Skip to content

CMA Foundation · Fundamentals of Business Mathematics and Statistics · Statistical Representation of Data

A histogram is drawn for monthly electricity bills of households in Pune. The class 200-300 has frequency 30 and a bar of height 30 units. The class 300-500 (width twice as large) has frequency 50. What should be the height of the bar for the class 300-500, if bar heights are frequency densities relative to a width of 100?

The bar height should be 25 units. With unequal class widths, the height must be the frequency density, which is frequency divided by width. The width of 300-500 is two units of 100, so 50 divided by 2 gives 25, keeping area proportional to frequency.

  1. A50 units
  2. B25 unitsCorrect
  3. C100 units
  4. D20 units

Explanation

For unequal class widths, the height is frequency divided by width in units of 100. The class 300-500 has width 200 = 2 units, so height = 50/2 = 25. Plotting 50 directly ignores the unequal width and distorts the area.

Did you get it right without looking?

One question tells you little. A timed set on Statistical Representation of Data shows your real accuracy, how long you take and where you lose marks.

Practise Statistical Representation of DataFree diagnostic mock

More Statistical Representation of Data questions