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:
If wrong item is omitted.
Number of observations (n) $=20$
Incorrect mean $=10$
Incorrect standard deviation $=2$
$\bar x = \frac{1}{n}\sum\limits_{i = 1}^{20} {{x_i}} $
$10 = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}} $
$ \Rightarrow \sum\limits_{i = 1}^{20} {{x_i}} = 200$
That is, incorrect sum of observations $=200$
Correct sum of observations $=200-8=192$
$\therefore$ Correct mean $=\frac{\text { correct sum }}{19}=\frac{192}{19}=10.1$
Standard deviation $\sigma = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {{x_i}^2 - \frac{1}{{{n^2}}}{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} } $
$ = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - {{\left( {\bar x} \right)}^2}} } $
$ \Rightarrow 2 = \sqrt {\frac{1}{{20}}Incorrect\sum\limits_{i = 1}^n {x_i^2 - {{\left( {10} \right)}^2}} } $
$ \Rightarrow 4 = \frac{1}{{20}}Incorrect\sum\limits_{i = 1}^n {x_i^2 - 100} $
$ \Rightarrow Incorrect\sum\limits_{i = 1}^n {x_i^2 = 2080} $
$\therefore Correct\,\,\sum\limits_{i = 1}^n {x_i^2 = \,} Incorrect\,\,\sum\limits_{i = 1}^n {x_i^2 - {{\left( 8 \right)}^2}} $
$=2080-64$
$=2016$
$\therefore$ Correct standard deviation $=\sqrt{\frac{\text { Correct } \sum x_{i}^{2}}{n}-(\text { Correct mean })^{2}}$
$=\sqrt{\frac{2016}{19}-(10.1)^{2}}$
$=\sqrt{1061 \cdot 1-102 \cdot 1}$
$=\sqrt{4.09}$
$=2.02$
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$X_i$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
$f_i$ | $k+2$ | $2k$ | $K^{2}-1$ | $K^{2}-1$ | $K^{2}-1$ | $k-3$ |
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