Question

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. One 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(ii)     If it is replaced by 12.

Answer 1

`bar x = (sum x_i)/n = 10 = (sumx_i)/20`
`sumx_i= 200`
for 19 terms `= 200-8 = 192 = sumx_i`
`sigma = sqrt(1/n*sumx_i^2 - ( bar x )^2)`
let incorrect sum be `e sum` `2 = sqrt(1/n* e sum x_i^2 - 100)`
`4 = 1/20 e sumx_i^2 - 100`
`e sum x_i^2= 104*20 = 2080`
correct `sum x_i^2` for 19 terms = `2080 - 64 = 2016`
(i)`bar x = 192/19 = 10.1 `
`sigma = sqrt( 1/19*2016 - (10.1)^2)`
`= sqrt(106.1 - 102) = sqrt(4.1) = 2.024`
(ii) n = 20 `sumx_i= 204`
`sum x_i^2 = 2016 + 144 = 2160`
`bar x = 204/20 = 10.2`
`sigma = sqrt(1/20*2160 - (10.2)^2)`
`= sqrt(108 - 104.04) = sqrt (3.96) = 1.98`