find th smallest value of n in a binomial distribution for which we can assert:
P{ |(Xn/n) – p| < .1} ≥ .9
Well this translates into: find the smallest $\displaystyle n$ such that:
$\displaystyle P(p-0.1<\frac{X_n}{n} <p+0.1) \ge 0.9$
where the rv $\displaystyle X_n$ the number of successes in $\displaystyle n$ independent trials with probability of success in a single trial of $\displaystyle p$. Now given the title of this thread I assume we are supposed to use a normal approximation to the binomial to do this, in which case we have:
$\displaystyle \frac{X_n}{n} \sim N(p, p(1-p)/n)$.
$\displaystyle 90\%$ of the probabilty mass of a normal RV is contained within $\displaystyle \pm 1.645$ standard deviations of the mean, so if $\displaystyle p>0.1$ we have:
$\displaystyle 0.1 \ge 1.645 \sqrt{p(1-p)/n}$
which will allow you to find the smallest $\displaystyle n$ for which this is true.
Now you can do the case where $\displaystyle p<0.1$ yourself.
CB