13.4 Median: The Middle of the Road

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Key Formulas

Solved Examples

Step 1

  1. Data Conversion: The data is given as 'Less than' cumulative frequency. We must convert it to standard class intervals to define frequency ff.
  • Below 140: f=4f=4
  • 140-145: 11−4=711 - 4 = 7
  • 145-150: 29−11=1829 - 11 = 18
  • 150-155: 40−29=1140 - 29 = 11
  • 155-160: 46−40=646 - 40 = 6
  • 160-165: 51−46=551 - 46 = 5

Step 2

  1. Identify Median Class: Total N=51N = 51. Target Position = N/2=25.5N/2 = 25.5. Look at the cumulative frequency (given in question). The value just greater than 25.5 is 29. This corresponds to the class interval 145 – 150.

Step 3

  1. Identify Parameters:
  • l=145l = 145 (Lower limit)
  • f=18f = 18 (Frequency of this class)
  • cf=11cf = 11 (Cumulative frequency of the previous class, i.e., <145)
  • h=5h = 5 (Class width)

Step 4

  1. Calculate: Median =145+(25.5−1118)×5= 145 + \left( \frac{25.5 - 11}{18} \right) \times 5 =145+(14.518)×5= 145 + \left( \frac{14.5}{18} \right) \times 5 =145+72.518= 145 + \frac{72.5}{18} =145+4.03= 145 + 4.03

Step 5

  1. Final Result: 149.03 cm. Interpretation: 50% of the girls are shorter than 149.03 cm, and 50% are taller.
Final Answer

149.03 cm

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