Let $X\sim Bin(30, 0.60)$ and $f(x)=x^2$
(A) Calculate the probablitiy P(X-1=30) and P(29>=X).
My solution:
Since $np=30*0.60=18>10$ and $n(1-p)=30(0.40)=12>10$, we can use the following approach: \begin{align} & Z = \frac{X-np}{\sqrt{np(1-p)}}\\ \end{align} \begin{align} & P(x-1=30) \Rightarrow P(x = 31)=\frac{31-18}{\sqrt{12}}=\frac{13}{2\sqrt{3}}=\frac{13\sqrt{3}}{6}=3.75\\ \end{align}
For $P(x>=29)$ I solved it like this:
\begin{align} &P(29>=X)=1-P(X<=28)=1-\sum_{i=1}^{28}{P(x_i)} \end{align} Since $n$ is large and $p$ is small, we can use an approach which is very similar to $Poisson$ distribtution with $\lambda = np$:
\begin{align} & P(X=k)=\frac{\lambda^k}{k!} * e^{-\lambda} \\ \end{align}
\begin{align} & P(29>=X)=1-\left(\frac{\lambda^1}{1!}*e^{-\lambda}+\frac{\lambda^2}{2!}*e^{-\lambda}+...+\frac{\lambda^{28}}{28!}*e^{-\lambda}\right) \\ & \ P(29>=X) = 1 - 0.98969 = 0.01031 \end{align} Is this correct (I do not have access to the solutions for these exercises)?
(B) Calculate the expected value $E(X)$ and the variance $V(X)$ to $X$ and $f(x)$. You can use the fact that $E(X^4)=1.19*10^5$
My solution:
\begin{align} & E(X)=np=30*0.60=18 \\ & V(X)=np(1-p)=18(0.40)=7.2 \\ \end{align}
\begin{align} E(f(x))=E(X^2) = V(X)-\left(E(X)\right)^2 = 7.2-(18^2)=-316.8 \end{align}
But this didn't seem right to me, because of the negative result, so I tried using the hint($E(X^4)=1.19*10^5$):
\begin{align} & E(X^4) = 1.19*10^5\Rightarrow E(X^2)=\sqrt{E(X^4)}=\sqrt{1.19*10^5}= 344.96 \end{align}
Is this correct? If not, how do I solve this correctly?