Miscellaneous
Last updated at May 8, 2024 by Teachoo
Misc 5 The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. Given that number of observations (n) = 20 Incorrect mean ( 𝑥) = 10, Incorrect standard deviation (σ) = 2 We know that 𝑥 = 1𝑛 10 = 120 10 × 20 = 200 = = 200 ∴ Incorrect sum of observations = 200 Finding correct sum of observations , incorrect observation 8 is removed So, Correct sum of observations = Incorrect sum – 8 = 200 – 8 = 192 Hence, Correct mean = Correct sum of observationsNumber of observations = 19220 − 𝟏 = 192𝟏𝟗 = 10.1 Now, Incorrect Standard deviation (σ) = 1N N ×(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) – 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 2 = 120 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) – 𝑰𝒏𝒄𝒐𝒓𝒓𝒆𝒄𝒕 xi2 2 × 20 = 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) – 𝟐𝟎𝟎2 40 = 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) – 2002 Squaring both sides (40)2 = 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 )– 20022 1600 = 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) – 2002 1600 = 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) – 40000 1600 + 40000 = 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) 41600 = 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) = 41600 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 = 120 × 41600 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 = 2080 Since 8 is removed So, (𝐶𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) = 2080 – (8)2 = 2080 – 64 = 2016 Now, Correct Standard deviation (σ) = 1N N ×(𝐶𝑜𝑟𝑟𝑒𝑐𝑡 xi2 )– 𝐶𝑜𝑟𝑟𝑒𝑐𝑡 xi2 = 1𝟏𝟗 𝟏𝟗 2016 − 1922 = 119 38304−36864 = 119 1440 = 119 144 ×10 = 119 144 × 10 = 119 122 × 10 = 119 × 12 × 𝟏𝟎 = 119 × 12 × 3.162 = 1.99 Hence, correct Standard deviation = 1.99 Misc 5 The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (ii) If it is replaced by 12. From part (i), Incorrect sum of observations = 200 Finding correct sum of observations , incorrect observation 8 is removed & 12 is added So, Correct sum of observations = Incorrect sum – 8 + 12 = 200 – 8 + 12 = 204 Hence, Correct mean = Correct sum of observationsNumber of observations = 20420 = 10.2 Finding correct Standard deviation From last part 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 xi2 = 2080 Since 8 is removed & 12 is added So, (𝐶𝑜𝑟𝑟𝑒𝑐𝑡 xi2 ) = 2080 – (8)2 + (12)2 = 2080 – 64 + 144 = 2160 Now, Correct Standard deviation (σ) = 1N N ×(𝐶𝑜𝑟𝑟𝑒𝑐𝑡 xi2 )– 𝐶𝑜𝑟𝑟𝑒𝑐𝑡 xi2 = 120 20 2160 − 2042 = 120 43200 −41616 = 120 1584 = 120 × 39.799 = 1.98 Hence, Correct Standard deviation = 1.98