Someone asks: “What’s a typical Cartwheel order?” You compute the mean, 228 dollars, and say so. Then the marketing team designs a free-shipping threshold around 228 dollars, and discovers that two thirds of orders never reach it. The mean was correct and the answer was wrong, because order values are not spread evenly around their average. Describing data honestly means describing its shape, not only its centre.
\nimport pandas as pd\n\nitems = pd.read_csv(\"order_items.csv\")\norders = pd.read_csv(\"orders.csv\", parse_dates=[\"order_date\"])\nitems[\"revenue\"] = items[\"quantity\"] * items[\"unit_price\"] * (1 - items[\"discount\"])\norders = orders.merge(items.groupby(\"order_id\", as_index=False)[\"revenue\"].sum(), on=\"order_id\", validate=\"one_to_one\")\norders = orders[orders[\"status\"] != \"cancelled\"]\nvalue = orders[\"revenue\"]\n\nprint(value.describe(percentiles=[0.1, 0.25, 0.5, 0.75, 0.9, 0.99]).round(2))\nprint(\"skew:\", round(value.skew(), 2))Read it from the middle out. The median (50%) is 147 dollars: half of all orders are smaller. The quartiles say the middle half of orders falls between 68 and 303 dollars. The 90th percentile is 505 and the 99th is 1,168, while the maximum is 2,117. The mean, 228, sits far above the median because of that long right tail, and the positive skew of 2.43 confirms it. For a typical value in skewed data, quote the median; for totals and forecasts, the mean is the right number, because mean × count = total.
\nstd, the standard deviation, is 241 dollars, larger than the median itself. That is another sign of a spread-out, skewed distribution; in that situation quantiles describe the spread far better than “mean plus or minus one standard deviation”.
value_counts with bins counts how many values fall in each range, which is a histogram in text form:
import pandas as pd\n\nitems = pd.read_csv(\"order_items.csv\")\norders = pd.read_csv(\"orders.csv\", parse_dates=[\"order_date\"])\nitems[\"revenue\"] = items[\"quantity\"] * items[\"unit_price\"] * (1 - items[\"discount\"])\norders = orders.merge(items.groupby(\"order_id\", as_index=False)[\"revenue\"].sum(), on=\"order_id\", validate=\"one_to_one\")\norders = orders[orders[\"status\"] != \"cancelled\"]\n\nprint(orders[\"revenue\"].value_counts(bins=[0, 50, 100, 200, 400, 800, 3000]).sort_index())Most orders sit between 50 and 400 dollars, and a thinning tail runs past 800. Notice the bands get wider as they go right; equal-width bands of 100 dollars would squash nearly everything into the first few bars.
\nA common rule of thumb flags values more than 1.5 interquartile ranges (IQR, the distance between the quartiles) beyond the quartiles:
\nimport pandas as pd\n\nitems = pd.read_csv(\"order_items.csv\")\norders = pd.read_csv(\"orders.csv\", parse_dates=[\"order_date\"])\nitems[\"revenue\"] = items[\"quantity\"] * items[\"unit_price\"] * (1 - items[\"discount\"])\norders = orders.merge(items.groupby(\"order_id\", as_index=False)[\"revenue\"].sum(), on=\"order_id\", validate=\"one_to_one\")\norders = orders[orders[\"status\"] != \"cancelled\"]\nvalue = orders[\"revenue\"]\n\nq1, q3 = value.quantile([0.25, 0.75])\nupper = q3 + 1.5 * (q3 - q1)\nbig = value > upper\nprint(f\"fence {upper:.2f}: {big.sum()} orders, {value[big].sum() / value.sum():.1%} of revenue\")125 orders are above 656.62 dollars, and together they bring in 25.9% of revenue. These are not errors to clean away: they are multi-item orders of desks, chairs and dumbbells, real and important. An outlier rule tells you where to look. Data entry errors, such as a total with a missing decimal point, get fixed; genuine extremes stay, and you check whether your conclusion survives with and without them.
\n\nDistributions can be compared across periods too, and this one carries a clue for the case study. If Q4 2025 revenue jumped because customers bought bigger baskets, the typical order should have grown:
\nimport pandas as pd\n\nitems = pd.read_csv(\"order_items.csv\")\norders = pd.read_csv(\"orders.csv\", parse_dates=[\"order_date\"])\nitems[\"revenue\"] = items[\"quantity\"] * items[\"unit_price\"] * (1 - items[\"discount\"])\norders = orders.merge(items.groupby(\"order_id\", as_index=False)[\"revenue\"].sum(), on=\"order_id\", validate=\"one_to_one\")\norders = orders[orders[\"status\"] != \"cancelled\"]\n\nquarter = orders[\"order_date\"].dt.to_period(\"Q\")\nprint(orders.groupby(quarter)[\"revenue\"].agg([\"count\", \"median\"]).round(1).tail(4))It did not. The median order in Q4 2025 was 151.70 dollars, slightly below Q3’s 159.20, while the number of orders doubled. Whatever happened in Q4, it was about how many orders arrived, not how big they were. One table of medians has already ruled out a popular explanation.
\ncorr() measures how strongly two numeric columns move together, from -1 (perfectly opposite) through 0 (no linear relation) to 1 (perfectly together). churn.csv has one row per customer and a churned flag:
import pandas as pd\n\nchurn = pd.read_csv(\"churn.csv\")\ncols = [\"tenure_days\", \"orders\", \"days_since_last_order\", \"support_tickets\", \"churned\"]\nprint(churn[cols].corr().round(2)[\"churned\"])\nprint(churn.groupby(\"churned\")[\"days_since_last_order\"].median())Customers who churned had gone far longer since their last order: a median of 214 days against 46, and a correlation of 0.45. That makes the gap a good warning signal. It does not show that the gap causes churn; a customer who has already lost interest stops ordering and churns, and both measurements reflect that one fact. Treat correlations as leads to investigate, and keep the word “causes” for experiments.
\nIn the exercise you describe Cartwheel’s order values yourself. Next you turn numbers like these into charts.
\n","toc":[{"id":"sec-1","text":"Mean, median and quantiles","level":2},{"id":"sec-2","text":"The shape, without a chart","level":2},{"id":"sec-3","text":"Outliers: flag, then decide","level":2},{"id":"sec-4","text":"The same lens over time","level":3},{"id":"sec-5","text":"Correlation: moving together is not causing","level":2}],"takeaways":["In skewed data such as order values, the mean sits well above the median; report the median (and a few quantiles) when you describe a typical value.","quantile([0.25, 0.75]) gives the quartiles; values beyond 1.5 times the interquartile range from them are worth a look, not automatic deletion.","value_counts(bins=[...]) is a quick text histogram that shows the shape of a distribution without a chart.","Compare groups with both their mean and median and their sizes, so one big order or one tiny group does not drive the conclusion.","corr() measures how two columns move together; a strong correlation is a lead to investigate, never proof that one causes the other."],"further":[{"title":"Descriptive statistics in pandas","url":"https://pandas.pydata.org/docs/user_guide/basics.html#descriptive-statistics"},{"title":"pandas.Series.quantile","url":"https://pandas.pydata.org/docs/reference/api/pandas.Series.quantile.html"},{"title":"pandas.DataFrame.corr","url":"https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.corr.html"}],"quiz":[{"q":"Cartwheel’s order values have a mean of 228 dollars and a median of 147. What does that tell you about the distribution?
","options":[{"text":"It is skewed to the right; a minority of large orders pulls the mean above the typical order.","why":"Correct. Half of all orders are below 147 dollars, and a long tail of big orders lifts the mean by 81 dollars."},{"text":"The data contains errors, because mean and median should be equal.","why":"They are equal only for symmetric distributions. Money, durations and counts are usually skewed, with no error involved."},{"text":"Most orders are close to 228 dollars.","why":"The median says half the orders are below 147, so 228 is not where most orders sit."},{"text":"There are more small orders than large ones, so the mean is too low.","why":"Many small orders and a few huge ones push the mean up, not down."}],"answer":0},{"q":"You find 125 orders above the 1.5 × IQR upper fence. What should you do with them?
","options":[{"text":"Delete them; outliers distort averages.","why":"These are real orders worth 26% of revenue. Deleting them would misrepresent the business."},{"text":"Replace them with the median.","why":"Replacing real sales with a typical value is inventing data, and it would erase most of that quarter of revenue."},{"text":"Inspect them, check they are genuine, and report results with and without them if they change the conclusion.","why":"Correct. The fence flags values for review. Genuine large orders are part of the story, often the most interesting part."}],"answer":2},{"q":"In churn.csv, days_since_last_order has a correlation of 0.45 with churned. Which statement is justified?
value holds the revenue of every non-cancelled order. Set median_value and p90_value (the 90th percentile), each rounded to 2 decimals. Set below_mean_share to the share of orders whose value is below the mean, rounded to 3 decimals. Compute upper_fence as Q3 + 1.5 × (Q3 − Q1) and n_outliers as the number of orders above it. Finally build by_channel, a DataFrame indexed by channel with the columns count, mean and median of order value, in that order.
value.median() and value.quantile(0.9); wrap each in round(..., 2).","A share is the mean of a boolean Series: (value < value.mean()).mean(). value.quantile([0.25, 0.75]) returns both quartiles, which you can unpack into two names.","orders.groupby("channel")["revenue"].agg(["count", "mean", "median"]) gives the three columns in that order."],"explanation":"The median order is 147.26 dollars, and 66.3% of orders are below the 227.77 mean: the clearest possible demonstration that the mean is not the typical order in skewed data. The fence at 656.62 flags 125 large orders, which you would inspect rather than delete. The channel table puts sizes next to both averages, so you can see that the means and the medians do not even rank the channels the same way, and that the marketplace group is the smallest at 208 orders.
\n","checkLabels":{"median":"median_value is 147.26","p90":"p90_value is the 90th percentile","below-mean":"66.3% of orders are below the mean","fence":"The fence is 656.62 with 125 orders above it","by-channel":"by_channel has count, mean and median per channel"}},"words":684,"readMinutes":3}