First you have to use the cdf of the binomial distribution:
$P(X \geq 20)=\sum_{x=20}^{130}{130 \choose x}\cdot \left(0.16\right)^x\cdot\left(0.84\right)^{130-x} $
Applying converse probability
$P(X \geq 20)=1-P(X \leq 19)=1-\sum_{x=0}^{19} {130 \choose x}\cdot \left(0.16\right)^x\cdot\left(0.84\right)^{1000-x} $
Approximation of the Binomial distribution by the Normal distribution:
$$P(X \geq 20)=1-\Phi\left(\frac{19+0.5-0.16\cdot 130}{\sqrt{130\cdot 0.16 \cdot 0.84}} \right)$$
0.5 is the continuity correction factor. And $\Phi(\cdot)$ is the cdf of the Standard Normal distribution. You can do the approximation, because $n\cdot p \cdot (1-p) >9$ (rule of thumb).
I think you can proceed from here.
You have percentage scores $X \sim \mathsf{Norm}(\mu = 65,\, \sigma=6)$ and you seek $c$ such that $P(X \le c) = .35.$ This can be solved by standardizing and using printed tables of the standard normal distribution:
$$.35 = P(X < 35) = p\left(\frac{X - \mu}{\sigma} < \frac{c - 65}{6}\right)
= P\left(Z < \frac{c - 65}{6}\right),$$
where $Z$ is standard nomal. From tables you should be able to find that $(c-65)/6 \approx -0.38,$ and then it is a simple matter to solve for $c,$
and round up to the nearest integer (percentage) and thus get the lowest passing score.
If you have access to statistical software or a statistical calculator, you
may be able to solve the question directly, without standardizing. In R statistical software qnorm is the quantile function (inverse CDF) of a
normal distribution. Thus the lowest passing score is 63.
qnorm(.36, 65, 6)
## 62.84925
I find that drawing a sketch is insurance against getting tangled up
reading printed normal tables and finding an "answer" that is obviously wrong.
By hand, you can't make a sketch as accurate as the plot below, but you
can make a sketch that is better than nothing.
Best Answer
You need to find $x$ such that $P(X\le x)=0.15$. Since $0.15<0.5$, we will take the upper $0.15$ instead. You have that \begin{align}0.15=P(X\le x)=P\left(Z\le \frac{x-9.5}{0.4}\right)=\Phi\left(\frac{x-9.5}{0.4}\right)=1-\Phi\left(\frac{9.5-x}{0.4}\right)\end{align} Hence, from the standardized normal distribution table, you find that $$\Phi(1.03643)=0.85$$ So, you need to solve $1.03643=\frac{9.5-x}{0.4}$.