The mean and standard deviation of a group of $100$ observations were found to be $20$ and $3,$ respectively. Later on it was found that three observations were incorrect, which were recorded as $21,21$ and $18 .$ Find the mean and standard deviation if the incorrect observations are omitted.
Number of observations $(n)=100$
Incorrect mean $(\bar{x})=20$
Incorrect standard deviation $(\sigma)=3$
$ \Rightarrow 20 = \frac{1}{{100}}\sum\limits_{i = 1}^{300} {{x_i}} $
$ \Rightarrow \sum\limits_{i = 1}^{100} {{x_i}} = 20 \times 100 = 2000$
Incorrect sum of observations $=2000$
$\Rightarrow$ Correct sum of observations $=2000-21-21-18=2000-60=1940$
The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each, $10$ items gave outcome $\frac {1}{2}$ each and the remaining $10$ items gave outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac {4}{3}$ then $\left| d \right|$ equals
The mean and the standard deviation $(s.d.)$ of five observations are $9$ and $0,$ respectively. If one of the observations is changed such that the mean of the new set of five observations becomes $10,$ then their $s.d.$ is?
Let $y_1$ , $y_2$ , $y_3$ ,..... $y_n$ be $n$ observations. Let ${w_i} = l{y_i} + k\,\,\forall \,\,i = 1,2,3.....,n,$ where $l$ , $k$ are constants. If the mean of $y_i's$ is is $48$ and their standard deviation is $12$ , then mean of $w_i's$ is $55$ and standard deviation of $w_i's$ is $15$ , then values of $l$ and $k$ should be
Let the mean and variance of $8$ numbers $x , y , 10$, $12,6,12,4,8$, be $9$ and $9.25$ respectively. If $x > y$, then $3 x-2 y$ is equal to $...........$.
If the variance of the frequency distribution is $160$ , then the value of $\mathrm{c} \in \mathrm{N}$ is
$X$ | $c$ | $2c$ | $3c$ | $4c$ | $5c$ | $6c$ |
$f$ | $2$ | $1$ | $1$ | $1$ | $1$ | $1$ |