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Last updated at May 29, 2023 by Teachoo
Example 16 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 ( 𝑥 ) = 40, Incorrect standard deviation (σ) = 5.1 We know that 𝑥 = 1𝑛 40 = 1100 40 × 100 = 4000 = = 4000 ∴ Incorrect sum of observations = 4000 Finding correct sum of observations , 50 is taken in place of 40 So, Correct sum of observations = Incorrect sum – 50 + 40 = 4000 – 50 + 40 = 3990 Hence, Correct mean = Correct sum of observationsNumber of observations = 3990100 = 39.9 Now, Incorrect Standard deviation (σ) = 1N N ×(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) – 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 5.1 = 1100 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) – 𝑰𝒏𝒄𝒐𝒓𝒓𝒆𝒄𝒕 xi2 5.1 × 100 = 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) – 𝟒𝟎𝟎𝟎2 510 = 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) – 40002 Squaring both sides (510)2 = 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 )– 400022 260100= 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 )– 40002 260100= 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) – 16000000 260100 + 16000000 = 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) 16260100 = 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) = 16260100 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 = 1100 × 16260100 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 = 162601 Since 50 is removed and 40 is added, So, (𝑪𝒐𝒓𝒓𝒆𝒄𝒕 xi2 ) = 162601 – (50)2 + (40)2 = 162601 – 2500 + 1600 = 161701 Now, Correct Standard deviation (σ) = 1N N ×(𝐶𝑜𝑟𝑟𝑒𝑐𝑡 xi2 )– 𝐶𝑜𝑟𝑟𝑒𝑐𝑡 xi2 = 1100 100 161701 − 39902 = 1100 16170100 −15920100 = 1100 250000 = 1100 5002 = 1100 × 500 = 5 Hence, Correct mean = 39.9 & Correct standard deviation = 5