Ok. I will show you an example.
$x$ is distributed normally with mean $a=34$ and standard deviation $\sigma=12$. Find $x_0$ such that $x>x_0$ with probability $0.95$.
Suppose $x_0=a-\delta$. We need to find $\delta$, such that $P(x>a-\delta)=0.95$ or, equivalently, $P(x-a\leq-\delta)=0.05$. First note, that normal distribution is symmetric. Therefore $P(x-a\geq\delta)=0.05$ and $P(|x-a|\geq\delta)=0.1$ or $P(|x-a|<\delta)=0.9$. Here you can find, that $\delta=n\sigma$, where $n=1.644853626951$ (just search for this number on the page to see the table, I've used to find it). Therefore the answer is
Edit: Your task is actually the same. The difference is that in my task we know $a$ and we are asked to find $a-\delta$ and in your case we know $a-\delta$ and we are asked to find $a$. That's simple. Just write $a= (a-\delta)+\delta$. I don't know Excel, but if your use of it is right, you can find the answer of your task by
= NORMINV(0.95, 55, 12). Actually, the answers in the following tasks are th same (here $a,\sigma,p$ are given; you can replace $a$, $\sigma$ by any positive numbers and $p$ by any number in $[0,1]$):
- $x$ is distributed normally with mean $a$ and standard deviation $\sigma$. Find $b$ such that $P(x\leq b)=p$.
- $y$ is distributed normally with standard deviation $\sigma$. It is known, that $P(y\geq a)=p$. Find $b$ --- the mean of the distribution of $y$.
That's because if $x$ is distributed normally with parameters $a$, $\sigma$, then distribution of $z=x-a$ depends only on $\sigma$ (it is independent of $a$) and symmetric (for each $\delta$ we have $p(z<\delta)=p(z>-\delta)$). You can see, that tasks are the same by denoting $x-a$ in the first task by $z$ (and replacing $x$ by $z+a$) and denoting $y-b$ in the second by $z$ (and replacing $y$ by $z+b$). Then both tasks have the following form
- $z$ is distributed normally with mean 0 and standard deviation $\sigma$. Find $b$ such that $P(z\geq b-a)=p$.