sampling and normal distribution The random variable x is normally distributed with mean 68 cm and sd 2.5 cm.What should be the size of the sample whose mean will not differ from the population mean by more than 1 cm with probability 0.95?[Given that area under standard normal curve to the right of the ordinate at 1.96 is 0.025]
 A: You have$X\sim \mathcal{N}(\mu,\sigma^2)$ with $\mu=68,\sigma=2.5$.
This means that the sample mean (for a sample of size $n$) follows a sampling distribution as $\overline{X}\sim \mathcal{N}(\mu,\sigma^2/n)$.
Your goal is to have $P( \mu-1 \leq \overline{X} \leq \mu+1) \geq p $, where $p=0.95$. That is, the chance of the sample mean being more than 1 away from the known mean is at least $p$.
Unless you are using a program to integrate the densities, we have to transform to the standard normal case. Let $Z=(\overline{X}-\mu)/\sqrt{\sigma^2/n}$. Note that $Z\sim \mathcal{N}(0,1)$.
Then the problem is equivalent to:
$$
P(-\sqrt{n}/\sigma \leq Z \leq \sqrt{n}/\sigma) \geq p
$$ $$
1-2 P(Z \geq \sqrt{n}/\sigma) \geq p
$$ $$
P(Z \geq \sqrt{n}/\sigma) \leq \frac{1-p}{2} = \delta
$$
using the symmetry of the Gaussian density and $\delta=0.025$. 
Now we will use the information you were given: 
$$
P(Z\geq \sqrt{n}/\sigma) \leq \delta\;\;\;\implies\;\;\; Z\geq1.96
$$
Therefore:
$$
\sqrt{n}/\sigma \geq 1.96
$$
So $n\geq 1.96^2\sigma^2$, meaning $n$ should be around 24 at least.
