The mean and variance of seven observations are $8$ and $16$,respectively. If $5$ of the observations are $2, 4, 10, 12, 14$,then the product of the remaining two observations is

Suppose that the mean and median of the non-negative numbers $21, 8, 17, a, 51, 103, b, 13, 67$ $(a > b)$ are $40$ and $21$, respectively. If the mean deviation about the median is $26$, then $2a$ is equal to:

The mean and variance of $7$ observations are $8$ and $16$ respectively. If the first five observations are $2, 4, 10, 12, 14$,then the absolute difference of the remaining two observations is

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 the correct data,then $(\alpha, \beta)$ 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 $.................$.

The sum of two positive integers is $100$. The probability that their product is greater than $1000$ is