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Last updated at March 16, 2023 by Teachoo

Example 19 The mean and standard deviation of 100 observations were calculated as 40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation? Given that number of observations (n) = 100 Incorrect mean ( š„ļ·Æ ) = 40, Incorrect standard deviation (Ļ) = 5.1 We know that š„ļ·Æ = 1ļ·®šļ·Æ 40 = 1ļ·®100ļ·Æ 40 Ć 100 = 4000 = = 4000 ā“ Incorrect sum of observations = 4000 Finding correct sum of observations , 50 is taken in place of 40 So, Correct sum of observations = Incorrect sum ā 50 + 40 = 4000 ā 50 + 40 = 3990 Hence, Correct mean = Correct sum of observationsļ·®Number of observations ļ·Æ = 3990ļ·®100ļ·Æ = 39.9 Now, Incorrect Standard deviation (Ļ) = 1ļ·®Nļ·Æ ļ·®N Ć(š¼šššššššš” ļ·®ļ·®xi2 ) ā š¼šššššššš” ļ·®ļ·®xiļ·Æļ·Æ2ļ·Æļ·Æ 5.1 = 1ļ·®100ļ·Æ ļ·®100(š¼šššššššš” ļ·®ļ·®xi2 ) ā š°šššššššš ļ·®ļ·®xiļ·Æļ·Æ2ļ·Æļ·Æ 5.1 Ć 100 = ļ·®100(š¼šššššššš” ļ·®ļ·®xi2 ) ā ššššļ·Æ2ļ·Æļ·Æ 510 = ļ·®100(š¼šššššššš” ļ·®ļ·®xi2 ) ā 4000ļ·Æ2ļ·Æļ·Æ Squaring both sides (510)2 = ļ·®100(š¼šššššššš” ļ·®ļ·®xi2 )ā 4000ļ·Æ2ļ·Æļ·Æļ·Æļ·®2ļ·Æ 260100= 100(š¼šššššššš” ļ·®ļ·®xi2 )ā 4000ļ·Æ2ļ·Æ 260100= 100(š¼šššššššš” ļ·®ļ·®xi2 ) āļ·Æ 16000000 260100 + 16000000 = 100(š¼šššššššš” ļ·®ļ·®xi2 ) ļ·Æ 16260100 = 100(š¼šššššššš” ļ·®ļ·®xi2 ) ļ·Æ 100(š¼šššššššš” ļ·®ļ·®xi2 ) ļ·Æ = 16260100 š¼šššššššš” ļ·®ļ·®xi2 ļ·Æ = 1ļ·®100ļ·Æ Ć 16260100 š¼šššššššš” ļ·®ļ·®xi2 ļ·Æ = 162601 Since 50 is removed and 40 is added, So, (šŖšššššš ļ·®ļ·®xi2 ) ļ·Æ = 162601 ā (50)2 + (40)2 = 162601 ā 2500 + 1600 = 161701 Now, Correct Standard deviation (Ļ) = 1ļ·®Nļ·Æ ļ·®N Ć(š¶šššššš” ļ·®ļ·®xi2 )ā š¶šššššš” ļ·®ļ·®xiļ·Æļ·Æ2ļ·Æļ·Æ = 1ļ·®100ļ·Æ ļ·®100 161701ļ·Æ ā 3990ļ·Æ2ļ·Æ = 1ļ·®100ļ·Æ ļ·®16170100 ā15920100ļ·Æ = 1ļ·®100ļ·Æ ļ·®250000ļ·Æ = 1ļ·®100ļ·Æ ļ·® 500ļ·Æ2ļ·Æ = 1ļ·®100ļ·Æ Ć 500 = 5 Hence, Correct mean = 39.9 & Correct standard deviation = 5