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Six readings were recorded as 9, 11, 14, 16, 18, 21. Later the reading 14 was found to be a mis-entry whose correct value is 30. What is the effect of the correction on the mean and on the median?

Competition practiceData & Probability
  • AThe mean rises by 8/3 and the median rises by 2
  • BThe mean rises by 16 and the median rises by 2
  • CThe mean rises by 8/3 and the median is unchanged
  • DBoth the mean and the median rise by 8/3

Step-by-Step Solution

Before the correction the six values total 89, so the mean is 89/6, and the median is the average of the middle pair, (14 + 16)/2 = 15. Replacing 14 by 30 adds 16 to the total, giving 105 and a mean of 105/6 = 17.5, a rise of 16/6 = 8/3. In order the corrected data read 9, 11, 16, 18, 21, 30, so the middle pair becomes 16 and 18 and the median rises to 17, a rise of 2. The whole rise of 16 does not go to the mean — it is shared among all six observations — and the median is not always immune to a correction, because here the corrected value crossed the middle of the data.
Answer: A) The mean rises by 8/3 and the median rises by 2

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