A machine produces items and the number of defective items produced behaves according to a Poisson process at the rate of 10 defective items per hour. Let X be the number of defective items produced within 2.5 hours of production. $Y = \frac{X}{\sqrt{V (X)}}$ has an approximate distribution:
I've done:
$X$ has distribution $P(25)$ with $E(X)=V(X)=25$
$\frac{X}{\sqrt{V(X)}} = \frac{X-E(X)}{\sqrt{V(X)}} + \frac{E(X)}{\sqrt{V(X)}} = \frac{X-25}{5} + 5$
I know that $\frac{X-25}{5}$ has distribution $\mathcal{N}(0,1)$ but I don't know the distribution of $\frac{X-25}{5} + 5$
Thinking logically it could also be $\mathcal{N}(5,1)$ or $\mathcal{N}(0,6)$ or even $\mathcal{N}(5,6)$
The solution is $\mathcal{N}(5,1)$ but I can't understand why...
If $Y$ have normal $\mathcal N(\mu,\sigma^2)$ distribution, then $\mu=\mathbb E[Y]$ and $\sigma^2=\text{Var}(Y)$. For this case $Y+a$ also have normal distribution with parameters $\mathbb E[Y+a]=\mathbb E[Y]+a$, $\text{Var}(Y+a)=\text{Var}(Y)$.
So the parameters of approximate normal distribution for $\frac{X-25}{5}+5$ are $0+5$ and $1$.