Variance and Standard Deviation

Indication in a way that variance of y-variable is explained by x-variable is shown as

Let the mean and variance of $12$ observations be $\frac{9}{2}$ and $4$ respectively. Later on, it was observed that two observations were considered as $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $m+n$ are coprime, then $m+n$ is equal to

The mean and standard deviation of the marks of $10$ students were found to be $50$ and $12$ respectively. Later, it was observed that two marks $20$ and $25$ were wrongly read as $45$ and $50$ respectively. Then the correct variance is

Let the positive numbers $a_1, a_{2,} a_3, a_4$ and $a_5$ be in a G.P. Let their mean and variance be $\frac{31}{10}$ and $\frac{m}{n}$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{10}$ and $a_3+a_4+a_5=14$, then $m+n$ is equal to ____________.

Let the mean of 6 observations $1,2,4,5,x$ and $y$ be $5$ and their variance be $10$. Then their mean deviation about the mean is equal to

Two dice $A$ and $B$ are rolled. Let the numbers obtained on $A$ and $B$ be $\alpha$ and $\beta$ respectively. If the variance of $\alpha -\beta$ is $\frac{p}{q},$ where $p$ and $q$ are co-prime, then the sum of the positive divisors of $p$ is equal to

Let the mean of the data

be $5.$ If $m$ and ${\sigma }^2$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3\alpha }{m+{\sigma }^2}$ is equal to $\_\_\_\_\_\_\_.$

Let sets $A$ and $B$ have $5$ elements each. Let the mean of the elements in sets $A$ and $B$ be $5$ and $8$ respectively and the variance of the elements in sets $A$ and $B$ be $12$ and $20$ respectively. A new set $C$ of $10$ elements is formed by subtracting $3$ from each element of $A$ and adding $2$ to each element of $B$. Then the sum of the mean and variance of the elements of $C$ is

The mean and standard deviation of $10$ observations are $20$ and $8$ respectively. Later on, it was observed that one observation was recorded as $50$ instead of $40$. Then the correct variance is

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 $3x-2y$ is equal to $\_\_\_\_\_\_\_$

If the mean and variance of the frequency distribution

are $9$ and $15.08$ respectively, then the value of ${\alpha }^2+{\beta }^2-\alpha \beta$ is _____.

Let the mean and variance of $12$ observations be $\frac{9}{2}$ and $4$ respectively. Later on, it was observed that two observations were considered as $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $m+n$ are coprime, then $m+n$ is equal to

Find the mean, variance and standard deviation for the following frequency distribution:

Let $\mu$  be the mean and $\sigma$ be the standard deviation of the distribution

where $\Sigma f_i=62$. If $\left[x\right]$ denotes the greatest integer $\le x$, then$\left[{\mu }^2 + {\sigma }^2\right]$  is equal to

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

Find the mean, variance and standard deviation for the following frequency distribution:

The mean and variance of $15$ observations is $20$ and $64$ respectively. If $55$ is wrongly read as $40$ as one of the observation, then corrected variance is 

In a given data set, mean of $40$ observations is $50$ and standard deviation is $12$. Two readings which were $20 \& 25$ mistakenly taken as $40$ and $45$. Find correct variance of data set.

$5$ positive numbers $a_1, a_2,..., a_5$ are in geometric progression. Their mean and variance are $\frac{31}{10}$ and $\frac{m}{n}$. The mean of the reciprocals is $\frac{31}{40}$, then $m+n$ is _____

Anita made observations of the maximum temperature, $t°C$, on $50$ days. Her results are summarised by $\sum t=910$ and $\sum (t-t{)}^2=876$, where $t$ denotes the mean of the $50$ observations. Calculate $t$ and the standard deviation of the observations.