The arithmetic mean and standard deviation of series of 20 items were calculated by a student as 20 cm and 5 cm respectively. But while calculating them an item 15 was misread as 30 . Find the correct standard deviation.
Given: \(n = 20\), wrong mean = 20, wrong SD = 5.
Correct sum: wrong sum = \(20 \times 20 = 400\). Correct sum = \(400 - 30 + 15 = 385\), so the correct mean = \(385/20 = 19.25\).
Correct sum of squares: \(\sum x^2 = n(\sigma^2 + \bar{x}^2) = 20(25 + 400) = 8500\). Correct \(\sum x^2 = 8500 - 30^2 + 15^2 = 8500 - 900 + 225 = 7825\).
Correct variance = \(\frac{7825}{20} - (19.25)^2 = 391.25 - 370.5625 = 20.6875\)
Correct SD = \(\sqrt{20.6875} \approx 4.54\)
Answer: (C) 4.54
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