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:

1. If wrong item is omitted.
2. If it is replaced by 12.

Answer 1

`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