Question 154857
Assuming a normal distribution and using the old standard deviation as a measure of the standard deviation, 
you can use the confidence interval formula,
{{{z*sigma/sqrt(N)=4}}}
For 95% confidence, z=1.96.
{{{1.96*10/sqrt(N)=4}}}
{{{sqrt(N)=1.96*10/4}}}
{{{N=(19.6/4)^2}}}
{{{N=24.01}}}
To be sure, we'll round up to N=25.