Answer:
np.mean(skin['Price'] >= 100) or equivalent expressions such as skin['Price'].ge(100).mean() or (skin['Price'] >= 100).sum() / skin.shape[0].

To calculate the proportion of products in skin that are expensive (price at least 100), we first evaluate the condition skin['Price'] >= 100. This generates a Boolean series where:

• Each True represents a product with a price greater than or equal to 100.

• Each False represents a product with a price below 100.

When we call np.mean on this Boolean series, Python internally treats True as 1 and False as 0. The mean of this series therefore gives the proportion of True values, which corresponds to the proportion of products that meet the “expensive” criteria.

Alternatively, we can: 1. Count the total number of “expensive” products using (skin['Price'] >= 100).sum(). 2. Divide this count by the total number of products in the dataset (skin.shape[0]).

Both approaches yield the same result.

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Difficulty: ⭐️

The average score on this problem was 90%.

Answer:
(i): "Brand"
(ii): filter
(iii): lambda x: (x['Price'] >= 100).sum() < 5

The expression filters the grouped DataFrame to include only groups where the number of expensive products (those costing at least $100) is fewer than 5. Here’s how the expression works:

1. groupby('Brand'): Groups the skin DataFrame by the unique values in the "Brand" column, creating subgroups for each brand. Now, when we apply the next step, it will perform the action on each Brand group.
2. filter: We use filter instead of agg (or other options) because filter is designed to evaluate a condition on the entire group and decide whether to include or exclude the group in the final result. In this case, we want to include only groups (brands) where the number of expensive products is fewer than 5.
3. (lambda x: (x['Price'] >= 100).sum() < 5): This lambda function runs on each group (in this case, x is a dataframe). It calculates the number of products in the group where the price is greater than or equal to $100 (i.e., expensive products). If this count is less than 5, the group is kept. Otherwise, the group is excluded from our final result.

Since we have an .nunique() at the end of our original expression, we will return the number (n in nunique) of unique brands.

Alternative forms for (iii) include:
- lambda x: np.sum(x['Price'] >= 100) < 5
- lambda x: x[x['Price'] >= 100].shape[0] < 5
- lambda x: x[x['Price'] >= 100].count() < 5

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: (i): HAVING (ii): WHERE

An important distinction to make for part (ii) is that we’re asking about the keyword for querying. SELECT is incorrect, as SELECT is for specifying a subset of columns, whereas querying is for specifying a subset of rows.

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: 4

vc computes the frequency of each value in small_prices, sorts the counts in descending order, and stores the result in vc. The resulting vc will look like this, where 36, 1, 18, 100 are the indices and 4, 3, 2, 1 are the values:

• iloc[0] retrieves the value at integer position 0 in vc. Remember that position can be different than index for a Series: for this Series, the indices are 36, 1, 18, 100, whereas the integer positions are 0, 1, 2, 3.
• The result is 4.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer: Error

• loc[0] attempts to retrieve the value associated with the index 0 in vc.
• Since vc has indices 36, 1, 18, 100, and 0 is not a valid index, this operation results in an error.

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Difficulty: ⭐️⭐️

The average score on this problem was 83%.

Answer: 36

• index[0] retrieves the first index in vc, which corresponds to the value 36.`

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 69%.

Answer: 3

• iloc[1] retrieves the value at integer position 1 in vc, which is 3.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

Answer: 3

• loc[1] retrieves the value associated with the index 1 in vc, which corresponds to the frequency of the value 1 in small_prices.
• The result is 3.

---

Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer: 1

• index[1] retrieves the positionally second index in vc.
• The result is 1.

---

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 64%.

Answer: 10

The expression clinique.merge(fresh, on="Type", how="inner") performs an inner join on the Type column between the clinique and fresh DataFrames. An inner join includes only rows where the Type values are present in both DataFrames.

From the provided type_pivot table:

• The overlapping Type values between CLINIQUE and FRESH are:

  • Face Mask (3.0 for CLINIQUE, 4.0 for FRESH)

  • Moisturizer (3.0 for CLINIQUE, 3.0 for FRESH)

However, we must notice that the aggfunc for our original pivot table is lambda s: s.shape[0] + 1. Thus, we know that the values in the pivot table include an additional count of 1 for each group. This means the raw counts are actually:

• Face Mask: 2 rows in CLINIQUE, 3 rows in FRESH

• Moisturizer: 2 rows in CLINIQUE, 2 rows in FRESH

The inner join will match these rows based on Type. Since there are 2 rows in CLINIQUE for both Face Mask and Moisturizer, and 3 and 2 rows in FRESH respectively, the result will have:

• Face Mask: 2 (from CLINIQUE) × 3 (from FRESH) = 6 rows

• Moisturizer: 2 (from CLINIQUE) × 2 (from FRESH) = 4 rows

The total number of rows in the resulting DataFrame is 6 + 4 = 10.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 42%.

Answer: 31

An outer join includes all rows from all DataFrames, filling in missing values (NaN) for any Type not present in one of them. From the table, the counts are calculated as follows:

Remember that our pivot_table aggfunc is calculating lambda s: s.shape[0] + 1 (so we need to subtract one from the pivot table entry to get the actual number of rows).

• Cleanser: 5 · 1
  • 5 rows in CLINIQUE
  • 1 row in BOSCIA
• Eye cream: 3 · 1
  • 3 rows in CLINIQUE
  • 1 row in BOSCIA
• Face Mask: 2 · 3 · 3
  • 2 rows in CLINIQUE
  • 3 rows in FRESH
  • 3 rows in BOSCIA
• Moisturizer: 2 · 2
  • 2 rows in CLINIQUE
  • 2 rows in FRESH
• Sun protect: 1
  • 1 row in CLINIQUE

Final Calculation:

5 · 1 + 3 · 1 + 2 · 3 · 3 + 2 · 2 + 1 = 31

---

Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 27%.

Answer:

• All 60 product names contain the term drops.
  
• None of the 59 other product names contain the term drops.
  
• There are 25 terms in the product name above in total.

The TF-IDF score of a term is calculated as: \text{TF-IDF} = \text{TF} \cdot \text{IDF} Where:

• \text{TF}: Term Frequency in a document (the ratio of the term’s occurrences to the total number of terms in the document).

• \text{IDF}: Inverse Document Frequency, which is given by: \text{IDF} = \log_2\left(\frac{N}{1 + n}\right) Here:

  • N: Total number of documents (60 in this case).

  • n: Number of documents containing the term.

The TF-IDF of drops in the product name is given as \frac{2}{3}.

---

Step 1: Calculating Term Frequency (TF) The problem states that the term drops appears in the product name. If the total number of terms in the product name is T, then: \text{TF} = \frac{5}{T}

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Step 2: Calculating Inverse Document Frequency (IDF) Substitute the known values into the IDF formula: \text{IDF} = \log_2\left(\frac{60}{1 + n}\right) Where n is the number of other product names (out of 59) that contain drops.

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Step 3: Combine TF and IDF to Match TF-IDF The TF-IDF is given as \frac{2}{3}. Substituting the expressions for TF and IDF: \frac{5}{T} \cdot \log_2\left(\frac{60}{1 + n}\right) = \frac{2}{3}

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Step 4: Analyze Each Option

If all 60 product names contain drops, then n = 59. Substituting into the IDF formula: \text{IDF} = \log_2\left(\frac{60}{1 + 59}\right) = \log_2(1) = 0 Since \text{IDF} = 0, the TF-IDF score would also be 0, which contradicts the given TF-IDF of \frac{2}{3}.

This is NOT possible.

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If 14 other product names contain drops, then n = 14. Substituting into the IDF formula: \text{IDF} = \log_2\left(\frac{60}{1 + 14}\right) = \log_2\left(\frac{60}{15}\right) = \log_2(4) = 2 Substituting \text{IDF} = 2 into the TF-IDF equation: \frac{5}{T} \cdot 2 = \frac{2}{3} Solving for T: \frac{10}{T} = \frac{2}{3} \implies T = 15 This within the problem’s constraints.

This is possible.

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If no other product names contain drops, then n = 0. Substituting into the IDF formula: \text{IDF} = \log_2\left(\frac{60}{1 + 0}\right) = \log_2(60) Approximating \log_2(60) \approx 5.91, substituting into the TF-IDF equation: \frac{5}{T} \cdot 5.91 = \frac{2}{3} Solving for T: \frac{29.55}{T} = \frac{2}{3} \implies T \approx 44.33 Since T must be an integer, this is not consistent with the problem’s constraints.

This is NOT possible.

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If T = 15, substituting into the TF equation: \text{TF} = \frac{5}{15} = \frac{1}{3} Substituting \text{TF} = \frac{1}{3} into the TF-IDF equation: \frac{1}{3} \cdot \text{IDF} = \frac{2}{3} \implies \text{IDF} = 2 Substituting \text{IDF} = 2 into the IDF formula: 2 = \log_2\left(\frac{60}{1 + n}\right) 2 = \log_2(4) \implies \frac{60}{1 + n} = 4 \implies n = 14 This is consistent with the problem’s constraints.

This is possible.

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If T = 25, substituting into the TF equation: \text{TF} = \frac{5}{25} = \frac{1}{5} Substituting \text{TF} = \frac{1}{5} into the TF-IDF equation: \frac{1}{5} \cdot \text{IDF} = \frac{2}{3} \implies \text{IDF} = \frac{10}{3} Substituting \text{IDF} = \frac{10}{3} into the IDF formula: \frac{10}{3} = \log_2\left(\frac{60}{1 + n}\right) This results in a value for n that is not an integer.

This is NOT possible.

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Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer: Option 2

Explanation:

The given code:

prods = [row.get("prod") for row in soup.find_all("row", class_="thing")]

retrieves all <row> elements with the class "thing" and extracts the value of the "prod" attribute for each.

Option 1: In this structure, the product names appear as text content within the <row> tags, not as an attribute. row.get("prod") would return None.

Option 2: Here, the product names are stored as the value of the "prod" attribute. row.get("prod") successfully retrieves these values.

Option 3: In this structure, the "prod" attribute contains the value "thing", and the product names are stored as the value of the "class" attribute. soup.find_all("row", class_="thing") will not retrieve any of these tags since there are no tags with class="thing".

Option 4: In this structure, the product names are part of the text content inside the <row> tags. row.get("prod") would return None.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: The mode of the data

The minimizer of empirical risk for the constant model when using zero-one loss is the mode.

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Difficulty: ⭐️

The average score on this problem was 100%.

Answer: The median of the data

The minimizer of empirical risk for the constant model when using absolute loss is the median.

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Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer: None of the above

The minimizer of empirical risk for R_1(h_1^*) is the sum of absolute deviations from the median divided by n.

R_1(h_1^*) represents the sum of absolute deviations from the median divided by n:

R_1(h_1^*) = \frac{1}{n} \sum_{i=1}^n |y_i - h_1^*|

---

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 44%.

Answer: The mean of the data

The minimizer of empirical risk for the constant model when using squared loss is the mean.

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: The variance of the data

The minimizer of empirical risk for R_2(h_2^*) is the mean squared error, which is equivalent to the variance of the data when h_2^* equals the mean.

R_2(h_2^*) represents the mean squared error:

R_2(h_2^*) = \frac{1}{n} \sum_{i=1}^n (y_i - h_2^*)^2

---

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 70%.

Answer: h_\text{U}^* \geq M

Since we’re multiplying by y_i, the Ulta loss function skews the optimal prediction higher to reduce the impact of larger y_i values on the loss.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 42%.

Answer: h^* = \frac{\sum_{i = 1}^n y_i^2}{\sum_{i = 1}^n y_i + n\lambda}

To minimize the regularized average Ulta loss, we solve the equation by setting the derivative to zero and solving for h.

Set the derivative to zero:

\frac{\partial R_\lambda(h)}{\partial h} = 0

-2 \left( \frac{1}{n} \sum_{i = 1}^n y_i (y_i - h) \right) + 2\lambda h = 0

Simplify:

\frac{1}{n} \sum_{i = 1}^n y_i^2 - \frac{1}{n} \sum_{i = 1}^n y_i h - \lambda h = 0

Combine terms:

\frac{1}{n} \sum_{i = 1}^n y_i^2 = h \left( \frac{1}{n} \sum_{i = 1}^n y_i + \lambda \right)

Solve for h:

h^* = \frac{\frac{1}{n} \sum_{i = 1}^n y_i^2}{\frac{1}{n} \sum_{i = 1}^n y_i + \lambda}

h^* = \frac{\sum_{i = 1}^n y_i^2}{\sum_{i = 1}^n y_i + n\lambda}

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 71%.

Answer: 16

The predicted number of ingredients for Victors’ Veil is calculated using the regression model: \hat{y} = w_0^* + w_1^* x

Substituting w_0^* = 11, w_1^* = \frac{1}{10}, and x = 40: \hat{y} = 11 + \frac{1}{10} \cdot 40 = 11 + 4 = 15

The squared loss is then:

L = (\hat{y} - y)^2 = (15 - 11)^2 = 4^2 = 16

---

Difficulty: ⭐️⭐️

The average score on this problem was 86%.

Answer: Yes; the values of w_0^* and w_1^* don’t impact the answer to part (a).

To answer part (a), we only need to know the average values \bar{x} and \bar{y}, as the regression line will always pass through (\bar{x}, \bar{y}).

---

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 56%.

Answer: \begin{bmatrix} 15 & -600 \\ -600 & 30000 \end{bmatrix}

The design matrix U is defined as: U = \begin{bmatrix} 1 & -x_1 \\ 1 & -x_2 \\ \vdots & \vdots \\ 1 & -x_n \end{bmatrix} where 1 represents the intercept term, and -x_i represents the negative price of product i. Thus:

U^T U = \begin{bmatrix} 1 & 1 & \cdots & 1 \\ -x_1 & -x_2 & \cdots & -x_n \end{bmatrix} \begin{bmatrix} 1 & -x_1 \\ 1 & -x_2 \\ \vdots & \vdots \\ 1 & -x_n \end{bmatrix}

When we multiply this out, we get: U^T U = \begin{bmatrix} n & -\sum x_i \\ -\sum x_i & \sum x_i^2 \end{bmatrix}

Conceptually, this matrix represents: \begin{bmatrix} \text{Number of elements} & \text{Sum of the negative values of the data} \\ \text{Sum of the negative values of the data} & \text{Sum of squared values of the data} \end{bmatrix}.

We know that the sum of all elements in a series is equal to the mean of the series multiplied by the number of elements in the series. \Sigma x_i = n \cdot \bar{x}

\bar{x} is equal to 40. The first element in the matrix, 15, represents the number of elements in the series. Thus, \Sigma x_i = 40 \times 15 = 600, and the solution must be as follows:

\begin{bmatrix} n & -\sum x_i \\ -\sum x_i & \sum x_i^2 \end{bmatrix} = \begin{bmatrix} 15 & -600 \\ -600 & 30000 \end{bmatrix}

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

Answer:

Columns: 22

Model A includes an intercept, the price feature, and a one-hot encoding of the 20 brands without dropping the first category. The intercept adds 1 column, the price adds 1 column, and the one-hot encoding for brands adds 20 columns.

---

Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer:

Columns: 2

Rank: 2

Model B includes an intercept and the price feature, but no one-hot encoding for the brands or product types. The intercept and price are linearly independent, resulting in both the number of columns and the rank being 2.

---

Difficulty: ⭐️⭐️

The average score on this problem was 82%.

Answer:

Columns: 8

Rank: 7

Model C includes an intercept, the price feature, and a one-hot encoding of the 6 product types without dropping the first category. The intercept adds 1 column, the price adds 1 column, and the one-hot encoding for the product types adds 6 columns. However, one column is linearly dependent because we didn’t drop one of the one-hot encoded columns, reducing the rank to 7.

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: Model D

For residuals to sum to zero in a linear regression model, the design matrix must include either an intercept term (Models A - C) or equivalent redundancy in the encoded variables to act as a substitute for the intercept (Model E). Models that lack both an intercept and equivalent redundancy, such as Model D, are not guaranteed to have residuals sum to zero.

One-hot encoding without dropping any category can behave like an intercept because it introduces a column for each category, allowing the model to adjust its predictions as if it had an intercept term. The sum of the indicator variables in such a setup equals 1 for every row, creating a similar effect to an intercept, which is why residuals may sum to zero in these cases.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 60%.

We have a grid with 2 features: L, which we use to iterate our \lambda value in ridge regression, and D, which we use to iterate the polynomial degree for one of our hyperparameters.

Answer: N \times \frac{K-1}{K}

In K-fold cross-validation, the data is split into K folds. Each time a model is trained, one fold is used for validation, and the remaining K-1 folds are used for training. Since N is evenly split among K folds, each fold contains \frac{N}{K} points. The training data, therefore, consists of N - \frac{N}{K} = N \times \frac{K-1}{K} points.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: LD \times (K-2)

In K-fold cross-validation, X_{\text{train}}.\text{iloc}[1] and X_{\text{train}}.\text{iloc}[-1] are in different folds and can only be used together for training when neither of their respective folds is the validation set. This happens in (K-2) folds for each combination of polynomial degree (D) and regularization hyperparameter (L). Since there are L \times D total combinations in the grid search, the two points are used together for training LD \times (K-2) times.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 53%.

Answer: Model complexity and model variance both decrease.

As \lambda increases, the L_1 regularization penalizes the model more heavily for including features, effectively reducing the number of non-zero coefficients in the model. This decreases model complexity. Additionally, by reducing complexity, the model becomes less sensitive to the training data, leading to lower variance.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 64%.

Answer: The AVMSE of an unregularized multiple linear regression model.

Point A represents the case where \lambda = 0, meaning no regularization is applied. This corresponds to an unregularized multiple linear regression model.

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Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer: The AVMSE of the \lambda we’d choose to use to train a model.

Point B is the minimum point on the graph, indicating the optimal \lambda value that minimizes the AVMSE. This is the \lambda we’d choose to train the model.

---

Difficulty: ⭐️

The average score on this problem was 91%.

Answer: The AVMSE of the constant model.

Point C represents the AVMSE when \lambda is very large, effectively forcing all coefficients to zero. This corresponds to the constant model.

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Difficulty: ⭐️⭐️

The average score on this problem was 83%.

Answer: Logistic Regression

Logistic regression creates a linear divide between classes.

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Difficulty: ⭐️

The average score on this problem was 93%.

Answer: Decision tree with \text{max depth} = 15

We know that Decision Boundaries 2 and 5 are decision trees, since the decision boundaries are parallel to the axes. Decision boundaries for decision trees are parallel to axes because the tree splits the feature space based on one feature at a time, creating partitions aligned with that feature’s axis (e.g., “Price ≤ 100” results in a vertical split).

Decision Boundary 2 is more complicated than Decision Boundary 5, so we can assume there are more levels.

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Difficulty: ⭐️⭐️

The average score on this problem was 82%.

Answer: k-nearest neighbors with k = 3

k-nearest neighbors decision boundaries don’t follow straight lines like logistic regression or decision trees. Thus, boundaries 3 and 4 are k-nearest neighbors.

Boundary 3 seems to be more fit to the data than boundary 4, so we can assume that we used a decision tree with \text{max depth} = 3.

For example, if in our training data we had 3 points in the group represented by the light-shaded region at (300, 150), the decision boundary for k-nearest neighbors with k = 3 would be light-shaded at (300, 150), whereas it may be dark-shaded for k-nearest neighbors with k = 100.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 59%.

Answer: k-nearest neighbors with k = 100

k-nearest neighbors decision boundaries don’t follow straight lines like logistic regression or decision trees. Thus, boundaries 3 and 4 are k-nearest neighbors.

Boundary 3 seems to be less fit to the data than boundary 4, so we can assume that we used a decision tree with \text{max depth} = 100.

---

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 56%.

In order to solve this problem, we need to use the sigmoid function to cast our result into a probability between 0-1. As a reminder, the sigmoid function \frac{1}{1 + e^{-x}} trends towards 0 when the input becomes more negative, and towards 1 as the input becomes a higher positive.

Wolfcare:

z = -1 + \frac{1}{5}(15) - \frac{3}{5}(20) + 0 = -1 + 3 - 12 = -10 P = \frac{1}{1 + e^{-(-10)}} = \frac{1}{1 + e^{10}} \approx 0

Decision: No, as P < 0.5

Go Blue Glow:

z = -1 + \frac{1}{5}(25) - \frac{3}{5}(5) + 0 = -1 + 5 - 3 = 1 P = \frac{1}{1 + e^{-1}} \approx \frac{1}{1 + 0.3679} \approx 0.73

Decision: Yes, as P \geq 0.5.

DataSPF:

z = -1 + \frac{1}{5}(50) - \frac{3}{5}(15) + 0 = -1 + 10 - 9 = 0 P = \frac{1}{1 + e^{-0}} = \frac{1}{2} = 0.5

Decision: Yes, as P \geq 0.5.

Maize Mist: z = -1 + \frac{1}{5}(0) - \frac{3}{5}(1) + 0 = -1 - 0.6 = -1.6 P = \frac{1}{1 + e^{-(-1.6)}} = \frac{1}{1 + e^{1.6}} \approx \frac{1}{1 + 4.953} \approx 0.17 Decision: No, as P < 0.5

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Difficulty: ⭐️⭐️

The average score on this problem was 85%.

Answer: \boxed{\frac{35}{40}} The precision of a classification model is calculated as:

\text{Precision} = \frac{\text{True Positives (TP)}}{\text{True Positives (TP)} + \text{False Positives (FP)}}

From the confusion matrix for threshold A:

\begin{array}{|l|c|c|} \hline & \text{Pred.} \: 0 & \text{Pred.} \: 1 \\ \hline \text{Actually} \: 0 & 40 & 5 \\ \hline \text{Actually} \: 1 & 35 & 35 \\ \hline \end{array}

The number of true positives (TP) is 35, and the number of false positives (FP) is 5. Substituting into the formula:

\text{Precision} = \frac{35}{35 + 5} = \frac{35}{40}

Thus, the precision for threshold A is:

\boxed{\frac{35}{40}}

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Difficulty: ⭐️⭐️

The average score on this problem was 85%.

Answer: 60

From the confusion matrix for threshold B:

\begin{array}{|l|c|c|} \hline & \text{Pred.} \: 0 & \text{Pred.} \: 1 \\ \hline \text{Actually} \: 0 & 5 & 40 \\ \hline \text{Actually} \: 1 & 10 & ??? \\ \hline \end{array}

The sum of all entries in a confusion matrix must equal the total number of samples. For thresholds A and C, the total number of samples is:

40 + 5 + 35 + 35 = 115

Using this total, for threshold B, the entries provided are:

5 (\text{TN}) + 40 (\text{FP}) + 10 (\text{FN}) + ??? (\text{TP}) = 115

Solving for the missing value (???):

115 - (5 + 40 + 10) = 60

Thus, the missing value for threshold B is:

\boxed{60}

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Difficulty: ⭐️

The average score on this problem was 90%.

Thresholds can be arranged based on their strictness. Higher thresholds result in more predictions for class 0 (more products classified as not designed for sensitive skin).

From the confusion matrices: - Threshold A predicts the least as 0 and Threshold B predicts the most as 0.

\boxed{A > C > B}

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

A false positive is worse, as it would mean we predict a product is safe when it isn’t; someone with sensitive skin may use it and it could harm them.

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Difficulty: ⭐️

The average score on this problem was 91%.

Answer: \nabla Q(\vec{x}) = \begin{bmatrix} 2(x_1 - x_2) \\ -2x_1 + 6x_2 \end{bmatrix}

The gradient of Q(\vec{x}) is given by: \nabla Q(\vec{x}) = \begin{bmatrix} \frac{\partial Q}{\partial x_1} \\ \frac{\partial Q}{\partial x_2} \end{bmatrix}

First, compute the partial derivatives:

1. Partial derivative with respect to x_1: Q(x_1, x_2) = x_1^2 - 2x_1x_2 + 3x_2^2 - 1 \frac{\partial Q}{\partial x_1} = 2x_1 - 2x_2

2. Partial derivative with respect to x_2: \frac{\partial Q}{\partial x_2} = -2x_1 + 6x_2

Thus, the gradient is: \nabla Q(\vec{x}) = \begin{bmatrix} 2(x_1 - x_2) \\ -2x_1 + 6x_2 \end{bmatrix}

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Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Final Answer: \frac{5}{4}

In gradient descent, the update rule is: \vec{x}^{(k+1)} = \vec{x}^{(k)} - \alpha \nabla Q(\vec{x}^{(k)})

We are given: \vec{x}^{(0)} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad \vec{x}^{(1)} = \begin{bmatrix} 1 \\ -4 \end{bmatrix}

1. Compute \nabla Q(\vec{x}^{(0)}):

Substitute \vec{x}^{(0)} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} into the gradient: \nabla Q(\vec{x}^{(0)}) = \begin{bmatrix} 2(1 - 1) \\ -2(1) + 6(1) \end{bmatrix} = \begin{bmatrix} 0 \\ 4 \end{bmatrix}

2. Use the update rule to find \alpha:

From the update rule: \vec{x}^{(1)} = \vec{x}^{(0)} - \alpha \nabla Q(\vec{x}^{(0)})

Substitute the values: \begin{bmatrix} 1 \\ -4 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} - \alpha \begin{bmatrix} 0 \\ 4 \end{bmatrix}

Separate components: 1 = 1 - \alpha(0) \quad \text{(satisfied for any $\alpha$)}, -4 = 1 - \alpha(4)

Solve for \alpha: -4 = 1 - 4\alpha \implies 4\alpha = 5 \implies \alpha = \frac{5}{4}

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Difficulty: ⭐️⭐️

The average score on this problem was 86%.