The results of a certain experiment whose true value is about 10.0 cm, follow a normal distribution with standard deviation of 1.0 cm. How many measurements will we have to do so that our uncertainty about the true value at the $90\%$ confidence level is less than $2\%$? Assume that systematic uncertainty is negligible.
So I did this with the following formula $$Error=Z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=0.02\Rightarrow n=\frac{Z_{\alpha/2}^2\sigma^2}{0.02^2}$$ being $Z_{\alpha/2}$ the value on the normal distribution that leaves $\alpha/2$ on the right side of $Z$, $\sigma$ the standard deviation and $p=0.9$ the confidence level which then $\alpha$ is $p=1-\alpha$. But the book says I'm wrong, what's happening here? And also I haven't used the info "the results of a certain experiment whose true value is about 10cm". Is that where I'm missing?
The confidence interval is
$$\large{\left[\overline x-z_{(1-\frac{\alpha}{2})}\cdot \frac{\sigma}{\sqrt n} ; \ \overline x+z_{(1-\frac{\alpha}{2})}\cdot \frac{\sigma}{\sqrt n} \right]}$$
You have $\alpha=1-0.9=0.1$. Thus $1-\frac{\alpha}{2}=1-\frac{0.1}{2}=1-0.05=0.95$.
And the width of the interval (uncertainty) is 2%. Thus
$2\cdot z_{0.95}\cdot \frac{1.0}{\sqrt n}=0.02\Rightarrow 1.64485\cdot \frac{1.0}{\sqrt n}=0.01$