     1. Chapter 15 Class 11 Statistics (Term 1)
2. Serial order wise
3. Examples

Transcript

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 ( 𝑥﷯ ) = 40, Incorrect standard deviation (σ) = 5.1 We know that 𝑥﷯ = 1﷮𝑛﷯ 40 = 1﷮100﷯ 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 observations﷮Number of observations ﷯ = 3990﷮100﷯ = 39.9 Now, Incorrect Standard deviation (σ) = 1﷮N﷯ ﷮N ×(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) – 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi﷯﷯2﷯﷯ 5.1 = 1﷮100﷯ ﷮100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) – 𝑰𝒏𝒄𝒐𝒓𝒓𝒆𝒄𝒕 ﷮﷮xi﷯﷯2﷯﷯ 5.1 × 100 = ﷮100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) – 𝟒𝟎𝟎𝟎﷯2﷯﷯ 510 = ﷮100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) – 4000﷯2﷯﷯ Squaring both sides (510)2 = ﷮100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 )– 4000﷯2﷯﷯﷯﷮2﷯ 260100= 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 )– 4000﷯2﷯ 260100= 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) –﷯ 16000000 260100 + 16000000 = 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) ﷯ 16260100 = 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) ﷯ 100(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) ﷯ = 16260100 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ﷯ = 1﷮100﷯ × 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) = 1﷮N﷯ ﷮N ×(𝐶𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 )– 𝐶𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi﷯﷯2﷯﷯ = 1﷮100﷯ ﷮100 161701﷯ − 3990﷯2﷯ = 1﷮100﷯ ﷮16170100 −15920100﷯ = 1﷮100﷯ ﷮250000﷯ = 1﷮100﷯ ﷮ 500﷯2﷯ = 1﷮100﷯ × 500 = 5 Hence, Correct mean = 39.9 & Correct standard deviation = 5 