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

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

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

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

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

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

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

\(2080-64+144\)

\(=2160\)

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

\(=√\frac{2016}{20}-(10.2)^2\)

\(=√108-104.04\)

\(=√3.96\)

\(1.98\)