Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance
Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .
Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.
Statement $-1$ is false, Statement $-2$ is true;
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not acorrect explanation for Statement $-1$
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$
Statement $-1$ is true, Statement $-2$ is false
The mean and standard deviation of $10$ observations are $20$ and $84$ respectively. Later on, it was observed that one observation was recorded as $50$ instead of $40$. Then the correct variance is:
While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
The standard deviation of $25$ numbers is $40$. If each of the numbers is increased by $5$, then the new standard deviation will be
Let the mean and standard deviation of marks of class $A$ of $100$ students be respectively $40$ and $\alpha( > 0)$, and the mean and standard deviation of marks of class $B$ of $n$ students be respectively $55$ and $30-\alpha$. If the mean and variance of the marks of the combined class of $100+ n$ students are respectively $50$ and $350$,then the sum of variances of classes $A$ and $B$ is
Given that $\bar{x}$ is the mean and $\sigma^{2}$ is the variance of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ Prove that the mean and variance of the observations $a x_{1}, a x_{2}, a x_{3}, \ldots ., a x_{n}$ are $a \bar{x}$ and $a^{2} \sigma^{2},$ respectively, $(a \neq 0)$