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+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import numpy as np\n",
+    "from scipy.stats import norm\n",
+    "from IPython.display import display, Math, Markdown\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Opgaver\n",
+    "\n",
+    "Har lavet problem 1 på papir, men vil lave resten i python da det nok er lidt lettere."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Problem 2\n",
+    "\n",
+    "Går ud fra at PCB er i ppm, og kalder den $\\theta$.\n",
+    "\n",
+    "$$\n",
+    "X_i = \\theta + W_i\n",
+    "$$\n",
+    "hvor $W_i \\sim \\mathcal{N}(\\mu, \\sigma^2)$, $\\sigma = 0.08$.\n",
+    "Her går jeg ud fra at $\\mu = 0$.\n",
+    "\n",
+    "Derfor er: \n",
+    "$$\n",
+    "E[X] = \\theta\n",
+    "$$\n",
+    "og\n",
+    "$$\n",
+    "Var[X] = (0.08)^2\n",
+    "$$\n",
+    "\n",
+    "Kan sige at confidence level er:\n",
+    "$$\n",
+    "[\\bar{X} - z_{\\frac \\alpha 2} \\cdot \\frac \\sigma {\\sqrt{n}}, \\bar{X} + z_{\\frac \\alpha 2} \\cdot \\frac \\sigma {\\sqrt{n}}]\n",
+    "$$\n",
+    "with probability $1-\\alpha$.\n",
+    "Her er \n",
+    "$$\n",
+    "z_p = \\Phi^{-1}(1 - p)$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "samples = [11.2, 12.4, 10.8, 11.6, 12.5, 10.1, \n",
+    "           11.0, 12.2, 12.4, 10.6]\n",
+    "n = len(samples)\n",
+    "\n",
+    "X_bar = np.mean(samples)\n",
+    "sigma = 0.08"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/latex": [
+       "$\\displaystyle P(11.430416397415636 \\leq \\theta \\leq 11.529583602584365) = 0.95$"
+      ],
+      "text/plain": [
+       "<IPython.core.display.Math object>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "# Del A\n",
+    "alpha = 0.05\n",
+    "Z = norm.ppf(1 - alpha/2)\n",
+    "\n",
+    "dist = Z * sigma / np.sqrt(n)\n",
+    "lower = X_bar - dist\n",
+    "upper = X_bar + dist\n",
+    "\n",
+    "display(Math(f\"P({lower} \\\\leq \\\\theta \\\\leq {upper}) = {1 - alpha}\"))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/latex": [
+       "$\\displaystyle [-\\infty, 11.529583602584365]$"
+      ],
+      "text/plain": [
+       "<IPython.core.display.Math object>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "\n",
+    "# Okay så lower confidence level er åbenbart at man går fra -infty til upper limit\n",
+    "\n",
+    "display(Math(f\"[-\\\\infty, {upper}]\"))\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/latex": [
+       "$\\displaystyle [11.430416397415636, \\infty]$"
+      ],
+      "text/plain": [
+       "<IPython.core.display.Math object>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "# Del C\n",
+    "# Og igen for uppwer\n",
+    "\n",
+    "display(Math(f\"[{lower}, \\\\infty]\"))"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "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": "3.9.2"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 4
+}