A random sample of size n is obtained from normal distribution. Calculate how large n must be to ensure that the sample mean lies within 0.25s.d of the mean with probability of 95%.
I know the formula of a confidence interval, any tips of how to start this would be great!
If you're using a normal (z) distribution, the value is easy to calculate.
The 95% confidence interval around a sample mean $\displaystyle \mu_s$ is given by $\displaystyle (\mu_s - z\frac{s}{\sqrt{n}}, \mu_s + z\frac{s}{\sqrt{n}})$, where $\displaystyle \mu_s$ is the sample mean and $\displaystyle s$ is the sample standard deviation. $\displaystyle \frac{s}{\sqrt{n}}$ is the standard error of the mean. For large enough samples, you can assume that the sample standard deviation is a good estimator of the population standard deviation.
The confidence interval tells you that the population mean lies within it with a certain probability. When z = 1.96, that probability is 95%.
So the width of the confidence interval for that z-score should be less than or equal to 0.25 times the population std. deviation (which can be estimated by s), that is:
$$\displaystyle 1.96 \times 2 \times \frac{s}{\sqrt{n}} \leq 0.25s$$
Cancel out s and solve to get:
$$\displaystyle n \geq 245.86$$
So the minimum sample size you need is 246.