot P(\neg E) \] \[ = 0.45p + 0.10(1 - p) \] \[ = 0.45p + 0.10 - 0.10p = 0.35p + 0.10 \] 4. **Final Probability**: Using Bayes' theorem to find \( P(E | T_A \cap T_B) \): \[ P(E | T_A \cap T_B) = \frac{P(T_A \cap T_B | E) P(E)}{P(T_A \cap T_B)} = \frac{0.45p}{0.35p + 0.10} \] Thus, we have derived the formula for the probability that the event actually occurred given A and B's claims. If we assume \( P(E) = \frac{1}{2} \) (as a prior), we can substitute \( p = 0.5 \) into our formula: \[ P(E | T_A \cap T_B) = \frac{0.45 \times 0.5}{0.35 \times 0.5 + 0.10} = \frac{0.225}{0.175 + 0.10} = \frac{0.225}{0.275} \approx 0.8182 \] Thus, the probability that the event actually occurred is approximately: \[ \boxed{0.8182} \]

# Given Data - Number of observations: \( n = 100 \) - Incorrect mean of observations: \( \text{Mean} = 50 \) - Incorrect observations: \( 30 \) and \( 70 \) - Correct observations: \( 50 \) and \( 90 \) # Definition and Terms Used **Mean Calculation:** The mean is calculated as: \[ \text{Mean} = \frac{\text{Sum of observations}}{n} \] To find the correct mean, we will first calculate the total sum of the incorrect observations and then adjust it based on the corrections made. # Solution 1. **Calculate the Total Incorrect Sum:** \[ \text{Total Sum}_{\text{incorrect}} = \text{Mean} \times n = 50 \times 100 = 5000 \] 2. **Adjust the Total Sum:** - Remove the incorrect observations: \[ \text{Adjusted Sum} = 5000 - (30 + 70) = 5000 - 100 = 4900 \] - Add the correct observations: \[ \text{Corrected Sum} = 4900 + (50 + 90) = 4900 + 140 = 5040 \] 3. **Calculate the Correct Mean:** \[ \text{Correct Mean} = \frac{\text{Corrected Sum}}{n} = \frac{5040}{100} = 50.40 \] Thus, the correct mean is: \[ \boxed{50.40} \]

# Given Data - Number of observations: \( n = 100 \) - Incorrect mean of observations: \( \text{Mean} = 50 \) - Incorrect observations recorded: \( 30 \) and \( 70 \) - Correct observations should be: \( 50 \) and \( 90 \) # Definition and Terms Used **Mean Calculation:** The mean is defined as: \[ \text{Mean} = \frac{\text{Sum of observations}}{n} \] To find the correct mean, we first calculate the total sum of the observations based on the incorrect mean, then adjust it to reflect the correct values. # Solution 1. **Calculate the Total Incorrect Sum:** \[ \text{Total Sum}_{\text{incorrect}} = \text{Mean} \times n = 50 \times 100 = 5000 \] 2. **Adjust the Total Sum:** - Remove the incorrectly recorded observations: \[ \text{Adjusted Sum} = 5000 - (30 + 70) = 5000 - 100 = 4900 \] - Add the correct observations: \[ \text{Corrected Sum} = 4900 + (50 + 90) = 4900 + 140 = 5040 \] 3. **Calculate the Correct Mean:** \[ \text{Correct Mean} = \frac{\text{Corrected Sum}}{n} = \frac{5040}{100} = 50.40 \] Thus, the correct mean is: \[ \boxed{50.40} \]

# Given Data - Number of observations: \( n = 100 \) - Incorrect mean of observations: \( \text{Mean} = 50 \) - Incorrectly recorded values: \( 30 \) and \( 70 \) - Correct values should be: \( 50 \) and \( 90 \) # Definition and Terms Used **Mean Calculation:** The mean is given by: \[ \text{Mean} = \frac{\text{Sum of observations}}{n} \] To find the correct mean, we will determine the total incorrect sum and then adjust it to account for the correct values. # Solution 1. **Calculate the Total Incorrect Sum:** \[ \text{Total Sum}_{\text{incorrect}} = \text{Mean} \times n = 50 \times 100 = 5000 \] 2. **Adjust the Total Sum:** - Remove the incorrect observations: \[ \text{Adjusted Sum} = 5000 - (30 + 70) = 5000 - 100 = 4900 \] - Add the correct observations: \[ \text{Corrected Sum} = 4900 + (50 + 90) = 4900 + 140 = 5040 \] 3. **Calculate the Correct Mean:** \[ \text{Correct Mean} = \frac{\text{Corrected Sum}}{n} = \frac{5040}{100} = 50.40 \] Thus, the correct mean is: \[ \boxed{50.40} \]

# Given Data - Number of observations: \( n = 100 \) - Incorrect mean of observations: \( \text{Mean} = 50 \) - Incorrect observations recorded: \( 30 \) and \( 70 \) - Correct observations should be: \( 50 \) and \( 90 \) # Definition and Terms Used **Mean Calculation:** The mean is defined as: \[ \text{Mean} = \frac{\text{Sum of observations}}{n} \] To find the correct mean, we will first calculate the total sum of the observations based on the incorrect mean, then adjust it to reflect the correct values. # Solution 1. **Calculate the Total Incorrect Sum:** \[ \text{Total Sum}_{\text{incorrect}} = \text{Mean} \times n = 50 \times 100 = 5000 \] 2. **Adjust the Total Sum:** - Remove the incorrectly recorded observations: \[ \text{Adjusted Sum} = 5000 - (30 + 70) = 5000 - 100 = 4900 \] - Add the correct observations: \[ \text{Corrected Sum} = 4900 + (50 + 90) = 4900 + 140 = 5040 \] 3. **Calculate the Correct Mean:** \[ \text{Correct Mean} = \frac{\text{Corrected Sum}}{n} = \frac{5040}{100} = 50.40 \] Thus, the correct mean is: \[ \boxed{50.40} \]

# Given Data - Total number of observations: \( n = 100 \) - Incorrect mean of the observations: \( \text{Mean} = 50 \) - Mistakenly recorded values: \( 30 \) and \( 70 \) - Correct values should be: \( 50 \) and \( 90 \) # Definition and Terms Used **Mean Calculation:** The mean is computed as: \[ \text{Mean} = \frac{\text{Total Sum of Observations}}{n} \] To determine the accurate mean, we will first find the total sum based on the incorrect mean and then adjust it based on the correct values. # Solution 1. **Calculate the Total Incorrect Sum:** \[ \text{Total Sum}_{\text{incorrect}} = \text{Mean} \times n = 50 \times 100 = 5000 \] 2. **Adjust the Total Sum:** - Subtract the incorrect observations: \[ \text{Adjusted Sum} = 5000 - (30 + 70) = 5000 - 100 = 4900 \] - Add the correct observations: \[ \text{Corrected Sum} = 4900 + (50 + 90) = 4900 + 140 = 5040 \] 3. **Calculate the Correct Mean:** \[ \text{Correct Mean} = \frac{\text{Corrected Sum}}{n} = \frac{5040}{100} = 50.40 \] Thus, the revised mean is: \[ \boxed{50.40} \]

# Given Data - Mean of the frequency distribution: \( \mu = 25 \) - Standard deviation: \( \sigma = 5 \) - Interval of interest: \( [15, 35] \) # Definition and Terms Used **Chebyshev's Theorem:** This theorem states that for any dataset, at least \( \frac{1}{k^2} \) of the data values must lie within \( k \) standard deviations from the mean, where \( k \) is any positive integer greater than 1. The formula can be expressed as: \[ P(|X - \mu| < k\sigma) \geq 1 - \frac{1}{k^2} \] # Solution 1. **Calculate the distance from the mean to the boundaries:** - Lower boundary: \( 15 \) - Upper boundary: \( 35 \) The distance from the mean to the lower boundary: \[ d_1 = \mu - 15 = 25 - 15 = 10 \] The distance from the mean to the upper boundary: \[ d_2 = 35 - \mu = 35 - 25 = 10 \] 2. **Determine \( k \):** The distance from the mean to either boundary is \( 10 \). We find \( k \) using: \[ k = \frac{d}{\sigma} = \frac{10}{5} = 2 \] 3. **Apply Chebyshev's Theorem:** Now using \( k = 2 \): \[ P(|X - \mu| < 2\sigma) \geq 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \] 4. **Convert to Percentage:** The minimum percentage of observations that lie within the interval \( [15, 35] \) is: \[ 75\% \] Thus, using Chebyshev's theorem, the minimum percentage of observations lying between 15 and 35 is: \[ \boxed{75\%} \]