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 if the wrong item is omitted.

If the coefficients of variation of two distributions are $40$ and $20$ and their variances are $144$ and $64$ respectively,then the mean of their arithmetic means 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 co-prime,then $m + n$ is equal to

Let the mean and the variance of $5$ observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observations are $\frac{7}{2}$ and $a$ respectively,then $(4a + x_{5})$ is equal to

Let ${x_1}, {x_2}, ..., {x_n}$ be $n$ observations such that $\sum x_i^2 = 400$ and $\sum x_i = 80$. Then a possible value of $n$ among the following is:

Let $X = \{11, 12, 13, \ldots, 40, 41\}$ and $Y = \{61, 62, 63, \ldots, 90, 91\}$ be two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and $\sigma^2$ is the variance of all the observations in $X \cup Y$,then $|\bar{x} + \bar{y} - \sigma^2|$ is equal to $.................$.