# Step-by-Step Solution and Explanation Let's solve each part step by step, with calculations and explanations where needed. --- ## Part A: Identify the Type of Data for Each Variable | **Variable** | **Type of Data** | **Explanation** | |-----------------|-----------------------------------------------------|-----------------| | Customers | Quantitative (Discrete) | Count of customers (whole numbers) | | Day | Qualitative (Categorical) | Days of the week (Monday, Tuesday, etc.) are categories | | Browser | Qualitative (Categorical) | Types of browsers (Chrome, Firefox, etc.) are categories | | Time (min) | Quantitative (Continuous) | Time spent in minutes can take any value (decimals possible) | | Page viewed | Quantitative (Discrete) | Number of pages (whole numbers) | | Amount spend ($)| Quantitative (Continuous) | Amount in dollars (can be decimal values) | --- ## Part B: Most Appropriate Variable for Ungrouped Frequency Distributions **Qualitative or discrete variables** are best for ungrouped frequency distribution. - **Day** is most appropriate because it is qualitative with a limited number of categories (Monday, Tuesday, etc.). - **Why?** Because you can easily count how many transactions happened on each day. --- ## Part C: Constructing an Ungrouped Frequency Distribution for "Day" Let’s say you have the following sample data for transactions by day: | Day | Frequency | |----------|-----------| | Monday | 10 | | Tuesday | 12 | | Wednesday| 9 | | Thursday | 14 | | Friday | 15 | | Saturday | 20 | | Sunday | 8 | ### **Relative Frequency Calculation** Relative frequency = (Frequency for Day) / (Total Transactions) Assume total = 10 + 12 + 9 + 14 + 15 + 20 + 8 = **88** | Day | Frequency | Relative Frequency | Cumulative Frequency | |----------|-----------|-------------------|---------------------| | Monday | 10 | 10/88 = .114 | 10 | | Tuesday | 12 | 12/88 = .136 | 22 | | Wednesday| 9 | 9/88 = .102 | 31 | | Thursday | 14 | 14/88 = .159 | 45 | | Friday | 15 | 15/88 = .171 | 60 | | Saturday | 20 | 20/88 = .227 | 80 | | Sunday | 8 | 8/88 = .091 | 88 | ### **Histogram** - Plot days on the x-axis, frequency on the y-axis. - Each bar represents one day. --- ## Part D: Grouped Frequency, Histogram, Boxplot for "Time (min)" ### **Step 1: Create Intervals ("Bins") for Time** Suppose times range from to 60 minutes. Bins could be: –10, 11–20, ..., 51–60. | Time (min) | Frequency | |------------|-----------| | –10 | 8 | | 11–20 | 16 | | 21–30 | 20 | | 31–40 | 15 | | 41–50 | 9 | | 51–60 | 5 | ### **Step 2: Histogram** - x-axis = time intervals, y-axis = frequency. - Bar heights = number of shoppers in each interval. ### **Step 3: Boxplot** - Shows median, quartiles, outliers. - Calculate Q1, median, Q3, minimum, and maximum from the data. ### **Step 4: Comment on Shape** - If most shoppers spend 21–40 minutes, distribution is slightly right-skewed. - Outliers may indicate some customers spending unusually long or short times. --- ## Part E: Descriptive Statistics for "Time (min)" Suppose you have these values for times (in minutes): | Time (min) | |------------| | 8, 12, 15, 22, 25, 30, 32, 40, 48, 55 | - **Mean** = (Sum of times) / (Number of shoppers) = (8 + 12 + 15 + 22 + 25 + 30 + 32 + 40 + 48 + 55) / 10 = 287 / 10 = **28.7 min** - **Median** = Middle value (or average of two middle values) Ordered: 8, 12, 15, 22, 25, 30, 32, 40, 48, 55 Median = (25 + 30) / 2 = 27.5 min - **Mode** = Value that appears most (none here, so no mode) - **Standard deviation**: Calculate using formula: \[ SD = \sqrt{\frac{\sum (x_i - \text{mean})^2}{n-1}} \] - **Conclusion:** Most shoppers spend ~28 minutes on the site, but there is some variation. --- ## Part F: Grouped Frequency, Histogram, Boxplot for "Amount Spent ($)" Suppose amounts range from $ to $100. Bins could be: $–$20, $21–$40, ..., $81–$100. | Amount ($) | Frequency | |------------|-----------| | –20 | 10 | | 21–40 | 18 | | 41–60 | 25 | | 61–80 | 12 | | 81–100 | 5 | - **Histogram**: Plot intervals vs. frequency. - **Boxplot**: Calculate quartiles, median, etc. **Comment:** If most values are in the $41–$60 range, the distribution is likely centered there. --- ## Part G: Descriptive Statistics for "Amount Spent ($)" Suppose amounts are: | Amount ($) | |------------| | 12, 22, 30, 35, 45, 50, 55, 60, 80, 90 | - **Mean** = (12 + 22 + 30 + 35 + 45 + 50 + 55 + 60 + 80 + 90) / 10 = 479 / 10 = **47.9** - **Median** = (45 + 50) / 2 = **47.5** - **Mode** = None - **Standard deviation**: Calculate as above. **Conclusion:** The average customer spends about $48 per transaction. The distribution may be right-skewed if there are some high spenders. --- # Summary Table | Part | Variable | Type/Analysis | Example Calculation/Comment | |------|--------------|-------------------------|--------------------------------------------| | A | All | Data type | See table above | | B | Day | Ungrouped freq. dist. | Easiest to categorize/count | | C | Day | Freq. table & histogram | See above, with sample numbers | | D | Time | Grouped, boxplot, hist. | Use intervals, shape comments | | E | Time | Descriptive stats | Mean ~28.7 min, median ~27.5 min | | F | Amount | Grouped, boxplot, hist. | Use $ bins, comment on center/shape | | G | Amount | Descriptive stats | Mean ~$48, median ~$47.5 | --- **If you have real data, substitute with your actual numbers for calculations!**