Answer
No, the condition $n \gt 9\left(\dfrac{p}{q}\right) = 9\left(\dfrac{0.7}{0.3}\right) = 21$ is not satisfied as $n=10$ only.
Work Step by Step
No, the normal approximation is not appropriate.
Note that $P \left[ \dfrac{3.5-10(0.7)}{\sqrt{10(0.7)(0.3)}}\le Z\le \dfrac{8.5-10(0.7)}{\sqrt{10(0.7)(0.3)}}\right]$ is a normal approximation (with a correction factor) to $P(4\le X\le 8)$. This is so since, based on the given pdf, $X\sim \text{Binomial}(n=10,p=0.7)$, so that $\mu_X = np = 10(0.7)$ and $\sigma_X = npq = 10(0.7)(0.3)$.
For a normal approximation to be appropriate for approximating such a probability, both of the following two conditions have to be satisfied (see Comment, p. 240):
(i) $n \gt 9\left(\dfrac{p}{q}\right) = 9\left(\dfrac{0.7}{0.3}\right) = 21$, and
(ii) $n \gt 9\left(\dfrac{q}{p}\right) = 9\left(\dfrac{0.3}{0.7}\right) \gt 3$.
Now, $n=10 > 3$ but $10 \not\gt 21$, so that condition (ii) is satisfied but not condition (i).
