The marks obtained by 10 students in a Statistics test are 24, 47, 18, 32, 19, 15, 21, 35, 50 and 41.
Understanding the Problem
The question asks us to calculate the Mean Deviation for the largest five observations from a given set of marks obtained by 10 students in a Statistics test. The marks are: 24, 47, 18, 32, 19, 15, 21, 35, 50, and 41.
Identifying the Largest Five Observations
First, we need to identify the largest five observations from the list. To do this, we should arrange the marks in ascending order:
15, 18, 19, 21, 24, 32, 35, 41, 47, 50
From the sorted list, the largest five observations are:
32, 35, 41, 47, 50
Calculating the Mean of the Largest Five Observations
Next, we calculate the mean (average) of these five observations. Let's denote these observations as \(x_1, x_2, x_3, x_4, x_5\).
The mean (\(\bar{x}\)) is calculated using the formula:
\(\bar{x} = \frac{\sum_{i=1}^{5} x_i}{5}\)Substituting the values:
\(\bar{x} = \frac{32 + 35 + 41 + 47 + 50}{5}\) \(\bar{x} = \frac{205}{5}\) \(\bar{x} = 41\)So, the mean of the largest five observations is 41.
Calculating Mean Deviation
The Mean Deviation (MD) is the average of the absolute differences between each observation and the mean.
The formula for Mean Deviation is:
\(MD = \frac{\sum_{i=1}^{5} |x_i - \bar{x}|}{5}\)Now, we calculate the absolute difference of each of the largest five observations from their mean (which is 41):
- \(|32 - 41| = |-9| = 9\)
- \(|35 - 41| = |-6| = 6\)
- \(|41 - 41| = |0| = 0\)
- \(|47 - 41| = |6| = 6\)
- \(|50 - 41| = |9| = 9\)
Finally, we calculate the Mean Deviation by averaging these absolute differences:
\(MD = \frac{9 + 6 + 0 + 6 + 9}{5}\) \(MD = \frac{30}{5}\) \(MD = 6\)Conclusion
The Mean Deviation of the largest five observations (32, 35, 41, 47, 50) from the given dataset is 6.
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