Sorry in advance, the question is quite long. At first I had no idea how to even start this question but as I was reading through textbook, I've found that the sample size necessary for the CI to have a width $w$ is $$n=\left(2z_{\alpha/2}\frac{\sigma}{w}\right)^2$$
and the half width of the 95% CI is called the bound on the error or estimation and is denoted as $$1.96\frac{\sigma}{\sqrt{n}}$$
and if its normally distributed, do I have to consider CLT(central limit thm) as well?? any help would be appreciated!!
(a) $67.3\%$.
Letting $x$ be the sample height of the plant after $8$ weeks (i.e. the plant that belongs to the researcher),
$u$ be the population mean of plant sizes after $8$ weeks,
$\sigma$ be the standard deviation of an observation,
$n$ being the sample size: $\dfrac{x-u}{\dfrac{\sigma}{\sqrt{n}}}$ ~$N(0,1)$ asymptotically (by CLT).\
The width is therefore $1.96\cdot\dfrac{\sigma}{\sqrt{n}}$. Halving the width: $0.98\cdot\dfrac{\sigma}{\sqrt{n}}$.
Since $P(Z>|0.98|)=0.673$, this is the answer (refer to std normal tables); just as $P(Z>|1.96|)=0.95$.
(b) $n' = 4\cdot\dfrac{n}{\sigma^2}$. Just solve for $1.96\cdot\dfrac{\sigma}{\sqrt{n^\prime}} = \dfrac{1.96}{2}\cdot\dfrac{\sigma}{\sqrt{n}}$.