How to find a standard deviation determined by a Normal distribution probability? The question is

A liquid drug is marketed in phials containing a nominal 1.5ml but the amounts can vary slightly. The volume in each phial may be modeled by a normal distribution with the mean 1.55ml and standard deviation $\sigma$ ml. The phials are sold in packs of 5 randomly chosen phials . It is required that in less than 0.5% of the packs will the total volume of the drug be less than 7.5ml. Find the greatest possible value of $\sigma$.

I need to find the greatest possible value of the standard deviation ($\sigma$).
I worked out the following:
$$\mu= 1.55*5 = 7.75.$$
We are asked to find value of $\sigma$ such that probability of (total volume of $5$ packs $\lt 7.5)\lt0.5\%$
$$P(X\lt7.5)\lt0.005.$$
After standardizing,
$$P(X\le\frac{7.5-7.75}{\sigma/5})<0.005$$ and I found $\sigma=0.2170.$ However, the answer provided is $0.0434.$
Please assist.
 A: Among the objectives of good introductory statistics courses is learning how to think about the Normal distribution.  This question provides a nice example.
The key is to use units of measurement that are adapted to the distribution.  That is, let the mean be the zero point and let the standard deviation be one unit.  This is what a "Z score" measures.
In light of this, let's parse the question.  To do so, I will use two fundamental facts: expectations add ("linearity of expectation") and variances of independent variables also add:


*

*The mean volume of one pack is 1.55 ml, whence the mean volume of five packs must be five times as large, or 7.75 ml: this is the zero point.

*Since the unknown variance of a single pack is $\sigma^2,$ the variance of the sum of five independent packs is $5\sigma^2.$  Therefore the standard deviation of the sum--the unit of measurement we must adopt--is $\sqrt{5\sigma^2} = \sigma\sqrt{5}.$
The question stipulates that in less than 0.5% of cases should the total be less than 7.5 ml.  For the (standard) Normal distribution we remember (or can compute) that exactly 0.5% of cases are $2.57\ldots$ or more less than the mean.  An example of this computation is
qnorm(0.5/100)

in R or
=NORMSINV(0.5/100)

in Excel, for instance.
One aim of the introductory course is to help you reach the point where such considerations are automatic: you can do them in your head correctly, apart (perhaps) from the arithmetical calculations.

This preliminary work enables us to rephrase the question like this:

What unit of measurement, given by $\sigma\sqrt{5}$ for a five-pack of drugs, will re-express an amount of $7.5$ ml as being $2.57$ less than $7.75$ ml?

The solution obviously is
$$\sigma\sqrt{5} = (7.75 - 7.5)/2.57\ldots = 0.097\ldots,$$
implying
$$\sigma = \frac{0.097\ldots}{\sqrt{5}} = 0.0433797\ldots$$

Comparing this result to the question shows that the work in the question was entirely correct up to the point where "$\sigma/5$" appeared: the square root was lost.  This suggests remembering to think in terms of variances rather than standard deviations.
Comparing this result to the older answers that were posted also shows how they were basically moving in the correct direction but made mistakes along the way, too.  Because arithmetical mistakes are easy to make, when one has the chance it's a good idea to check probabilistic calculations with simulations.  For instance, the following R statement generates a large number of five-packs of drugs as described in the question (using the answer I obtained) and, to check my answer, computes the fraction with totals less than 7.5 ml:
mean(colSums(matrix(rnorm(5*1e6, 1.55, -0.25/qnorm(0.5/100) / sqrt(5)), nrow=5)) < 7.5)

(You can see all the data from the question embedded in this expression, along with the value 1e6 giving the number of five-packs to simulate.)  When I run and re-run this code (which takes less than a second each time), I consistently obtain results between 0.0048 (0.48%) and 0.0052 (0.52%), in satisfactory agreement with the intended 0.5% target.
A: \begin{align}
5\times 1.55 &= 7.75  \\
5\times SD   &= 5SD
\end{align}
Problem statement: $P(X<7.5)<0.005$
\begin{align}
\frac{(7.5-7.75)}{(5SD^2)^{1/2}}  &< 0.005  \\[8pt]
\frac{-0.25}{2.576} &= (5SD^2)^{1/2}  \\[5pt]
&-0.0970 / 5^{(1/2)}
\end{align}
$0.0434$ as the standard deviation can never me negative, take the mod value. 
A: I think your understanding of the variance of the sum is mistaken. The variance of the 5-pack sum is 25 times the variance of the single pack.
