# Find mean of normal distrubution such that it fits a condition

The strength of glue is known to follow a normal distribution with a mean of 59 and a standard deviation of 3.4.

Find the mean strength needed such that the probability that the glue is weaker than 55 is less than 5%.

To solve this, I just guessed and checked in Excel:

=NORMDIST(55, 50.1, 3.4, 1)

=NORMDIST(55, 50.2, 3.4, 1)

=NORMDIST(55, 50.3, 3.4, 1)

until the cell was close to .05. The answer I got was 60.6, which appears to work.

That probably isn't the proper way to solve the problem. I searched my book for a problem like it but didn't find any examples. I did find this "Percentiles of an Arbitrary Normal Distribution" formula:

(100p)th percentile for normal(mu, signma) = mu + [(100p)th for standard normal] * sigma

But I'm not sure how to apply it (or maybe that formula doesn't help at all).

So what's the smart way to solve the problem?

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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 $$x_0=a-\delta=a-n\sigma=14.3$$

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 $[0,1]$):

1. $x$ is distributed normally with mean $a$ and standard deviation $\sigma$. Find $b$ such that $P(x\leq b)=p$.
2. $y$ 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

1. $z$ is distributed normally with mean 0 and standard deviation $\sigma$. Find $b$ such that $P(z\geq b-a)=p$.
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I don't think that is the same type of problem. I can find that answer just by =NORMINV(0.05,34,12) in Excel. – user1977 Oct 5 '10 at 1:04
These problems are essentially the same. I've added the explanation in the answer. – Fiktor Oct 5 '10 at 6:03
Ah, now I get it. Thank you! – user1977 Oct 6 '10 at 3:18