From 43a2793f43735141d3583bf7d2f46ef1f891a271 Mon Sep 17 00:00:00 2001
From: singhshakti182 <112379803+singhshakti182@users.noreply.github.com>
Date: Mon, 16 Oct 2023 14:02:12 +0530
Subject: [PATCH 1/4] Update theory.md
---
experiment/theory.md | 4 ++--
1 file changed, 2 insertions(+), 2 deletions(-)
diff --git a/experiment/theory.md b/experiment/theory.md
index 6a8e3f8..49fc220 100644
--- a/experiment/theory.md
+++ b/experiment/theory.md
@@ -1,2 +1,2 @@
-
-
+
+
From 95232d2f3a3281d1ede9717a69540d126b8d1f4d Mon Sep 17 00:00:00 2001
From: singhshakti182 <112379803+singhshakti182@users.noreply.github.com>
Date: Mon, 16 Oct 2023 14:02:52 +0530
Subject: [PATCH 2/4] Update theory.md
From 14fccbab48382e4b5d3b401433e2b2242c6328f5 Mon Sep 17 00:00:00 2001
From: singhshakti182 <112379803+singhshakti182@users.noreply.github.com>
Date: Mon, 16 Oct 2023 14:03:46 +0530
Subject: [PATCH 3/4] Update aim.md
---
experiment/aim.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/experiment/aim.md b/experiment/aim.md
index 3eb0dae..4a0ecab 100644
--- a/experiment/aim.md
+++ b/experiment/aim.md
@@ -1 +1 @@
-The aim of this experiment is to investigate the probability distributions of discrete and continuous random variables. Specifically, the focus is on calculating probabilities associated with discrete and continuous random variables and understanding the graphical representation of Probability Mass Functions and Probability Density Functions.
+This experiment aims to investigate the probability distributions of discrete and continuous random variables. Specifically, the focus is on understanding the graphical representation of Probability Mass Functions and Probability Density Functions.
From 7e5d9cb91f112079b7ea6b60537f87bcdfe74556 Mon Sep 17 00:00:00 2001
From: singhshakti182 <112379803+singhshakti182@users.noreply.github.com>
Date: Mon, 16 Oct 2023 14:09:07 +0530
Subject: [PATCH 4/4] Update posttest.json
---
experiment/posttest.json | 40 ++++++++++++++++++++--------------------
1 file changed, 20 insertions(+), 20 deletions(-)
diff --git a/experiment/posttest.json b/experiment/posttest.json
index 6b6cba7..4358185 100644
--- a/experiment/posttest.json
+++ b/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"