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Question
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:
- If wrong item is omitted.
- If it is replaced by 12.
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Solution
`overline x = (sumx_i)/n` or 10 = `(sumx_i)/20`
⇒ `sumx_i = 10 xx 20 = 200`
Standard deviation σ = `1/nsqrt(nsumx_i^2 - (sumx_i)^2)`
∴ `nσ = sqrt(nsumx_i^2 - (sumx_i)^2)`
or `n sumx_i^2 = n^2 σ^2 + (sumx_i)^2`
or `sumx_i^2 = (n^2 σ^2 + (sumx_i)^2)/n`
i. (a) When an observation 8 is excluded.
Addition of new observations = 200 − 8 = 192
New mean = `192/19 = 10.11`
(b) `sumx_i^2 = ((20)^2 xx 4 + (200)^2)/20` .....`[∵ sum = 2, sumx_i = 200]`
= 80 + 10 × 200
= 2080
New `sumx_i^2 = 2080 - 8^2`
= 2080 − 64
= 2016
∴ New Standard Deviation = `1/19 sqrt(19 xx 2016 - (192)^2)`
= `1/19 xx sqrt(38304 - 36864)`
= `1/19 xx sqrt1440`
= 1.997
ii. New `sumx_i = 200 - 8 + 12`
= 204
∴ New mean = `204/20`
= 10.2
`sumx_i^2 = 2080`
New `sumx_i^2 = 2080 - 64 + 144`
= 2160
∴ New (corrected) standard deviation = `1/20 sqrt(20 xx 2160 - (204)^2)`
= `1/20 sqrt(43200 - 41616)`
= `sqrt1584/20`
= 1.99
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