The marks obtained by 10 students in a Statistics test are 24, 47, 18, 32, 19, 15, 21, 35, 50 and 41.
This problem requires calculating the variance of a specific subset of data – the largest five observations from a list of student marks in a Statistics test. Variance measures how spread out the data points are from their average value (mean).
The marks obtained by 10 students are:
24, 47, 18, 32, 19, 15, 21, 35, 50, 41
First, we need to sort the marks in ascending order to easily identify the largest values:
15, 18, 19, 21, 24, 32, 35, 41, 47, 50
The largest five observations are:
32, 35, 41, 47, 50
To find the variance, we first need the mean (\(\bar{x}\)) of these five numbers.
Mean (\(\bar{x}\)) = Sum of observations / Number of observations
\(\bar{x} = \frac{32 + 35 + 41 + 47 + 50}{5}\)
\(\bar{x} = \frac{205}{5}\)
\(\bar{x} = 41\)
Variance measures the average squared difference of each data point from the mean. Since we are considering these five observations as the complete set of interest for this calculation, we use the formula for population variance (\(\sigma^2\)):
Population Variance (\(\sigma^2\)) = \(\frac{\sum_{i=1}^{N}(x_i - \bar{x})^2}{N}\)
Where:
Let's calculate the squared difference for each observation:
Now, sum these squared differences:
Sum of squared differences = \(81 + 36 + 0 + 36 + 81 = 234\)
Finally, calculate the variance by dividing the sum of squared differences by the number of observations (\(N=5\)):
\(\sigma^2 = \frac{234}{5}\)
\(\sigma^2 = 46.8\)
Therefore, the variance of the largest five observations is 46.8.
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