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Difficulty: MediumBasic Tools of Economic Analysis

An agricultural officer recorded the annual yield of palm oil (in metric tons) produced by a commercial farm over a 5-year period as follows: 1212, 1616, 1818, 2020, and 2424. What is the standard deviation of the annual palm oil yield in metric tons?

Answer: 4 metric tons

Answer

The standard deviation of the palm oil yield over the 5-year period is 4 metric tons4\text{ metric tons}.
The standard deviation is calculated by determining the mean yield (18 metric tons18\text{ metric tons}), taking the sum of the squared deviations from the mean (8080), dividing by the total number of observations (55) to obtain a variance of 1616, and then taking the square root of 1616, which equals 4 metric tons4\text{ metric tons}.

Step-by-Step Solution

1
Calculate the arithmetic mean (μ\mu) of the data set.
μ=12+16+18+20+245=18 metric tons\mu = \frac{12 + 16 + 18 + 20 + 24}{5} = 18\text{ metric tons}.
The mean is required to determine the deviations of individual data values.
2
Determine the squared deviations from the mean for each yield value.
(12−18)2=36(12-18)^2 = 36, (16−18)2=4(16-18)^2 = 4, (18−18)2=0(18-18)^2 = 0, (20−18)2=4(20-18)^2 = 4, and (24−18)2=36(24-18)^2 = 36. The sum of these squared deviations is 8080.
Squaring deviations eliminates negative values so that dispersion around the mean can be aggregated accurately.
3
Compute the variance (σ2\sigma^2) of the dataset.
σ2=805=16\sigma^2 = \frac{80}{5} = 16.
Variance represents the average of the squared deviations from the mean.
4
Compute the standard deviation (σ\sigma) by taking the square root of the variance.
σ=16=4 metric tons\sigma = \sqrt{16} = 4\text{ metric tons}.
Standard deviation measures dispersion in the original units of the data.

Key Concept

Standard Deviation of Ungrouped Data
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Topics

Basic Tools of Economic Analysis

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