For a binomial distribution over [tex]n[/tex] trials and with success probability [tex]p[/tex], the mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] ar such that
[tex]\begin{cases}\mu=np\\\sigma=\sqrt{np(1-p)}\end{cases}[/tex]
[tex]6=\sqrt{np(1-p)}\implies np-np^2=36[/tex]
[tex]54=np\implies54-54p=36\implies p=\dfrac13[/tex]
[tex]54=\dfrac n3\implies n=162[/tex]
