4.3 Histograms & Frequency Distributions in Civil Service Reporting

Histograms vs. Bar Charts: The Continuous Data Distinction

Histograms group a numerical variable into intervals. The variable can be continuous, such as measured duration, or discrete numerical data grouped for display, such as ages recorded in whole years. Bar charts instead compare categories; a year-by-year bar chart can also display time observations.

• Spacing: Adjacent histogram bins follow the numerical scale. A zero-frequency interval has zero height; do not insert arbitrary gaps between nonempty adjacent bins.
• Equal widths: With frequency on the vertical axis, heights show counts and areas are proportional to counts.
• Frequency-density scale: Height is frequency divided by class width, so area equals frequency in the axis units. This gives a fair area comparison when widths differ.
• Relative-density scale: If height is frequency divided by total observations and width, areas show proportions and sum to one. Always read the axis label before interpreting height or area.

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Class Intervals & Equal vs. Unequal Bin Widths

Continuous variables are grouped into class intervals (or bins). Use non-overlapping intervals with explicit boundary conventions. Let $L$ and $U$ denote the lower and upper boundaries, not just the printed smallest and largest integer labels. The class width ($w$) is the difference between these boundaries: $w = U - L$

Equal Class Widths

When all intervals share an identical width (e.g., age groups $20\text{--}29$, $30\text{--}39$, $40\text{--}49$, using boundaries 19.5–29.5, 29.5–39.5 and 39.5–49.5 for whole-year records, each of width 10 years), the vertical axis can display raw frequency directly. Because widths are identical, a bar with twice the frequency has twice the height and twice the area.

Unequal Class Widths: The Distortion Trap

A practice display may use unequal intervals. For example, administrative earnings data might group salaries into €10,000 bands at lower pay scales, but widen to €20,000 or €30,000 bands at senior management levels.

If raw frequency were plotted on the vertical axis for unequal intervals:

• A wide interval covering €60,000 to €90,000 ($w = 30$) might contain 1,500 employees.
• A narrower interval covering €45,000 to €60,000 ($w = 15$) might contain 1,200 employees.
• Plotting raw frequency would make the €60k–€90k bar taller (1,500 vs. 1,200), falsely suggesting higher staff concentration. In reality, employees are far more densely clustered in the narrower €45k–€60k band (80 staff per €1k width vs. 50 staff per €1k width).

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The Frequency Density Principle

To eliminate visual and mathematical distortion when class widths vary, a density histogram scales the vertical axis using frequency density:

$\text{Frequency Density} = \frac{\text{Class Frequency}}{\text{Class Width}} = \frac{f}{w}$

Rearranging this formula confirms that the physical area of the bar equals the frequency: $\text{Frequency (Area)} = \text{Frequency Density} \times \text{Class Width} = \left(\frac{f}{w}\right) \times w = f$

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Identifying the Modal Interval & Estimating the Median Class

Two fundamental measures of central tendency tested in public service frequency distributions are the modal interval and the median class.

1. The Modal Interval (Mode)

For a density histogram, call the interval with the highest frequency density the modal interval. This describes the tallest density bar, rather than necessarily the interval containing the most observations. Grouped counts do not establish the exact mode within a bin:

• For equal class intervals, it is simply the interval with the highest raw frequency.
• For unequal class intervals, it is the interval with the highest frequency density, NOT necessarily the highest raw frequency. A wide bin with a large raw frequency may have a low density, disqualifying it as the modal group.

2. The Median Class & Continuous Linear Interpolation

The median is a 50th-percentile measure. For raw data with an even number of observations, the sample median is the average of the two middle ordered values. Grouped counts generally cannot recover those exact values.

To estimate a grouped median, use the cumulative-frequency level $N/2$. Locate where that level falls, then assume observations are spread uniformly within that bin. Linear interpolation gives:

$\text{Estimated Median}=L+\left(\frac{N/2-cf_{\text{prev}}}{f_m}\right)\times w$

This is an estimate under a within-class assumption, not a precise raw-data median. If the cumulative halfway level is exactly a class boundary, interpolation reaches that boundary; do not infer that both central raw observations belong to the same class.

Where:

• $L$ = Lower boundary of the median class interval
• $N$ = Total number of observations across all classes
• $cf_{\text{prev}}$ = Cumulative frequency of the class immediately preceding the median class
• $f_m$ = Frequency of the median class itself
• $w$ = Class width of the median class interval

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Cumulative Frequency Distributions & Ogives

A cumulative frequency distribution records the running total of frequencies up to the upper boundary of each class interval. When plotted on a graph, this gives a non-decreasing cumulative graph known as an ogive. Its shape need not be S-shaped. Between known boundaries, a straight segment implies uniform within-class interpolation.

Reading Quantiles from an Ogive

• Median ($Q_2$): Read across horizontally from the $50\%$ cumulative frequency level ($0.50 \times N$) to intersect the curve, then drop vertically to read the median value on the horizontal axis.
• Lower Quartile ($Q_1$): Read across from the $25\%$ level ($0.25 \times N$).
• Upper Quartile ($Q_3$): Read across from the $75\%$ level ($0.75 \times N$).
• Interquartile Range ($IQR$): Calculated as $IQR = Q_3 - Q_1$, measuring the dispersion of the central $50\%$ of the distribution, less sensitive than the mean to the magnitudes of extreme tail values. Quantile estimates from grouped data depend on the grouping and interpolation method.

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Case Study: Civil Service Passport Application Processing Times

The following hypothetical dataset contains 1,200 processing durations, recorded to the nearest whole working day. For histogram interpolation, use boundaries 0.5–5.5, 5.5–10.5, 10.5–15.5, 15.5–25.5, 25.5–40.5 and 40.5–65.5 days. Thus the printed 1–5 group has width 5, rather than 5 - 1. These are exercise data, not a claim about the Passport Service’s actual performance.

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Worked Walkthrough: Modal Group, Median & SLA Compliance

1. Determining the Modal Class

Comparing frequency densities across all classes:

• 1–5 days: 36.0 cases/day
• 6–10 days: 68.0 cases/day
• 11–15 days: 84.0 cases/day
• 16–25 days: 16.0 cases/day
• 26–40 days: 5.0 cases/day
• 41–65 days: 1.0 cases/day

The modal class is 11 to 15 days, as it possesses the highest frequency density (84.0 cases per day).

2. Calculating Estimated Median via Linear Interpolation

• Total applications $N = 1{,}200$. The grouped interpolation level is $N/2=600$. The exact sample median would average ordered observations 600 and 601.
• Both central observations lie within the 11 to 15 days class (since cumulative frequency reaches 520 at 10 days and 940 at 15 days).
• Variables: $L = 10.5$ (continuous lower boundary), $cf_{\text{prev}} = 520$, $f_m = 420$, $w = 5$. $\text{Estimated Median} = 10.5 + \left(\frac{600 - 520}{420}\right) \times 5 = 10.5 + \left(\frac{80}{420}\right) \times 5 \approx 10.5 + 0.95238 \approx 11.45\text{ working days}$

3. Evaluating Service Level Agreement (SLA) Compliance

Scenario: A hypothetical service target requires that at least $75.0\%$ of complex renewals must be resolved within 15 working days.

• Cumulative applications resolved within 15 days $= 940$.
• Percentage achieved: $\left(\frac{940}{1{,}200}\right) \times 100 \approx 78.33\%$.
• Conclusion: The Department exceeded the stipulated target by $3.33$ percentage points ($78.33\% - 75.0\%$).