Normal distribution: standard deviation given as a percentage. Book question: The concentration of salt in a cell, $X$, can be modelled by a normal distribution with mean $\mu$ and standard deviation $2$%. Find the value of $\alpha$ such that $P(\mu-\alpha&lt; X &lt; \mu+\alpha) = 0.9 $.
What does it mean, standard deviation $2$%? I assumed it meant $X \sim N(\mu, 0.02^2)$, and I then did:
$P(X&lt;\mu-\alpha) = 0.05$. So the $Z$-value is: $z=\frac{x - \mu}{\sigma} = \frac{(\mu-\alpha) - \mu}{\sigma} = \frac{-\alpha}{0.02} = -50\alpha$, where $Z \sim N(0,1)$ is the standard normal deviation. And using the inverse normal distribution function, I get: $-50-\alpha = -1.6448... \implies \alpha =0.0329\ (3sf)$.
But the answer in the back is $3.29$.
Ah but this is $100$ times more than my answer, so perhaps the % sign is just a typo in the question? Standard deviation $2$ instead of $0.02$ would get me the correct answer I think.
The textbook answer is correct. There's just a misunderstanding because the concentration is always misured in %. Thus the result is 3.29 (%, obviously).
Your error is due to the fact that you transformed in % a number that originally was already expressed in %
The correct solution was
$$\mathbb{P}[Z&lt;\frac{a}{2}]=0.95 \rightarrow a\approx3.29$$
Anyway also your reasoning is correct. It is enough that you realize that your result is a % espressed with respect to 1 and not to 100
It means the standard deviation is $0.02 \mu$.  So you should be looking at $N(\mu, (0.02 \mu)^2)$.