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Calculate the mean rainfall (X).

Mathematics

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4 parts
a

∑(X−Xˉ)2=1203\sum (X - \bar{X})^2 = 1203

b(i)

10.01 cm\text{10.01 cm}

b(iii)

74.17%\text{74.17\%}

d(i)

36 cm\text{36 cm}

Step-by-step solution

6 steps

Part a· 2 steps

  1. Calculate the mean rainfall (Xˉ\bar{X}).

    The sum of rainfall (∑X\sum X) is:

    ∑X=0+1+9+19+27+15+9+5+18+36+11+12=162\sum X = 0 + 1 + 9 + 19 + 27 + 15 + 9 + 5 + 18 + 36 + 11 + 12 = 162

    The number of months (nn) is 12.
    The mean rainfall (Xˉ\bar{X}) is:

    Xˉ=∑Xn=16212=13.5 cm\bar{X} = \frac{\sum X}{n} = \frac{162}{12} = 13.5 \text{ cm}

  2. Complete the table by calculating (X−Xˉ)(X - \bar{X}) and (X−Xˉ)2(X - \bar{X})^2 for each month.

    MonthRainfall (cm) (XX)Mean Rainfall (Xˉ\bar{X})(X−Xˉ)(X - \bar{X})(X−Xˉ)2(X - \bar{X})^2
    January013.5-13.5182.25
    February113.5-12.5156.25
    March913.5-4.520.25
    April1913.55.530.25
    May2713.513.5182.25
    June1513.51.52.25
    July913.5-4.520.25
    August513.5-8.572.25
    September1813.54.520.25
    October3613.522.5506.25
    November1113.5-2.56.25
    December1213.5-1.52.25
    Total16201203

    The sum of (X−Xˉ)2(X - \bar{X})^2 is ∑(X−Xˉ)2=1203\boxed{\sum (X - \bar{X})^2 = 1203}.

Part b(i)

10.01 cm\text{10.01 cm}

  1. Calculate the Standard Deviation (σ\sigma).

    Using the formula σ=∑(X−Xˉ)2n\sigma = \sqrt{\frac{\sum (X-\bar{X})^2}{n}}:

    σ=120312\sigma = \sqrt{\frac{1203}{12}} σ=100.25\sigma = \sqrt{100.25} σ≈10.01249\sigma \approx 10.01249

    Rounding to two decimal places:
    The standard deviation is 10.01 cm\boxed{\text{10.01 cm}}.

    Significance: The standard deviation of 10.01 cm indicates the average amount of variation or dispersion of the monthly rainfall data around the mean rainfall of 13.5 cm. A higher value would imply greater variability in rainfall, while a lower value would suggest rainfall amounts are more consistent from month to month.

Part b(ii)

Two advantages of using Standard Deviation are:

  • It considers every observation in the dataset, making it a comprehensive measure of dispersion.
  • It is expressed in the same units as the original data, which makes it easy to interpret and compare.

Part b(iii)

74.17%\text{74.17\%}

  1. Calculate the Coefficient of Variation (CV).

    Using the formula CV=Standard DeviationMean×100CV = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100:

    CV=10.0124913.5×100CV = \frac{10.01249}{13.5} \times 100 CV≈0.741666×100CV \approx 0.741666 \times 100 CV≈74.17%CV \approx 74.17\%

    Rounding to two decimal places:
    The Coefficient of Variation is \boxed{\text{74.17%}}.

Part c(i)

To represent the information on monthly rainfall using simple bar graphs:

  • Draw a horizontal axis labeled "Month" and a vertical axis labeled "Rainfall (cm)".
  • For each month (January to December), draw a vertical bar.
  • The height of each bar should correspond to the rainfall value for that specific month as given in the table.
  • Ensure all bars are of equal width and are separated by equal spaces.
  • Provide a clear title for the graph, such as "Monthly Rainfall in Sangmelima".

Part c(ii)

The distribution of rainfall shows a bimodal pattern. There is very low rainfall at the beginning of the year (January-February), followed by an increase to a first peak in May (27 cm). Rainfall then decreases in July-August before rising to a second, higher peak in October (36 cm). Rainfall then declines towards the end of the year (November-December). This indicates two distinct rainy seasons.

Part d(i)

36 cm\text{36 cm}

  1. Determine the maximum and minimum rainfall values.

    Maximum rainfall = 36 cm (October)
    Minimum rainfall = 0 cm (January)

  2. Calculate the annual rainfall range.

    Range = Maximum Rainfall - Minimum Rainfall
    Range = 36 cm−0 cm=36 cm36 \text{ cm} - 0 \text{ cm} = 36 \text{ cm}
    The annual rainfall range is 36 cm\boxed{\text{36 cm}}.

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