Let x_i (1 $$ \le $$ i $$ \le $$ 10) be ten observations of a random variable X. If
$$\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3$$ and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9$$
where 0 $$ \ne $$ p $$ \in $$ R, then the standard deviation of these observations is :

For the frequency distribution :
Variate (x) :      x₁   x₂   x₃ ....  x₁₅
Frequency (f) : f₁    f₂   f₃ ...... f₁₅
where 0 < x₁ < x₂ < x₃ < ... < x₁₅ = 10 and
$$\sum\limits_{i = 1}^{15} {{f_i}} $$ > 0, the standard deviation cannot be :

Let X = {x $$ \in $$ N : 1 $$ \le $$ x $$ \le $$ 17} and
Y = {ax + b: x $$ \in $$ X and a, b $$ \in $$ R, a > 0}. If mean
and variance of elements of Y are 17 and 216
respectively then a + b is equal to :

Let the observations x_i (1 $$ \le $$ i $$ \le $$ 10) satisfy the
equations, $$\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} $$ = 10 and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $$ = 40.
If $$\mu $$ and $$\lambda $$ are the mean and the variance of the
observations, x₁ – 3, x₂ – 3, ...., x₁₀ – 3, then
the ordered pair ($$\mu $$, $$\lambda $$) is equal to :