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:
1.If wrong item is omitted
2.If it is replaced by 12

Answer 1

Hint: As we know, the mean is the average of the data. In this question, we need to calculate the mean of random ungrouped data. To calculate mean of ungrouped data given, we have formula $\overline x = \dfrac{1}{n}\sum\limits_{i = 1}^{20} {{x_i}} $ . We have to put the values in the above formula to get the mean value. Now, the standard deviation basically represents the measure of how spread out our data is or how spreads out numbers are. It is represented by a geek symbol $\sigma $ . Now, we can calculate the standard deviation by taking square root of variance. It can also be represented as formula that shows standard deviation $\sigma = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - \dfrac{1}{{{n^2}}}{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} } = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - {{\left( {\overline x } \right)}^2}} } $ . Now, we can calculate for correct and incorrect data as per the question.

Complete step-by-step answer:
According to the question, number of observations $n = 20$
Incorrect mean = 10
Incorrect standard deviation = 2
Using formula for mean, we get:
$
  \overline x = \dfrac{1}{n}\sum\limits_{i = 1}^{20} {{x_i}} \\
  10 = \dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}} \\
  \sum\limits_{i = 1}^{20} {{x_i}} = 200 \\
 $
So, the incorrect sum of 20 observations is 200.
Now, standard deviation according to incorrect data is calculated by:
$
  \Rightarrow \sigma = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - \dfrac{1}{{{n^2}}}{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} } \\
  \Rightarrow \sigma = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - {{\left( {\overline x } \right)}^2}} } \\
\Rightarrow 2 = \sqrt {\dfrac{1}{{20}}\sum\limits_{i = 1}^n {x_i^2 - {{\left( {10} \right)}^2}} } \\
\Rightarrow 4 = \dfrac{1}{{20}}\sum\limits_{i = 1}^n {x_i^2 - 100} \\
  \Rightarrow \sum\limits_{i = 1}^n {x_i^2} = 2080 \\
 $
So, incorrect \[\sum\limits_{i = 1}^n {x_i^2} = 2080\]

If wrong item is omitted
Correct sum of observations = incorrect sum of observations – 8
The correct sum of observations is $200 - 8 = 192$
Therefore, correct mean = correct sum / total number of observations
Correct mean $ = \dfrac{{192}}{{19}} = 10.1$
Now, incorrect \[\sum\limits_{i = 1}^n {x_i^2} = 2080\]
So, correct \[\sum\limits_{i = 1}^n {x_i^2} \] = Incorrect \[\sum\limits_{i = 1}^n {x_i^2} - {\left( 8 \right)^2}\]
Correct \[\sum\limits_{i = 1}^n {x_i^2} \] $ = 2080 - 64 = 2016$
Therefore, correct standard deviation will be:
$
\Rightarrow \sigma = \sqrt {\dfrac{{\sum {x_i^2} }}{n} - {{\left( {\overline x } \right)}^2}} \\
\Rightarrow \sigma = \sqrt {\dfrac{{2016}}{{19}} - {{\left( {10.1} \right)}^2}} \\
\Rightarrow \sigma = \sqrt {4.09} \\
\Rightarrow \sigma = 2.02 \\
 $
Hence, correct mean is 10.1 and correct standard deviation is 2.02.
If 8 is replaced by 12
Incorrect sum of observations is 200.
If 8 is replaced by 12,
Correct sum of observations $200 - 8 + 12 = 204$
So, corrected mean will be = correct sum / total number of observations
$\overline x = \dfrac{{204}}{{20}} = 10.2$
Now, we have incorrect \[\sum\limits_{i = 1}^n {x_i^2} = 2080\]
So, correct \[\sum\limits_{i = 1}^n {x_i^2} \] = Incorrect \[\sum\limits_{i = 1}^n {x_i^2} - {\left( 8 \right)^2} + {\left( {12} \right)^2}\]
Correct \[\sum\limits_{i = 1}^n {x_i^2} = 2080 - 64 + 144 = 2160\]
Therefore, correct standard of deviation can be calculated by:
$
\Rightarrow \sigma = \sqrt {\dfrac{{\sum {x_i^2} }}{n} - {{\left( {\overline x } \right)}^2}} \\
\Rightarrow \sigma = \sqrt {\dfrac{{2160}}{{20}} - {{\left( {10.2} \right)}^2}} \\
\Rightarrow \sigma = \sqrt {108 - 104.4} \\
\Rightarrow \sigma = \sqrt {3.96} \\
\Rightarrow \sigma = 1.98 \\
 $
Hence, correct mean is 10.2 and correct standard deviation is 1.98.

Note:As we know, attention is needed to do these types of questions because we have to take a lot of values in calculation and we cannot make mistakes in the values. The calculation of mean and standard deviation should also be rechecked to avoid any mistakes in these types of questions.