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:

• Count the total number of “expensive” products using (skin['Price'] >= 100).sum().
• Divide this count by the total number of products in the dataset (skin.shape[0]).

Both approaches yield the same result.

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

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.

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, and 100 are the indices and 4, 3, 2, and 1 are the values:

• iloc[0] retrieves the value at integer position 0 in vc. The result is 4.

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, 10, and 0 is not a valid index, this operation results in an error.

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.

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.

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, we know that 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.

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 \cdot 1
  • 5 rows in clinique
  • 1 row in BOSCIA
• Eye cream: 3 \cdot 1
  • 3 rows in clinique
  • 1 row in BOSCIA
• Face Mask: 2 \cdot 3 \cdot 3
  • 2 rows in clinique
  • 3 rows in fresh
  • 3 rows in BOSCIA
• Moisturizer: 2 \cdot 2
  • 2 rows in clinique
  • 2 rows in fresh
• Sun protect: 1
  • 1 row in clinique
    

Final Calculation: 5 \cdot 1 + 3 \cdot 1 + 2 \cdot 3 \cdot 3 + 2 \cdot 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}(t,d) = \text{tf}(t,d) \cdot \text{idf}(t) = \frac{\# \text{ of occurrences of } t \text{ in } d }{\text{total } \# \text{ of terms in } d} \cdot \log_2\left(\frac{\text{total } \# \text{ of documents}}{\# \text{ of documents in which } t \text{ appears}}\right)

We know:

• t appears 5 times in d

• Total number of documents = 60

• \text{TF-IDF} = \frac{2}{3}

Let T be the total number of terms in d (d refers to the sentence at the top) and n refer to number of documents in which t appears, then substituting this into our formula gives us: \frac{5}{T} \cdot \log_2\left(\frac{60}{n}\right) = \frac{2}{3}

The problem then boils down to checking whether the above equation holds true in each of the five options

• Option 1: All 60 product names contain the term drops.

  • If n = 60, then \log_2(60/60) = 0
  • This makes TF-IDF = 0, contradicting \frac{2}{3}
    

  This is NOT possible.

• Option 2: 14 other product names contain the term drops.

  • If n = 15 (original document + 14 others), \log_2(60/15) = 2
  • \frac{5}{T} \cdot 2 = \frac{2}{3} implies T = 15
    

  This is possible.

• Option 3: None of the 59 other product names contain the term drops.

  • If n = 1, \log_2(60/1) \approx 5.91
  • This leads to a non-integer total terms T
  • Since T must be an integer, this is not consistent with the problem’s constraints.
    

  This is NOT possible.

• Option 4: There are 15 terms in the product name above in total. If T = 15, substituting into the TF equation, we get:

  \text{TF} = \frac{5}{15} = \frac{1}{3} $ = = 2 $ Substituting \text{IDF} = 2 into the IDF formula: 2 = \log_2\left(\frac{60}{n}\right) 2 = \log_2(4) \implies \frac{60}{n} = 4 \implies n = 14 This is consistent with the problem’s constraints.

  This is possible.

• Option 5: There are 25 terms in the product name above in total. 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}{n}\right) This results in a value for n that is not an integer.

  This is NOT possible.

Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer: Option 2

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.

Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: The mode of the data

R_0(h) counts the number of incorrect predictions for each prediction, so choosing the most frequent value (mode) minimizes this count.

Difficulty: ⭐️

The average score on this problem was 100%.

Answer: The median of the data

As seen in lecture, the minimizer of empirical risk for the constant model when using absolute loss is the median.

Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer: None of the above

As we discussed in the previous subpart, h_1^*, the minimizer of mean absolute error, is the median. So, R_1(h_1^*) is the mean absolute difference (or deviation) between all values and the median. This is not any of the answer choices, and in general, is not a quantity we studied in detail in class.

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.

Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: The variance of the data

R_2(h) represents the mean squared error of the constant prediction, h. In the previous part, we discussed that h_2^*, the minimizer of R_2(h), is the mean, \bar{y}. So, R_2(h_2^*) is then:

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

This is the definition of the variance of y.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 70%.

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

Minimizing the average Ulta loss means minimizing the empirical risk:

R_U(h) = \frac{1}{n} \sum_{i = 1}^n y_i (y_i - h)^2.

This resembles minimizing mean squared error, except each y_i is given a weight of y_i. Since larger y_i values contribute more to the loss, the minimizer is pulled toward higher values to reduce their impact. This results in h_\text{U}^* being greater than the mean of the data.

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 for h by setting
\frac{\partial R_\lambda(h)}{\partial h} = 0.

Step 1: Compute the derivative and set it to zero

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

Step 2: Expand and simplify

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

Rearrange terms:

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

Step 3: Solve for h

Divide both sides by \frac{1}{n} \sum_{i = 1}^n y_i + \lambda:

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

Step 4: Multiply by \frac{n}{n}

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

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 71%.

Answer: 16

Using the equation of the regression model we have seen in class:
H(x) = w_0^* + w_1^* x

Substitute w_0^* = 11, w_1^* = \frac{1}{10}, and x = 40:
H(40) = 11 + \frac{1}{10} \cdot 40 = 11 + 4 = 15

The squared loss is:
L = (y - H(x_i))^2

Substituting y = 11 (actual) and H(x_i) = 15 (predicted):
L = (11 - 15)^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).

A fact from lecture is that the regression line always passes through the point (\bar{x}, \bar{y}), meaning that for an average x, we always predict an average y. We’re given that \bar{x} = 40 and \bar{y} = 15, meaning that for a product that costs $40 we will predict that it has 15 ingredients, no matter what the slope and intercept end up being.

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_{i = 1}^n x_i \\ -\sum_{i = 1}^n x_i & \sum_{i = 1}^n 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. \sum_{i = 1}^n 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,
\displaystyle \sum_{i = 1}^n x_i = 40 \cdot 15 = 600
and the solution must be as follows:

\begin{bmatrix} n & -\sum_{i = 1}^n x_i \\ -\sum_{i = 1}^n x_i & \sum_{i = 1}^n x_i^2 \end{bmatrix} = \begin{bmatrix} 15 & -600 \\ -600 & 30000 \end{bmatrix}

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

Answer:

Number of columns in X: 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. The rank is 21 because one of the brand categories can be predicted by the others (i.e., the columns are linearly dependent).

The rank is 21 because when encoding categorical variables (such as brands), dropping the first category avoids introducing perfect multicollinearity. By not dropping a category, we increase the rank by 1, but this still maintains linear dependence among the columns.

Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer:

Number of columns in X: 2
Rank of X: 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:

Number of columns in X: 8
Rank of X: 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.

This happens because if we have 6 product types, one of the columns can always be written as a linear combination of the other five. Thus, we lose one degree of freedom in the matrix, and the rank is 7.

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.

For a linear model, the residual vector \vec{e} is chosen such that it is orthogonal to the column space of the design matrix X. If one of the columns of X is a column of all ones (as in the case with an intercept term), it ensures that the sum of the residuals is zero. This is because the error vector \vec{e} is orthogonal to the span of X, and hence X^T \vec{e} = 0. If there isn’t a column of all ones, but we can create such a column using a linear combination of the other columns in X, the residuals will still sum to zero. Model D lacks this structure, so the residuals are not guaranteed to sum to zero.

In contrast, 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.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 60%.

Answer: LD

There are L possible values of \lambda and D possible polynomial degrees, giving LD total combinations of hyperparameters.

Answer: N \cdot \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 \cdot \frac{K-1}{K} points.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: LD \times (K-2)

In K-fold cross-validation, X_train.iloc[1] and X_train.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.

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 larger coefficient values more heavily, forcing the model to forget the details in the training data and focus on the bigger picture. This effectively reduces the number of non-zero coefficients, decreasing model complexity. Since model complexity and model variance are the same thing, variance also decreases.

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.

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 because we don’t typically regularize the intercept term’s coefficient, w_0^*.

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.

Difficulty: ⭐️⭐️

The average score on this problem was 83%.

Answer: Logistic regression

Logistic regression is a linear classification technique, meaning it creates a single linear decision boundary between classes. Among the five decision boundaries, only Decision Boundary 1 is linear, making it the correct answer.

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.

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.

A lower value of k means the model is more sensitive to local variations in the data. Since Decision Boundary 3 has more complex and irregular regions than Decision Boundary 4, it corresponds to k = 3, where the model closely follows the training data.

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.

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.

k-nearest neighbors decision boundaries adapt to the training data, but increasing k makes the model less sensitive to local variations.

Since Decision Boundary 4 is smoother and has fewer regions than Decision Boundary 3, it corresponds to k = 100, where predictions are averaged over a larger number of neighbors, making the boundary less sensitive to individual data points.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 56%.

Using the sigmoid function, we can predict the probabliity using this general formula -

P(y = 1 | \vec{x}) = \sigma(\vec{w}^T \vec{x})

where \sigma(x) = \frac{1}{1 + e^{-x}} is the sigmoid function.

We calculate z = \vec{w}^T \vec{x} for each product and then apply the sigmoid function to find the probability P(y = 1 | \vec{x}). If the probability is greater than or equal to 0.5, the product is predicted to be designed for sensitive skin.

Wolfcare:

z = \begin{bmatrix} -1 \\ \frac{1}{5} \\ -\frac{3}{5} \\ 0 \end{bmatrix}^T \cdot \begin{bmatrix} 1 \\ 15 \\ 20 \\ 4.5 \end{bmatrix} 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 = \begin{bmatrix} -1 \\ \frac{1}{5} \\ -\frac{3}{5} \\ 0 \end{bmatrix}^T \cdot \begin{bmatrix} 1 \\ 25 \\ 5 \\ 4.9 \end{bmatrix} 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 = \begin{bmatrix} -1 \\ \frac{1}{5} \\ -\frac{3}{5} \\ 0 \end{bmatrix}^T \cdot \begin{bmatrix} 1 \\ 50 \\ 15 \\ 3.6 \end{bmatrix} 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 = \begin{bmatrix} -1 \\ \frac{1}{5} \\ -\frac{3}{5} \\ 0 \end{bmatrix}^T \cdot \begin{bmatrix} 1 \\ 0 \\ 1 \\ 5.0 \end{bmatrix} 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

Difficulty: ⭐️⭐️

The average score on this problem was 85%.

Answer: \frac{35}{40}

The precision of a binary 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}

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 + 40 + 10 + ??? = 115

Solving for the missing value (???):

115 - (5 + 40 + 10) = 60

Thus, the missing value for threshold B is:

60

Difficulty: ⭐️

The average score on this problem was 90%.

Answer: A > C > B

Thresholds can be arranged based on their strictness. The higher a model’s threshold is, the fewer positive predictions and the more negative predictions it has. Model A made 40 negative predictions, Model B made 5 negative predictions and Model C made 10 negative predictions. Sorting these in descending order with more negative predictions implying a stricter threshold we get -

A > C > B

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: False positive

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.

Difficulty: ⭐️

The average score on this problem was 91%.

Answer: -2x_1 + 6x_2

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, calculate the partial derivative: Partial derivative with respect to x_1 is provided to us in the question: 2x_1 - 2x_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}

Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer: \frac{5}{4}

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

From the question we know that: \vec{x}^{(0)} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad \vec{x}^{(1)} = \begin{bmatrix} 1 \\ -4 \end{bmatrix}

We can use \vec{x}^{(0)} to first calculate what is the gradient of Q at \vec{x}^{(0)} \nabla Q(\vec{x}^{(0)}) = \begin{bmatrix} 2(1 - 1) \\ -2(1) + 6(1) \end{bmatrix} = \begin{bmatrix} 0 \\ 4 \end{bmatrix}

We can now substitute this value for the gradient of Q at \vec{x}^{(0)} into the update rule. This gives us - \begin{bmatrix} 1 \\ -4 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} - \alpha \begin{bmatrix} 0 \\ 4 \end{bmatrix}

Next, we can solve each separate component to find the value of \alpha - 1 = 1 - \alpha(0) \quad \text{(satisfied for any $\alpha$)}, -4 = 1 - \alpha(4)

Solving for \alpha gives us : -4 = 1 - 4\alpha \implies 4\alpha = 5 \implies \alpha = \frac{5}{4}

Difficulty: ⭐️⭐️

The average score on this problem was 86%.