2022-I-11

$\begin{array}{rcl}
\text{the median} &= & \dfrac{1}{2}(\text{the $10$th datum} +\text{the $11$th datum)} \\
31 & = & \dfrac{1}{2}(30+b +30 +b) \\
31 & = & 30 +b \\
b & = & 1
\end{array}$

And

$\begin{array}{rcl}
\text{the inter-quartile range} & = & Q_3 -Q_1 \\
\text{the inter-quartile range} & = & \dfrac{1}{2} (\text{the $15$th datum} +\text{the $16$th datum}) -\dfrac{1}{2}(\text{the $5$th datum}+\text{the $6$th datum}) \\
14 & = & \dfrac{1}{2}(36+36) -\dfrac{1}{2}(20+a+20+a) \\
14 & = & 36-20-a \\
a & = & 2
\end{array}$

1. Note that the mode is $36$ and the highest frequency is $4$. Note also that the second highest frequency is $2$. If the age of the left player is $36$, then the frequency of $36$ becomes $3$, which is still the highest. Therefore, the mode remains unchanged.
2. If the range of the distribution is decreased, then the possible ages of the left player are $17$ and $43$.

   For the age of the left player is $17$, the standard deviation is $7.162\ 537\ 194$.

   For the age of the left player is $43$, the standard deviation is $7.132\ 307\ 207$.

   Therefore, the greatest possible standard deviation of the distribution is $7.16$.