Merge pull request #111 from gperdrizet/dev

Updated lesson 12, 13 and 14 solution notebooks

Author: gperdrizet

notebooks/notebooks.yml

@@ -133,4 +133,6 @@ units:
           - name: "In class demo"
             file: "Lesson_14_demo.ipynb"
           - name: "Activity"
-            file: "Lesson_14_activity.ipynb"
+            file: "Lesson_14_activity.ipynb"
+          - name: "Activity solution"
+            file: "Lesson_14_activity_solution.ipynb"

notebooks/unit2/lesson_12/Lesson_12_activity_solution.ipynb

@@ -190,9 +190,17 @@
     "   - **Bonus**: Calculate the standard deviation and explain what it tells you about the variability in monthly rainfall patterns"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "c0fa8e75",
+   "metadata": {},
+   "source": [
+    "### Probability of rain calculation"
+   ]
+  },
   {
    "cell_type": "code",
-   "execution_count": 7,
+   "execution_count": null,
    "id": "996e854d",
    "metadata": {},
    "outputs": [
@@ -205,14 +213,21 @@
     }
    ],
    "source": [
-    "# 1. Calculate probability of rain\n",
     "rainy_days = len(df[df['weather_condition'] == 'Rainy'])\n",
     "total_days = len(df)\n",
     "p_rain = rainy_days / total_days\n",
     "\n",
     "print(f\"Based on our data, there's a {p_rain*100:.1f}% chance of rain on any given day\")"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "32786f54",
+   "metadata": {},
+   "source": [
+    "### Binomial distribution calculation"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 8,
@@ -226,6 +241,14 @@
     "probabilities = stats.binom.pmf(k_values, n_days, p_rain)"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "22f972af",
+   "metadata": {},
+   "source": [
+    "### Binomial distribution plot"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 9,
@@ -270,9 +293,17 @@
     "**Probability of 15+ rainy days**: We can calculate this using the cumulative distribution function."
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "0ecbe5a9",
+   "metadata": {},
+   "source": [
+    "### Probability of >= 15 rainy days"
+   ]
+  },
   {
    "cell_type": "code",
-   "execution_count": 13,
+   "execution_count": null,
    "id": "5bc3a33e",
    "metadata": {},
    "outputs": [
@@ -285,12 +316,19 @@
     }
    ],
    "source": [
-    "# Calculate probability of 15 or more rainy days\n",
     "prob_15_or_more = 1 - stats.binom.cdf(14, n_days, p_rain)\n",
     "\n",
     "print(f\"Probability of 15+ rainy days: {prob_15_or_more:.4f} ({prob_15_or_more*100:.2f}%)\")"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "981ac816",
+   "metadata": {},
+   "source": [
+    "### Extra: Binomial cumulative distribution function (CDF) visualization"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 14,
@@ -401,9 +439,17 @@
     "   - **Bonus**: Repeat the experiment with different sample sizes (n=5, n=10, n=50). How does sample size affect the spread and normality of the sampling distribution?"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "c879505c",
+   "metadata": {},
+   "source": [
+    "### Population distribution"
+   ]
+  },
   {
    "cell_type": "code",
-   "execution_count": 16,
+   "execution_count": null,
    "id": "da2ffd4c",
    "metadata": {},
    "outputs": [
@@ -427,7 +473,6 @@
     }
    ],
    "source": [
-    "# 1. Examine population distribution\n",
     "population_mean = df['rainfall_inches'].mean()\n",
     "population_std = df['rainfall_inches'].std()\n",
     "\n",
@@ -449,14 +494,21 @@
     "The population distribution is highly right-skewed with many zero values (no rain) and a long tail of higher rainfall amounts."
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "d2b5ce7c",
+   "metadata": {},
+   "source": [
+    "### Sampling"
+   ]
+  },
   {
    "cell_type": "code",
-   "execution_count": 17,
+   "execution_count": null,
    "id": "9b1cacd5",
    "metadata": {},
    "outputs": [],
    "source": [
-    "# 2. Create sampling distribution\n",
     "n_samples = 1000\n",
     "sample_size = 30\n",
     "sample_means = []\n",
@@ -468,9 +520,17 @@
     "sample_means = np.array(sample_means)"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "af68a64a",
+   "metadata": {},
+   "source": [
+    "### Sampling distribution plot"
+   ]
+  },
   {
    "cell_type": "code",
-   "execution_count": 24,
+   "execution_count": null,
    "id": "6dd93618",
    "metadata": {},
    "outputs": [
@@ -486,7 +546,6 @@
     }
    ],
    "source": [
-    "# 3. Visualize sampling distribution\n",
     "standard_error = population_std / np.sqrt(sample_size)\n",
     "\n",
     "# Normal curve\n",
@@ -502,6 +561,14 @@
     "plt.show()"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "ebf6dae2",
+   "metadata": {},
+   "source": [
+    "### Sampling distribution versus population comparison"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 23,