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.
(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
∴ \(Correct\,mean=\frac{Correct\,sum}{20}=\frac{204}{20}=10.2\)
\(Standard\,\, deviation\,σ=√\frac{1}{n}\sum_{i=1}^nx_i^2-\frac{1}{n^2}(\sum_{i=1}^nx_i)^2=√\frac{1}{n}\sum_{i=1}^2-(\bar{x})^2\)
\(⇒2=√\frac{1}{20}incorrect\,\sum_{i=1}^{n}x_i^2-(10)^2\)
\(⇒4=\frac{1}{20}\,incorrect\,\sum_{i=1}^{n}x_i^2-100\)
⇒ \(incorrect\,\sum_{i=1}^{n}x_i^2=2080\)
∴ \(Correct\,\sum_{i=1}^{n}x_i^2=incorrect\,\sum_{i=1}^{n}x_i^2-(8)^2+(12)^2\)
\(2080-64+144\)
\(=2160\)
∴ \(Correct \,standard \,deviation=\,√\frac{Correct\,\sum{x_i^2}}{n}-(Correct\,mean)^2\)
\(=√\frac{2016}{20}-(10.2)^2\)
\(=√108-104.04\)
\(=√3.96\)
\(1.98\)