Choosing the Right Average for a Skewed Salary Survey
A small Kabul workshop employs 11 people. The owner records the monthly wages (in Afghanis): 9,000; 9,000; 9,500; 9,500; 10,000; 10,000; 10,000; 11,000; 11,000; 12,000; 55,000. The owner wants to advertise a single 'typical monthly wage' for the workshop. A new worker argues that the mean is unfair because one figure is far larger than all the others. The owner also computes the median and the mode before deciding what to print.
1. Compute the mean, median, and mode of the wage data, and state which measure best represents a 'typical' wage here. Justify with the shape of the data.
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Sum = 9,000+9,000+9,500+9,500+10,000+10,000+10,000+11,000+11,000+12,000+55,000 = 156,000. Mean = 156,000 / 11 ≈ 14,182.
With 11 values the median is the 6th: 10,000. The mode is 10,000 (appears 3 times).
The data is strongly right-skewed: ten wages cluster between 9,000 and 12,000, but one value (55,000) pulls the mean above every wage except the top one. The mean (≈14,182) is higher than 9 of the 11 actual wages, so it is misleading. The median (10,000) is robust to the outlier and matches the mode. Median (or mode) best represents a typical wage.
2. The owner drops the 55,000 figure (the manager's wage) and recomputes. By how much does the mean fall, and does the median change? Explain why.
Read the worked answer
New sum = 156,000 − 55,000 = 101,000 over 10 values. New mean = 10,100.
Fall in mean = 14,182 − 10,100 ≈ 4,082 Afghani.
The median of the remaining 10 values is the average of the 5th and 6th: (10,000 + 10,000)/2 = 10,000 - unchanged. Removing one extreme value shifts the mean sharply but leaves the median almost untouched, confirming the median's resistance to outliers.