$Y_1,...,Y_k$ be a sample of size n for each $Y_i$ ~N(m,$s^2$ ) normal distribution for each $i= \{1,...,k\}$ and $m$ and $s^2$ are unknown parameters.
let $m_l= \bar{Y_k}$ and $s_l^2=1/k \sum_{i=1}^k(Y_i-\bar{Y_k})^2$
And let
$P(|m_l - m|\le s/5 \ \ \ \text{and} \ \ \ |s_l- s |\le s/5 ) \ge 1/2$
Determine the smallest possible sample size k in order that this above probability holds.
Since sample mean and sample variance are independent for normal r.v.'s, we have that $$\mathbb P(|m_l - m|\le s/5 \, , |s_l- s |\le s/5 ) =\mathbb P(|m_l - m|\le s/5 )\times\mathbb P( |s_l- s |\le s/5 ) $$ Furthermore, by Chebyshev's inequality, we have $$\begin{align}\mathbb P(|m_l - m|\le s/5 ) &= 1-\mathbb P(|m_l - m|\ge s/5 )\\ &\ge1-\frac{25\mathrm{Var}(m_l)}{s^2}\\ &= 1 - \frac{25}{n}\end{align} $$ and $$\begin{align}\mathbb P(|s_l - s|\le s/5 ) &= 1-\mathbb P(|s_l - s|\ge s/5 )\\ &\ge1-\frac{25\mathrm{Var}(s_l)}{s^2}\\ &= 1 - \frac{25}{s^2}\frac{(n-1)(n-3)s^2}{n^3}\\ &= 1 - \frac{25(n-1)(n-3)}{n^3}\end{align} $$ (See here for a derivation of the variance of $s_l$).
Thus, to find $n$ such that $\mathbb P(|m_l - m|\le s/5 \, , |s_l- s |\le s/5 ) \ge 1/2 $, it suffices to find $n$ such that $$\left(1 - \frac{25}{n}\right)\times\left(1 - \frac{25(n-1)(n-3)}{n^3}\right) \ge \frac{1}{2} $$ Also note that this will most likely not give you the smallest $n$ such that the inequality holds, as Chebyshev's inequality tends to give quite loose bounds. If you want something better, you might need to look at some results from large deviations theory which can give much better (exponential) bounds.
