A company has $200$ employees. Assume that each employee invites $2$ guests, independently, to attend a promotion seminar with a probability of $0.8$. How many seats should be provided if the company desires to be $99\%$ confident of seating everyone? (Using Normal approximation to binomial distribution)
I am doing it with: $$ \begin{array}{rcl} \mathrm{E}(x) =& (400)(0.8) &= 320 \\ \mathrm{Var}(x) =& (400)(0.8)(0.2) &= 64 \end{array} $$
How should I determine the number of seats then?
The formula for the normal approximation is
$P(X \leq x)=\Phi\left( \frac{x+0.5-\mu}{\sigma} \right)$
$=\Phi\left( \frac{x+0.5-320}{\sqrt{64}} \right)=0.99$
$\Phi(\cdot )$ is the cdf of the standard normal distribution. $+0.5$ in the numerator is the continuity correction factor.
Taking the inverse function to solve for x.
$\frac{x+0.5-320}{\sqrt{64}}=\Phi ^{-1}\left( 0.99\right)$
$\sqrt{64}=8$ and to get the value of $\Phi ^{-1}\left( 0.99\right)$ you look at a table of the standard normal distribution. You´ll see that the corresponding value is $2.33$. The equation becomes
$\frac{x+0.5-320}{8}=2.33$
Solve for x. The result should be rounded up to a whole number.