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Additional rain gauges for 5% error, seven stations

Short answer

Mean is 130.43 cm, sample standard deviation 22.545 cm, Cv=17.29%. Optimum number N=(17.29/5)²≈12. With 7 existing stations, 5 additional are required.

A catchment has seven raingauge stations. In a year the annual rainfall in cm recorded by the gauges are as follows: 130, 142.1, 118.2, 108.5, 165.2, 102.1, 146.9. For a 5% error in the estimation of the mean rainfall, calculate the minimum number of additional stations required to be established in the catchment.

The optimum number of rain gauges for a permissible error ε\varepsilon in the mean rainfall 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): 130, 142.1, 118.2, 108.5, 165.2, 102.1, 146.9; existing stations n=7n = 7.

  • Mean: Pˉ=913/7=130.43\bar P = 913/7 = 130.43 cm
  • Standard deviation (sample): σn−1=22.545\sigma_{n-1} = 22.545 cm
  • Coefficient of variation: Cv=100×22.545130.43=17.29%C_v = \dfrac{100 \times 22.545}{130.43} = 17.29\%

For ε=5%\varepsilon = 5\%:

N=(17.295)2=11.95≈12N = \left(\frac{17.29}{5}\right)^2 = 11.95 \approx 12

Additional stations =12−7=5= 12 - 7 = 5.

Answer: Total 12 stations are needed, so at least 5 additional rain gauges must be established.

✓ Answer checked by Claude Sonnet 5.5

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