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Optimum rain gauges for 10% error, six stations

Short answer

Mean rainfall = 237.77 cm, standard deviation = 23.38 cm, Cv = 9.83%. Optimum number N = (9.83/10)² = 0.97 ≈ 1; the existing 6 stations are adequate for 10% error.

A catchment has six rain gauge stations. In a year, the annual rainfall in cm recorded by the gauges are as follows: 240, 252.2, 228.3, 218.6, 275.3, 212.2. For a 10% error in the estimation of the mean rainfall, calculate the optimum number of stations in the catchment.

The optimum number of gauges for a permissible error ε\varepsilon is

N=(Cvε)2,Cv=100 σn−1PˉN = \left(\frac{C_v}{\varepsilon}\right)^2, \qquad C_v = \frac{100\,\sigma_{n-1}}{\bar P}

Data (cm): 240, 252.2, 228.3, 218.6, 275.3, 212.2; n=6n = 6.

  • Mean: Pˉ=1426.6/6=237.77\bar P = 1426.6/6 = 237.77 cm
  • Standard deviation: σn−1=23.38\sigma_{n-1} = 23.38 cm
  • Cv=100×23.38237.77=9.83%C_v = \dfrac{100\times23.38}{237.77} = 9.83\%

For ε=10%\varepsilon = 10\%:

N=(9.8310)2=0.97≈1N = \left(\frac{9.83}{10}\right)^2 = 0.97 \approx 1

Answer: The optimum number is about 1 gauge. The existing 6 stations are more than enough for a 10% error, so no additional gauge is needed.

✓ Answer checked by Claude Sonnet 5.5

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