Add general feedback for individual coursework. (#390)

_posts/2025-01-13-general-feedback-for-individual-coursework.md

@@ -0,0 +1,66 @@
+---
+layout: post
+title: "Individual Coursework Feedback"
+tags:
+  - about-the-course
+  - assessment
+---
+
+Thanks all for your efforts in doing the individual coursework!
+
+You can find the solution [here]({{site.baseurl}}/assets/assessment/2024-2025/ind/solution.ipynb).
+
+The performance was really good, below are some summary statistics of your
+marks:
+
+- mean: 77.55
+- standard deviation: 13.6
+- min: 2.00
+- 25th percentile: 71.00
+- 50th percentile: 78.00
+- 75th percentile: 85.00
+- max: 100.00
+
+Here is a plot of the distribution of the marks:
+
+![]({{site.baseurl}}/assets/assessment/2024-2025/mark_distribution.png)
+
+Here is a detailed summary of the various things I checked for for each question
+as well as the percentage of students who were correct:
+
+- Q1a correct value: 96.31 %
+- Q1 a use st correlation: 99.08 %
+- Q1b correct value: 96.31 %
+- Q1 b use st correlation: 99.08 %
+- Q1 c correct value: 96.31 %
+- Q1 c use st correlation: 99.08 %
+- Q1 d correct value: 96.31 %
+- Q1 d use st correlation: 99.08 %
+- Q1 e correct value: 96.31 %
+- Q1 e use st correlation: 99.08 %
+- Q2 a use integrate: 96.77 %
+- Q2 a correct value: 77.42 %
+- Q2 a correct derivative: 83.87 %
+- Q2 b use integrate: 97.24 %
+- Q2 b correct value: 77.88 %
+- Q2 b correct derivative: 83.87 %
+- Q3 a correct approach: 82.49 %
+- Q3 a correct answer: 41.94 %
+- Q3 b correct approach: 80.65 %
+- Q3 b correct answer: 100.00 %
+- Q4 a create equations: 96.77 %
+- Q4 a use dsolve: 90.78 %
+- Q4 a correct solution to system of differential equations: 35.02 %
+- Q4 b correct use of ics: 94.01 %
+- Q4 b correct solution to system of differential equations: 9.68 %
+- Q4 c correct value of t: 5.99 %
+- Q4 c correct value of winner: 37.33 %
+
+Overall performance was good and the main area of difficulty was question 4. A
+number of students did not use exact numbers for the constants
+which implied that the equations were not exact solutions to the differential
+equation. This question was marked generously and a lot of partial credit
+given.
+
+A less common mistake but one that happened more than once in question 3 was to
+compute permutations instead of combinations.

assets/assessment/2024-2025/ind/assignment.ipynb

@@ -1 +1 @@
-{"cells": [{"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "### Computing for Mathematics - 2024/2025 individual coursework\n\n**Important** Do not delete the cells containing: \n\n```\n### BEGIN SOLUTION\n\n\n### END SOLUTION\n```\n\nwrite your solution attempts in those cells.\n\nTo submit this notebook:\n\n- Change the name of the notebook from `main` to: `<student_number>`. For example, if your student number is `c1234567` then change the name of the notebook to `c1234567`.\n- **Write all your solution attempts in the correct locations**;\n- **Do not delete** any code that is already in the cells;\n- Save the notebook (`File>Save As`);\n- Follow the instructions given to submit."}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "#### Question 1 (30 marks)\n\n(__Hint__: This question is similar to [this section of the python for mathematics text](https://vknight.org/pfm/tools-for-mathematics/08-statistics/how/main.html#calculate-the-pearson-correlation-coefficient).)\n\nFor each of the following pairs of data $x$ and $y$ compute the Pearson correlation coefficient.\n\na. \n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 1, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-a"]}, "outputs": [], "source": "import statistics as st\nx = (1, 2, 3, 4, 5, 6, 7, 8, 9)\ny = (20, 10, 40, 30, 60, 50, 2, -1, -5)\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b.\n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 3, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-b"]}, "outputs": [], "source": "x = (1, 7, -1, 5, 10, -4)\ny = (20, 10, 40, 30, 60, 50)\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "c. \n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 5, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-c"]}, "outputs": [], "source": "x = (1, 7, 21)\ny = (1 / 3, 1, 3)\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "d. \n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 7, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-d"]}, "outputs": [], "source": "x = (1, 2, 3, 4, 5)\ny = (-1, 1, -1, 1, -1)\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "e.\n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 9, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-e"]}, "outputs": [], "source": "x = (1, 2, 3, 4, 5)\ny = (0, 0, 0, 0, -1)\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "### Question 2 (18 marks)\n\n\n(__Hint__: This question is similar to the [third exercise of the Calculus chapter of Python for mathematics](https://vknight.org/pfm/tools-for-mathematics/03-calculus/solutions/main.html#question-3))\n\nConsider the second derivative $f''(x)=7\\cos(x) + x ^ 2 - \\ln(x)$\n\na. Create a variable `first_derivative` which has value $f'(x)$ (you can assume a constant of integration 0).\n\navailable marks: 12"}, {"cell_type": "code", "execution_count": 11, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q2-a"]}, "outputs": [], "source": "import sympy as sym\n\nx = sym.Symbol(\"x\")\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b. Create a variable `expression` that has value $f(x)$ (you can assume a constant of integration 0):\n\navailable marks: 6"}, {"cell_type": "code", "execution_count": 13, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q2-b"]}, "outputs": [], "source": "### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "### Question 3 (16 marks)\n\n(__Hint__: This question is similar to the [second exercise of the Combinatorics chapter of Python for mathematics](https://vknight.org/pfm/tools-for-mathematics/05-combinations-permutations/solutions/main.html#question-2))\n\na. Create a variable `number_of_combinations` that gives the number of combinations of the letters of the alphabet of size 3. Do this by generating and counting them.\n\n_available marks: 8_"}, {"cell_type": "code", "execution_count": 15, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q3-a"]}, "outputs": [], "source": "import itertools\n\nalphabet = \"abcdefghijklmnopqrstuvwxyz\"\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b. Create a variable `number_of_combinations_by_direct_computation` that gives the number of combinations of the letters of the alphabet of size 3. Do this by computing them directly.\n\n_available marks: 8_"}, {"cell_type": "code", "execution_count": 17, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q3-b"]}, "outputs": [], "source": "import scipy.special\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "### Question 4 (36 marks)\n\n(__Hint__: This question is similar to the [approach describe in the Further information section of the Python for Mathematics chapter on differential equations](https://vknight.org/pfm/tools-for-mathematics/09-differential-equations/why/main.html#how-to-solve-a-system-of-differential-equations).)\n\nConsider the [Lanchaster battle equations](https://en.wikipedia.org/wiki/Lanchester%27s_laws) which can be used to model the size of two armies in battle.\n\n$$\n\\begin{cases}\n\\frac{dX(t)}{dt} = \\alpha Y(t)\\\\\n\\frac{dY(t)}{dt} = \\beta X(t)\n\\end{cases}\n$$\n\na. Obtain the general solution to this equation. Do this by:\n\n- creating a variable `X` and assigning it value the equation with right hand side the size of the army $X(t)$.\n- creating a variable `Y` and assigning it value the equation with right hand side the size of the army $Y(t)$.\n\n_Available marks: 10_"}, {"cell_type": "code", "execution_count": 19, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q4-a"]}, "outputs": [], "source": "x = sym.Function(\"x\")\ny = sym.Function(\"y\")\nt = sym.Symbol(\"t\")\n\nalpha = sym.Symbol(\"alpha\")\nbeta = sym.Symbol(\"beta\")\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b. Consider the initial sizes of the armies: $x(0) = 1000$ and $y(0) = 500$ and the fact that $\\alpha=-4/5$ and $\\beta=-11/10$. Output the particular solutions (in the same format as for question a.):\n\n_Available marks: 12_"}, {"cell_type": "code", "execution_count": 21, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q4-b"]}, "outputs": [], "source": "### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "c. Create the 2 following variables:\n\n1. `end_t_value`: the value of $t$ at which the battle ends.\n2. `does_X_win`: the boolean saying if the $X$ army wins or not.\n\n_Available marks: 14_"}, {"cell_type": "code", "execution_count": 23, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q4-c"]}, "outputs": [], "source": "### BEGIN SOLUTION\n\n\n### END SOLUTION"}], "metadata": {"celltoolbar": "Tags", "kernelspec": {"display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3"}, "language_info": {"codemirror_mode": {"name": "ipython", "version": 3}, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.13.0"}}, "nbformat": 4, "nbformat_minor": 4}
+{"cells": [{"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "### Computing for Mathematics - 2024/2025 individual coursework\n\n**Important** Do not delete the cells containing: \n\n```\n### BEGIN SOLUTION\n\n\n### END SOLUTION\n```\n\nwrite your solution attempts in those cells.\n\nTo submit this notebook:\n\n- Change the name of the notebook from `main` to: `<student_number>`. For example, if your student number is `c1234567` then change the name of the notebook to `c1234567`.\n- **Write all your solution attempts in the correct locations**;\n- **Do not delete** any code that is already in the cells;\n- Save the notebook (`File>Save As`);\n- Follow the instructions given to submit."}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "#### Question 1 (30 marks)\n\n(__Hint__: This question is similar to [this section of the python for mathematics text](https://vknight.org/pfm/tools-for-mathematics/08-statistics/how/main.html#calculate-the-pearson-correlation-coefficient).)\n\nFor each of the following pairs of data $x$ and $y$ compute the Pearson correlation coefficient.\n\na. \n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 1, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-a"]}, "outputs": [], "source": "import statistics as st\nx = (1, 2, 3, 4, 5, 6, 7, 8, 9)\ny = (20, 10, 40, 30, 60, 50, 2, -1, -5)\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b.\n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 4, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-b"]}, "outputs": [], "source": "x = (1, 7, -1, 5, 10, -4)\ny = (20, 10, 40, 30, 60, 50)\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "c. \n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 7, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-c"]}, "outputs": [], "source": "x = (1, 7, 21)\ny = (1 / 3, 1, 3)\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "d. \n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 10, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-d"]}, "outputs": [], "source": "x = (1, 2, 3, 4, 5)\ny = (-1, 1, -1, 1, -1)\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "e.\n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 13, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-e"]}, "outputs": [], "source": "x = (1, 2, 3, 4, 5)\ny = (0, 0, 0, 0, -1)\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "### Question 2 (18 marks)\n\n\n(__Hint__: This question is similar to the [third exercise of the Calculus chapter of Python for mathematics](https://vknight.org/pfm/tools-for-mathematics/03-calculus/solutions/main.html#question-3))\n\nConsider the second derivative $f''(x)=7\\cos(x) + x ^ 2 - \\ln(x)$\n\na. Create a variable `first_derivative` which has value $f'(x)$ (you can assume a constant of integration 0).\n\navailable marks: 12"}, {"cell_type": "code", "execution_count": 16, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q2-a"]}, "outputs": [], "source": "import sympy as sym\n\nx = sym.Symbol(\"x\")\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b. Create a variable `expression` that has value $f(x)$ (you can assume a constant of integration 0):\n\navailable marks: 6"}, {"cell_type": "code", "execution_count": 20, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q2-b"]}, "outputs": [], "source": "### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "### Question 3 (16 marks)\n\n(__Hint__: This question is similar to the [second exercise of the Combinatorics chapter of Python for mathematics](https://vknight.org/pfm/tools-for-mathematics/05-combinations-permutations/solutions/main.html#question-2))\n\na. Create a variable `number_of_combinations` that gives the number of combinations of the letters of the alphabet of size 3. Do this by generating and counting them.\n\n_available marks: 8_"}, {"cell_type": "code", "execution_count": 24, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q3-a"]}, "outputs": [], "source": "import itertools\n\nalphabet = \"abcdefghijklmnopqrstuvwxyz\"\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b. Create a variable `number_of_combinations_by_direct_computation` that gives the number of combinations of the letters of the alphabet of size 3. Do this by computing them directly.\n\n_available marks: 8_"}, {"cell_type": "code", "execution_count": 27, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q3-b"]}, "outputs": [], "source": "import scipy.special\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "code", "execution_count": 29, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "outputs": [], "source": "feedback_string = \"The output is not the expected expression.\"\n\nnbchkr.check_variable_has_expected_property(\n    variable_string=\"number_of_combinations_by_direct_computation\",\n    feedback_string=feedback_string,\n    property_check=check_answer_is_exactly_correct,\n    expected_value=2600.0\n)"}, {"cell_type": "code", "execution_count": 30, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q4-a"]}, "outputs": [], "source": "x = sym.Function(\"x\")\ny = sym.Function(\"y\")\nt = sym.Symbol(\"t\")\n\nalpha = sym.Symbol(\"alpha\")\nbeta = sym.Symbol(\"beta\")\n\n### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b. Consider the initial sizes of the armies: $x(0) = 1000$ and $y(0) = 500$ and the fact that $\\alpha=-4/5$ and $\\beta=-11/10$. Output the particular solutions (in the same format as for question a.):\n\n_Available marks: 12_"}, {"cell_type": "code", "execution_count": 32, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q4-b"]}, "outputs": [], "source": "### BEGIN SOLUTION\n\n\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "c. Create the 2 following variables:\n\n1. `end_t_value`: the value of $t$ at which the battle ends.\n2. `does_X_win`: the boolean saying if the $X$ army wins or not.\n\n_Available marks: 14_"}, {"cell_type": "code", "execution_count": 34, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q4-c"]}, "outputs": [], "source": "### BEGIN SOLUTION\n\n\n### END SOLUTION"}], "metadata": {"celltoolbar": "Tags", "kernelspec": {"display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3"}, "language_info": {"codemirror_mode": {"name": "ipython", "version": 3}, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.12.0"}}, "nbformat": 4, "nbformat_minor": 4}