CCNA Exploratory Data Analysis Questions

For high cardinality categorical features, grouping rare categories into an 'Other' category reduces cardinality and allows meaningful visualizations like box plots. Option A is wrong because removing the feature loses information. Option B is wrong because one-hot encoding creates too many columns and is not suitable for visualization.

Option D is wrong because visualizing millions of categories is not feasible. Option E is wrong because feature hashing is for modeling, not EDA visualization.

Option A (box plot) shows distribution of a numeric feature across categories. Option C (violin plot) combines box plot and density. Option B is wrong because bar charts are for categorical vs categorical.

Option D is wrong because histograms show distribution of a single variable. Option E is wrong because scatter plots are for two numeric variables.

Option B is correct because parallel coordinates plot can show feature patterns for minority vs majority class in high dimensions. Option A is wrong because scatter plot matrices become cluttered with many features. Option C is wrong because histograms are univariate and do not show interaction.

Option D is wrong because box plots are univariate.

Option C is correct because AWS Glue is a serverless ETL service. Option A is wrong because SageMaker is for ML modeling. Option B is wrong because Athena is for querying.

Option D is wrong because Lambda is for serverless functions, but not a full ETL service.

The Interquartile Range (IQR) method does not assume normality and uses percentiles. Option A (Z-score) assumes normality. Option C (mean ± 3σ) also assumes normality.

Option D (MAD) is robust but Z-score based; IQR is more common for non-normal.

Option A is correct because the error clearly states that a row has more fields than the header. Option B is wrong because the error is about field count mismatch, not encoding. Option C is wrong because the error mentions row number, but the issue is field count.

Option D is wrong because the header is present and read correctly.

Option A is correct because PCA is designed to find principal components that maximize variance. Option B is wrong because t-SNE is for visualization but not variance-based; it focuses on preserving local structure. Option C is wrong because LDA is supervised and requires labels.

Option D is wrong because K-means is clustering, not dimensionality reduction.

Amazon SageMaker Data Wrangler provides built-in transformations including correlation analysis and dimensionality reduction. Option A is wrong because QuickSight is for visualization, not feature reduction. Option B is wrong because Glue is for ETL but lacks automatic dimensionality reduction.

Option D is wrong because Athena is a query service.

Option C is correct because SageMaker Data Wrangler is a visual data preparation tool integrated into SageMaker Studio that provides summary statistics, histograms, and correlation matrices without code. Option A is wrong because SageMaker Autopilot automates model building, not EDA. Option B is wrong because SageMaker Clarify is for bias detection and explainability.

Option D is wrong because SageMaker Ground Truth is for labeling.

Option B is correct because the median should be computed on the training set only to avoid data leakage, then applied to both training and test sets. Option A is wrong because imputing with zero may not be appropriate. Option C is wrong because computing median on the entire dataset and then splitting causes data leakage.

Option D is wrong because dropping rows with missing values may discard useful data and is not imputation.

Scatter plots for numerical vs continuous target and box plots for categorical vs continuous target are standard. Option C (confusion matrix) is for classification. Option D (histogram of target) is univariate.

Option E (silhouette score) is for clustering.

Options B and C are correct. Option A is wrong because MICE is multivariate imputation, not necessarily safest. Option D is wrong because listwise deletion can introduce bias.

Option E is wrong because mean imputation reduces variance.

Option D is correct because t-SNE is designed for non-linear dimensionality reduction and visualization. Option A is wrong because PCA is linear. Option B is wrong because LDA is supervised.

Option C is wrong because ICA separates independent components, not for visualization.

Option C is correct because zero variance means the feature is constant and provides no information for modeling; it should be removed. Option A is wrong because scaling does not change constant values. Option B is wrong because imputation is for missing values, not constant.

Option D is wrong because interaction with a constant feature remains constant.

Option C is correct because SMOTE generates synthetic samples for the minority class, which is effective for imbalanced datasets. Option A is wrong because accuracy is not a good metric for imbalanced data. Option B is wrong because under-sampling discards majority class data.

Option D is wrong because using raw data without handling imbalance typically leads to poor minority class performance.

Option A is correct because correlation indicates a relationship, but does not imply causation. Option B is wrong because correlation does not imply causation. Option C is wrong because correlation does not provide a prediction model.

Option D is wrong because correlation does not guarantee linearity; it could be non-linear.

A histogram is the standard tool for showing the distribution of a single continuous variable. Option A is wrong because scatter plots compare two variables. Option B is wrong because bar charts are for categorical data.

Option D is wrong because box plots show summary statistics, not the full distribution shape.

Option A is correct because Random Forest feature importance is a well-known method for ranking features in classification tasks. Option B is wrong because PCA is unsupervised and does not use the target. Option C is wrong because K-means is clustering, not feature selection.

Option D is wrong because t-SNE is for visualization, not feature importance.

Options A, B, and D are correct. AWS Glue DataBrew can profile data, visualize distributions, and detect anomalies. Amazon SageMaker Data Wrangler provides interactive data preparation and visualization.

Amazon Athena can be used to run SQL queries for data quality checks. Option C is wrong because AWS Data Pipeline is for workflow orchestration, not EDA. Option E is wrong because Amazon Kinesis Data Analytics is for streaming data, not batch EDA.

Amazon SageMaker Data Wrangler provides built-in data analysis and visualization capabilities, including correlation analysis, making it suitable for this task. Option A (Amazon QuickSight) is a BI tool for dashboards, not embedded data wrangling. Option C (AWS Glue) is for ETL jobs.

Option D (Amazon Athena) is for querying data in S3.

Option D is correct because target encoding (or mean encoding) replaces each category with the mean of the target, which is suitable for high cardinality in linear models but requires careful validation to avoid overfitting. Option A is wrong because one-hot encoding would create 199 dummy variables, leading to high dimensionality. Option B is wrong because label encoding imposes an arbitrary ordinal relationship.

Option C is wrong because frequency encoding may not capture the relationship with the target.

Option D is correct because median is robust to skewness and preserves the central tendency without affecting the skewness as much as mean. Option A is wrong because mean is sensitive to outliers and would reduce skewness. Option B is wrong because mode is for categorical data.

Option C is wrong because dropping rows reduces sample size.

Options A and D are correct. A: A box plot grouped by target class shows distribution differences. D: A histogram with overlaid target classes shows how the feature distribution differs.

Option B is incorrect because a scatter plot is for two numerical variables. Option C is incorrect because a bar chart is for categorical features. Option E is incorrect because a pie chart is for proportions, not relationships.

The CLI command filters objects larger than 1000 bytes, but the scientist wants all files in the prefix. The Athena table definition should point to the entire prefix without size filtering. Option A is wrong because the command was run locally, not affecting Athena.

Option B is wrong because adding a WHERE clause in Athena only filters after scanning. Option D is wrong because the scientist can still create the table without size restrictions.

Option A is correct because a line plot over time clearly shows weekly patterns. Option B is wrong because histograms show distribution of values, not time patterns. Option C is wrong because autocorrelation measures correlation with lagged values, but visual inspection is more direct for confirming seasonality.

Option D is wrong because scatter plots show relationships between two variables.

Option A is correct because the table is partitioned by year and month. The first query filters on a partition column (year), so Athena prunes partitions and scans only the relevant data. The second query filters on a non-partition column (amount), so Athena scans all partitions.

Option B is wrong because the data format is text (CSV), not Parquet. Option C is wrong because compression is not mentioned. Option D is wrong because the query is not partitioned correctly; the second query does not use partition columns.

Option C is correct because pandas reads numeric columns as int64 or float64 by default, which uses 8 bytes per value. For 500,000 rows, even a single numeric column can consume significant memory. Option A is wrong because pandas does not automatically index columns.

Option B is wrong because memory usage is typically not due to loading all rows; pandas loads all rows. Option D is wrong because CSV files are plain text and do not have compression overhead.

Option D is correct because aggregating the categories into groups (e.g., top 10 levels) and then using a bar plot or box plot makes the visualization interpretable. Option A is wrong because scatter plots are not suitable for categorical data. Option B is wrong because a heatmap of pairwise chi-square tests is for categorical-categorical relationships, not categorical-continuous.

Option C is wrong because a correlation matrix with Pearson correlation is for numerical data.

Using domain knowledge to define valid age range (e.g., 0-120) and filtering out or capping outliers is the most appropriate approach. Option B is wrong because removing the entire column loses information. Option C is wrong because imputing with mean distorts the distribution if outliers are extreme.

Option D is wrong because standard scaling does not handle outliers; it will still be affected.

Option B is correct because the repeated NullPointerException suggests a recurring issue that needs immediate attention. Option A is wrong because connection timeout appears only once. Option C is wrong because there are multiple timestamps.

Option D is wrong because WARN level appears only once.

Option B is correct because Athena has a 100 MB per file limit for CSV queries without partitioning. Option A is wrong because gzip compression is supported. Option C is wrong because 1 GB is large but Athena can handle larger with proper partitioning.

Option D is wrong because CSV is supported.

Option C is correct because AWS Glue provides built-in transformations to parse timestamps and extract date/time features. Option A is wrong because Amazon Athena is a query service, not a transformation service. Option B is wrong because Amazon SageMaker Studio is an IDE, not a data transformation service.

Option D is wrong because AWS Data Pipeline is a workflow orchestration service, not a timestamp parsing tool.

Options B and D are correct. Box plots use the IQR to identify outliers as points outside 1.5*IQR from the quartiles. Z-scores identify outliers as points with |z| > 3 (assuming normal distribution).

Option A is wrong because mean and median are measures of central tendency, not outlier detection. Option C is wrong because histograms show distribution shape but do not explicitly identify outliers. Option E is wrong because pairwise scatter plots may show outliers but are not a systematic method.

Option B is correct because a skewness of 2.5 indicates right skew, and a log transformation is commonly used to reduce skewness. Option A is incorrect because standardization does not change distribution shape. Option C is incorrect because Min-Max scaling does not change skewness.

Option D is incorrect because a Box-Cox transformation requires positive data and is a more general solution, but log is simpler and often sufficient; however, Box-Cox is also valid. In the context of this question, log is the most direct answer.

A box plot shows the distribution of the target across categories. Option A is wrong because scatter plots are for two numerical features. Option B is wrong because histogram shows distribution of a single numerical feature.

Option D is wrong because heatmaps typically show correlation between numerical features.

AWS Glue DataBrew provides a missing data report that includes patterns and correlations of missingness, such as heatmaps and bar charts. This helps determine the type of missingness. Option B is wrong because SageMaker Data Wrangler does not have built-in missingness pattern analysis.

Option C is wrong because Athena is for querying, not pattern analysis. Option D is wrong because Glue ETL jobs require custom coding and are less efficient for exploratory analysis.

Using Data Wrangler's built-in visualizations for quick analysis and exporting the flow as a Jupyter notebook for reproducibility are best practices. Option B (QuickSight) is separate. Option C (EMR) is not needed.

Option D (Athena) is for queries, not for building into pipeline.

Option A is correct because feature correlation with target in training vs. test sets may indicate leakage. Option C is correct because time-based split (chronological) can reveal if future data leaks into training. Option D is correct because distribution differences between train and test sets can indicate leakage (e.g., train has future data).

Option B is wrong because clustering is not directly helpful for leakage detection. Option E is wrong because PCA is for dimensionality reduction, not leakage detection.

Options B and D are correct. B: Grouping rare categories into an 'Other' category reduces cardinality while preserving information. D: Using feature hashing can transform high-cardinality categorical features into a fixed-size vector.

Option A is incorrect because one-hot encoding creates many columns, which can be problematic. Option C is incorrect because dropping the feature may lose important information. Option E is incorrect because label encoding implies ordinality, which may not exist.

The Area Under the ROC Curve (AUC-ROC) is robust to class imbalance and measures the trade-off between true positive rate and false positive rate. Accuracy is misleading. F1 score is sensitive to imbalance but less so for threshold selection.

Precision and recall individually are less comprehensive.

Option B is correct. COUNT(col2) counts non-null values in col2: rows 1 and 3 have non-null, so 2. COUNT(*) counts all rows: 4.

COUNT(DISTINCT col1) counts distinct col1 values: A,B,C,D = 4. Option A is wrong because COUNT(col2) is 2, not 3. Option C is wrong because COUNT(*) is 4, not 3.

Option D is wrong because COUNT(DISTINCT col1) is 4, not 3.

Option B is correct because the query counts rows where col1 IS NULL, and the result '5000' is that count. Option A is wrong because the query is not selecting distinct values. Option C is wrong because the query counts NULL rows, not non-NULL.

Option D is wrong because total row count was not queried.

Option B is correct because plotting learning curves (training and validation loss vs. epochs) would show overfitting. Option A is wrong because confusion matrix is for classification. Option C is wrong because box plots show outliers but not overfitting.

Option D is wrong because residual plots are for linear regression assumptions.

Option A is correct because generating summary statistics helps identify missing counts and outliers via min/max. Option D is correct because visualizing distributions with histograms helps spot outliers and skewness. Option B is wrong because a correlation matrix does not directly show missing values or outliers.

Option C is wrong because a confusion matrix is for classification models, not for data exploration. Option E is wrong because a scatterplot matrix shows pairwise relationships, not missing values.

Options A, B, and D are correct. Z-score and IQR are standard statistical methods, and isolation forest is a machine learning algorithm for anomaly detection. Option C is wrong because PCA is for dimensionality reduction, not outlier detection, though it can be used in some contexts but is not common.

Option E is wrong because k-means clustering is for clustering, not specifically for outlier detection.

Option D is correct because Data Wrangler's data quality report provides summary statistics and missing value analysis, and the histogram visualization shows distributions. Scatter plots reveal relationships between variables. Option A is incorrect because Data Wrangler does not include correlation heatmaps directly.

Option B is incorrect because imputation and one-hot encoding are transformations, not EDA steps. Option C is incorrect because handling missing values is part of data preparation, not initial EDA.

Option D is correct because SageMaker needs s3:PutObject, but also needs s3:GetObject for some operations. However, the error is likely due to missing encryption permissions. Option A is wrong because s3:ListBucket is for listing.

Option B is wrong because kms:Decrypt is for reading. Option C is wrong because s3:GetObject is for reading.

Option A (box plot) identifies outliers as points beyond whiskers. Option C (Z-score) flags points with |Z| > 3. Option B is wrong because bar charts are for categorical data.

Option D is wrong because Pearson correlation measures linear relationship. Option E is wrong because heatmaps show correlations.

SageMaker Data Wrangler provides a visual interface to handle missing values. Option B is wrong because writing a custom script is less efficient. Option C is wrong because Glue is external.

Option D is wrong because running a notebook from the local machine is not efficient.

Option A is correct because mean imputation is appropriate for normally distributed data. Option B is wrong because median imputation is better for skewed data. Option C is wrong because mode imputation is for categorical data.

Option D is wrong because dropping rows reduces sample size.

Options B and D are correct. IQR-based outlier detection identifies points beyond 1.5*IQR from quartiles. Z-score method flags points beyond a threshold (e.g., 3) from mean.

Option A is wrong because clustering is multivariate. Option C is wrong because box plots visualize outliers but are not a detection technique per se; they use IQR. Option E is wrong because correlation is bivariate.

Option D is correct because hierarchical clustering on correlations groups correlated features; then one can select a representative from each cluster. Option A is wrong because PCA creates new features but does not remove original ones. Option B is wrong because Lasso performs feature selection but may not handle multicollinearity well.

Option C is wrong because mutual information does not capture pairwise redundancy directly.

Option B is correct because capping at percentiles (e.g., 99th) limits extreme values while keeping the data points. Option A is wrong because removing rows with any outlier may discard useful data. Option C is wrong because log transformation does not fix errors.

Option D is wrong because imputing with mean distorts the distribution.

Option A is correct because target encoding (mean encoding) captures the relationship between the category and the target, and is suitable for high-cardinality features. Option B is wrong because one-hot encoding would create too many dummy variables. Option C is wrong because label encoding implies ordinality which is not present.

Option D is wrong because frequency encoding may not capture price variation well.

Options A, C, and D are correct. Using partition filtering limits data scanned. Using COUNT(DISTINCT) is efficient but still scans; however, the question asks for reducing cost, so partition filtering is key.

Option B is wrong because SELECT * scans all columns. Option E is wrong because ORDER BY without LIMIT requires full scan and sort.

Decomposition separates time series into trend, seasonal, and residual components. Autocorrelation plot (ACF) helps identify seasonality. Moving average smooths to reveal trends.

Linear regression is not typical for seasonal decomposition. Box plot by month can show seasonal patterns but is less common for trend.

Option B is correct because clustering can identify natural groups, which can be treated as separate modeling tasks. Option A is wrong because log transformation works for skewed unimodal distributions. Option C is wrong because scaling does not change distribution shape.

Option D is wrong because removing outliers would not address bimodality.

Option C is correct because the error indicates the column 'age' is not found in the input data, which has columns [id, name, salary]. Option A is wrong because the error is about a missing column, not a mismatch. Option B is wrong because schema evolution would add column, not cause error.

Option D is wrong because there is no indication of data type issue.

Option D is correct because word frequency distribution (e.g., zipfian) helps decide vocabulary cutoff. Option A is wrong because TF-IDF is a transformation, not EDA. Option B is wrong because document length distribution is about length, not vocabulary.

Option C is wrong because bigram analysis is more advanced; basic frequency is first.

A time series line plot with marked intervals (e.g., weekly, monthly) can reveal seasonal patterns. Option A (histogram) shows distribution, not time patterns. Option B (scatter plot of visits vs day) is essentially a line plot but less effective for seasonality.

Option C (correlation matrix) does not show temporal patterns.