# 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 $$P(\mu-\alpha< X < \mu+\alpha) = 0.9$$.

What does it mean, standard deviation $$2$$%? I assumed it meant $$X \sim N(\mu, 0.02^2)$$, and I then did:

$$P(X<\mu-\alpha) = 0.05$$. 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

$$\mathbb{P}[Z<\frac{a}{2}]=0.95 \rightarrow a\approx3.29$$

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

• But then I would have thought they would write, "$\mu$ % and standard deviation $2$ %" in the question, which they didn't. Maybe just an English language thing... Commented Nov 11, 2020 at 17:46

It means the standard deviation is $$0.02 \mu$$. So you should be looking at $$N(\mu, (0.02 \mu)^2)$$.

• I was thinking that, but then I thought that would give me $\alpha$ in terms of $\mu$. Yeah, surely $\alpha$ depends on $\mu$ there... Commented Nov 2, 2020 at 23:37
• So the book answer is wrong? Commented Nov 5, 2020 at 16:13