Update posttest.json

experiment/posttest.json

@@ -10,10 +10,10 @@
         "d": "0.35"
       },
       "explanations": {
-        "a": "Incorrect. Please check your calculation.",
-        "b": "Incorrect. Please check your calculation.",
-        "c": "Incorrect. Please check your calculation.",
-        "d": "Correct! To calculate the variance of a discrete random variable X, you can use the formula: Var(X) = ∑(xi - μ)² * P(X = xi), where xi represents each possible value of X, μ is the mean of X, and P(X = xi) is the probability of X taking the value xi. In this case, μ = 1 * 0.3 + 2 * 0.5 + 3 * 0.2 = 2, and Var(X) = (1 - 2)² * 0.3 + (2 - 2)² * 0.5 + (3 - 2)² * 0.2 = 0.35."
+        "a": " To calculate the variance of a discrete random variable X, you can use the formula: Var(X) = ∑(xi - μ)² * P(X = xi), where xi represents each possible value of X, μ is the mean of X, and P(X = xi) is the probability of X taking the value xi. In this case, μ = 1 * 0.3 + 2 * 0.5 + 3 * 0.2 = 2, and Var(X) = (1 - 2)² * 0.3 + (2 - 2)² * 0.5 + (3 - 2)² * 0.2 = 0.35.",
+        "b": " To calculate the variance of a discrete random variable X, you can use the formula: Var(X) = ∑(xi - μ)² * P(X = xi), where xi represents each possible value of X, μ is the mean of X, and P(X = xi) is the probability of X taking the value xi. In this case, μ = 1 * 0.3 + 2 * 0.5 + 3 * 0.2 = 2, and Var(X) = (1 - 2)² * 0.3 + (2 - 2)² * 0.5 + (3 - 2)² * 0.2 = 0.35.",
+        "c": " To calculate the variance of a discrete random variable X, you can use the formula: Var(X) = ∑(xi - μ)² * P(X = xi), where xi represents each possible value of X, μ is the mean of X, and P(X = xi) is the probability of X taking the value xi. In this case, μ = 1 * 0.3 + 2 * 0.5 + 3 * 0.2 = 2, and Var(X) = (1 - 2)² * 0.3 + (2 - 2)² * 0.5 + (3 - 2)² * 0.2 = 0.35.",
+        "d": " To calculate the variance of a discrete random variable X, you can use the formula: Var(X) = ∑(xi - μ)² * P(X = xi), where xi represents each possible value of X, μ is the mean of X, and P(X = xi) is the probability of X taking the value xi. In this case, μ = 1 * 0.3 + 2 * 0.5 + 3 * 0.2 = 2, and Var(X) = (1 - 2)² * 0.3 + (2 - 2)² * 0.5 + (3 - 2)² * 0.2 = 0.35."
       },
       "correctAnswer": "d",
       "difficulty": "beginner"
@@ -27,10 +27,10 @@
         "d": "0.2"
       },
       "explanations": {
-        "a": "Incorrect. Please check your calculation.",
-        "b": "Correct! For a uniform distribution between a and b, the probability density function (PDF) is constant within this range. The total probability is the area under the PDF curve, which is the height (constant value) multiplied by the width (b - a). In this case, the width is 10 - 0 = 10, and the height is 1/10 (to maintain the total area as 1). So, the probability of Y being between 2 and 6 is: P(2 ≤ Y ≤ 6) = Height * Width = (1/10) * 4 = 0.4.",
-        "c": "Incorrect. Please check your calculation.",
-        "d": "Incorrect. Please check your calculation."
+        "a": " For a uniform distribution between a and b, the probability density function (PDF) is constant within this range. The total probability is the area under the PDF curve, which is the height (constant value) multiplied by the width (b - a). In this case, the width is 10 - 0 = 10, and the height is 1/10 (to maintain the total area as 1). So, the probability of Y being between 2 and 6 is: P(2 ≤ Y ≤ 6) = Height * Width = (1/10) * 4 = 0.4.",
+        "b": " For a uniform distribution between a and b, the probability density function (PDF) is constant within this range. The total probability is the area under the PDF curve, which is the height (constant value) multiplied by the width (b - a). In this case, the width is 10 - 0 = 10, and the height is 1/10 (to maintain the total area as 1). So, the probability of Y being between 2 and 6 is: P(2 ≤ Y ≤ 6) = Height * Width = (1/10) * 4 = 0.4.",
+        "c": " For a uniform distribution between a and b, the probability density function (PDF) is constant within this range. The total probability is the area under the PDF curve, which is the height (constant value) multiplied by the width (b - a). In this case, the width is 10 - 0 = 10, and the height is 1/10 (to maintain the total area as 1). So, the probability of Y being between 2 and 6 is: P(2 ≤ Y ≤ 6) = Height * Width = (1/10) * 4 = 0.4.",
+        "d": " For a uniform distribution between a and b, the probability density function (PDF) is constant within this range. The total probability is the area under the PDF curve, which is the height (constant value) multiplied by the width (b - a). In this case, the width is 10 - 0 = 10, and the height is 1/10 (to maintain the total area as 1). So, the probability of Y being between 2 and 6 is: P(2 ≤ Y ≤ 6) = Height * Width = (1/10) * 4 = 0.4."
       },
       "correctAnswer": "b",
       "difficulty": "beginner"
@@ -44,10 +44,10 @@
         "d": "0.15"
       },
       "explanations": {
-        "a": "Correct! The expected value (μ) is calculated as the sum of each possible value multiplied by its probability: μ = 1 * P(Z = 1) + 2 * P(Z = 2) + 4 * P(Z = 4) + 5 * P(Z = 5). Given μ = 3, we can substitute the other known values and solve for P(Z = 4): 3 = P(Z = 1) + 2 * P(Z = 2) + 4 * P(Z = 4) + 5 * P(Z = 5). Since P(Z = 1) + P(Z = 2) + P(Z = 4) + P(Z = 5) = 1, and P(Z = 1) = P(Z = 5) = 0, we have: P(Z = 4) = (3 - 2 * P(Z = 2)) / 4 = 0.25.",
-        "b": "Incorrect. Please check your calculation.",
-        "c": "Incorrect. Please check your calculation.",
-        "d": "Incorrect. Please check your calculation."
+        "a": " The expected value (μ) is calculated as the sum of each possible value multiplied by its probability: μ = 1 * P(Z = 1) + 2 * P(Z = 2) + 4 * P(Z = 4) + 5 * P(Z = 5). Given μ = 3, we can substitute the other known values and solve for P(Z = 4): 3 = P(Z = 1) + 2 * P(Z = 2) + 4 * P(Z = 4) + 5 * P(Z = 5). Since P(Z = 1) + P(Z = 2) + P(Z = 4) + P(Z = 5) = 1, and P(Z = 1) = P(Z = 5) = 0, we have: P(Z = 4) = (3 - 2 * P(Z = 2)) / 4 = 0.25.",
+        "b": " The expected value (μ) is calculated as the sum of each possible value multiplied by its probability: μ = 1 * P(Z = 1) + 2 * P(Z = 2) + 4 * P(Z = 4) + 5 * P(Z = 5). Given μ = 3, we can substitute the other known values and solve for P(Z = 4): 3 = P(Z = 1) + 2 * P(Z = 2) + 4 * P(Z = 4) + 5 * P(Z = 5). Since P(Z = 1) + P(Z = 2) + P(Z = 4) + P(Z = 5) = 1, and P(Z = 1) = P(Z = 5) = 0, we have: P(Z = 4) = (3 - 2 * P(Z = 2)) / 4 = 0.25.",
+        "c": " The expected value (μ) is calculated as the sum of each possible value multiplied by its probability: μ = 1 * P(Z = 1) + 2 * P(Z = 2) + 4 * P(Z = 4) + 5 * P(Z = 5). Given μ = 3, we can substitute the other known values and solve for P(Z = 4): 3 = P(Z = 1) + 2 * P(Z = 2) + 4 * P(Z = 4) + 5 * P(Z = 5). Since P(Z = 1) + P(Z = 2) + P(Z = 4) + P(Z = 5) = 1, and P(Z = 1) = P(Z = 5) = 0, we have: P(Z = 4) = (3 - 2 * P(Z = 2)) / 4 = 0.25.",
+        "d": " The expected value (μ) is calculated as the sum of each possible value multiplied by its probability: μ = 1 * P(Z = 1) + 2 * P(Z = 2) + 4 * P(Z = 4) + 5 * P(Z = 5). Given μ = 3, we can substitute the other known values and solve for P(Z = 4): 3 = P(Z = 1) + 2 * P(Z = 2) + 4 * P(Z = 4) + 5 * P(Z = 5). Since P(Z = 1) + P(Z = 2) + P(Z = 4) + P(Z = 5) = 1, and P(Z = 1) = P(Z = 5) = 0, we have: P(Z = 4) = (3 - 2 * P(Z = 2)) / 4 = 0.25."
       },
       "correctAnswer": "a",
       "difficulty": "beginner"
@@ -61,10 +61,10 @@
         "d": "0.6915"
       },
       "explanations": {
-        "a": "Correct! To find the probability that W is less than 45 in a standard normal distribution (Z), we need to calculate the z-score: Z = (X - μ) / σ = (45 - 50) / 10 = -0.5. Using a standard normal table or a calculator, we find that the area to the left of Z = -0.5 is approximately 0.1587.",
-        "b": "Incorrect. Please check your calculation.",
-        "c": "Incorrect. Please check your calculation.",
-        "d": "Incorrect. Please check your calculation."
+        "a": " To find the probability that W is less than 45 in a standard normal distribution (Z), we need to calculate the z-score: Z = (X - μ) / σ = (45 - 50) / 10 = -0.5. Using a standard normal table or a calculator, we find that the area to the left of Z = -0.5 is approximately 0.1587.",
+        "b": " To find the probability that W is less than 45 in a standard normal distribution (Z), we need to calculate the z-score: Z = (X - μ) / σ = (45 - 50) / 10 = -0.5. Using a standard normal table or a calculator, we find that the area to the left of Z = -0.5 is approximately 0.1587..",
+        "c": " To find the probability that W is less than 45 in a standard normal distribution (Z), we need to calculate the z-score: Z = (X - μ) / σ = (45 - 50) / 10 = -0.5. Using a standard normal table or a calculator, we find that the area to the left of Z = -0.5 is approximately 0.1587.",
+        "d": " To find the probability that W is less than 45 in a standard normal distribution (Z), we need to calculate the z-score: Z = (X - μ) / σ = (45 - 50) / 10 = -0.5. Using a standard normal table or a calculator, we find that the area to the left of Z = -0.5 is approximately 0.1587."
       },
       "correctAnswer": "a",
       "difficulty": "beginner"
@@ -78,10 +78,10 @@
         "d": "0.1755"
       },
       "explanations": {
-        "a": "Correct! In a Poisson distribution, the probability of receiving k events in an interval is given by the formula: P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the average rate. In this case, λ = 5 and k = 3. Substituting the values: P(X = 3) = (e^(-5) * 5^3) / 3! = 0.1008.",
-        "b": "Correct! In a Poisson distribution, the probability of receiving k events in an interval is given by the formula: P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the average rate. In this case, λ = 5 and k = 3. Substituting the values: P(X = 3) = (e^(-5) * 5^3) / 3! = 0.1008.",
-        "c": "Correct! In a Poisson distribution, the probability of receiving k events in an interval is given by the formula: P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the average rate. In this case, λ = 5 and k = 3. Substituting the values: P(X = 3) = (e^(-5) * 5^3) / 3! = 0.1008.",
-        "d": "Correct! In a Poisson distribution, the probability of receiving k events in an interval is given by the formula: P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the average rate. In this case, λ = 5 and k = 3. Substituting the values: P(X = 3) = (e^(-5) * 5^3) / 3! = 0.1008." 
+        "a": " In a Poisson distribution, the probability of receiving k events in an interval is given by the formula: P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the average rate. In this case, λ = 5 and k = 3. Substituting the values: P(X = 3) = (e^(-5) * 5^3) / 3! = 0.1008.",
+        "b": " In a Poisson distribution, the probability of receiving k events in an interval is given by the formula: P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the average rate. In this case, λ = 5 and k = 3. Substituting the values: P(X = 3) = (e^(-5) * 5^3) / 3! = 0.1008.",
+        "c": " In a Poisson distribution, the probability of receiving k events in an interval is given by the formula: P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the average rate. In this case, λ = 5 and k = 3. Substituting the values: P(X = 3) = (e^(-5) * 5^3) / 3! = 0.1008.",
+        "d": " In a Poisson distribution, the probability of receiving k events in an interval is given by the formula: P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the average rate. In this case, λ = 5 and k = 3. Substituting the values: P(X = 3) = (e^(-5) * 5^3) / 3! = 0.1008." 
       },
       "correctAnswer": "a",
       "difficulty": "beginner"