Prove that for (n p q geqslant 25) [P_{n}(m)=frac{1}{sqrt{2 n p q}} e^{-frac{z^{2}}{2}}left[1+frac{(q-p)left(z^{3}-3 z ight)}{6 sqrt{n p

Prove that for \(n p q \geqslant 25\)

\[P_{n}(m)=\frac{1}{\sqrt{2 n p q}} e^{-\frac{z^{2}}{2}}\left[1+\frac{(q-p)\left(z^{3}-3 z\right)}{6 \sqrt{n p q}}\right]+\Delta\]

where

\[z=\frac{m-n p}{\sqrt{n p q}},|\Delta|<\frac{0.15+0.25|p-q|}{\sqrt{(n p q)^{3}}}|z| e^{-\frac{3}{2} \sqrt{n p q}}\]