



Get live Maths 1-on-1 Classs - Class 6 to 12
Examples
Example 2
Example 3 Important
Example 4
Example 5 Important
Example 6 (Normal Method)
Example 6 (Shortcut Method) Important
Example 7 Important
Example 8
Example 9 Important
Example 10 Important
Example 11
Example 12 Important
Example 13 Deleted for CBSE Board 2023 Exams
Example 14 Important Deleted for CBSE Board 2023 Exams
Example 15 Important Deleted for CBSE Board 2023 Exams
Example 16
Example 17 Important
Example 18
Example 19 Important You are here
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” ﷮﷮xiļ·Æļ·Æ2ļ·Æļ·Æ 5.1 = 1ļ·®100ļ·Æ ļ·®100(š¼šššššššš” ļ·®ļ·®xi2 ) ā š°šššššššš ﷮﷮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 ﷮﷮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” ﷮﷮xi2 )ā š¶šššššš” ﷮﷮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