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 $=\sqrt{\frac{\text { Correct } \sum x_{i}^{2}}{n}-(\text { Correct mean })^{2}}$
$=\sqrt{\frac{2016}{19}-(10.1)^{2}}$
$=\sqrt{1061 \cdot 1-102 \cdot 1}$
$=\sqrt{4.09}$
$=2.02$
The variance of first $50$ even natural numbers is
The mean and standard deviation of marks obtained by $50$ students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
Subject | Mathematics | Physics | Chemistty |
Mean | $42$ | $32$ | $40.9$ |
Standard deviation | $12$ | $15$ | $20$ |
Which of the three subjects shows the highest variability in marks and which shows the lowest?
The mean and standard deviation of $20$ observations were calculated as $10$ and $2.5$ respectively. It was found that by mistake one data value was taken as $25$ instead of $35 .$ If $\alpha$ and $\sqrt{\beta}$ are the mean and standard deviation respectively for correct data, then $(\alpha, \beta)$ is :
The mean and variance of a set of $15$ numbers are $12$ and $14$ respectively. The mean and variance of another set of $15$ numbers are $14$ and $\sigma^2$ respectively. If the variance of all the $30$ numbers in the two sets is $13$,then $\sigma^2$ is equal to $.........$.
For two data sets, each of size $5$, the variances are given to be $4$ and $5$ and the corresponding means are given to be $2$ and $4$, respectively. The variance of the combined data set is