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:

(i) If wrong item is omitted.

(ii) If it is replaced by 12.

Answer 1

(i) Number of observations (n) = 20

Incorrect mean = 10

Incorrect standard deviation = 2

$\bar{x}=\frac{1}{n}\sum_{i=1}^{20}x_i$

$10=\frac{1}{20}\sum_{i=1}^{20}x_i$

⇒ $\sum_{i=1}^{20}x_i=200$

That is, incorrect sum of observations = 200

Correct sum of observations = 200-8 = 192

∴ $Correct\,mean=\frac{Correct\,sum}{19}=\frac{192}{19}=10.1$

$Standard\,\, deviation\,σ=√\frac{1}{n}\sum_{i=1}^nx_i^2-\frac{1}{n^2}(\sum_{i=1}^nx_i)^2=√\frac{1}{n}\sum_{i=1}^2-(\bar{x})^2$

$⇒2=√\frac{1}{20}incorrect\,\sum_{i=1}^{n}x_i^2-(10)^2$

$⇒4=\frac{1}{20}\,incorrect\,\sum_{i=1}^{n}x_i^2-100$

⇒ $incorrect\,\sum_{i=1}^{n}x_i^2=2080$

∴ $Correct\,\sum_{i=1}^{n}x_i^2=incorrect\,\sum_{i=1}^{n}x_i^2-(8)^2$

$=2080-64$

$=2016$

∴ $Correct \,standard \,deviation=\,√\frac{Correct\,\sum{x_i^2}}{n}-(Correct\,mean)^2$

$=√\frac{2016}{19}-(10.1)^2$

$=√106.1-102.01$

$=√4.09$

=2.02

(ii) When 8 is replaced by 12,

incorrect sum of observations = 2

∴ Correct sum of observations = 200-8+12 = 204

∴ $Correct\,mean=\frac{Correct\,sum}{20}=\frac{204}{20}=10.2$

$Standard\,\, deviation\,σ=√\frac{1}{n}\sum_{i=1}^nx_i^2-\frac{1}{n^2}(\sum_{i=1}^nx_i)^2=√\frac{1}{n}\sum_{i=1}^2-(\bar{x})^2$

$⇒2=√\frac{1}{20}incorrect\,\sum_{i=1}^{n}x_i^2-(10)^2$

$⇒4=\frac{1}{20}\,incorrect\,\sum_{i=1}^{n}x_i^2-100$

⇒ $incorrect\,\sum_{i=1}^{n}x_i^2=2080$

∴ $Correct\,\sum_{i=1}^{n}x_i^2=incorrect\,\sum_{i=1}^{n}x_i^2-(8)^2+(12)^2$

$2080-64+144$

$=2160$

∴ $Correct \,standard \,deviation=\,√\frac{Correct\,\sum{x_i^2}}{n}-(Correct\,mean)^2$

$=√\frac{2016}{20}-(10.2)^2$

$=√108-104.04$

$=√3.96$

$1.98$