I am doing a question which asks for an approximation of the distribution of X, which is the number of successes out of 50 independent trials with probability 0.4, using the central limit theorem. Then the question asks the probability that the number of successes is less than 10 using this approximate distribution.
So the average of the sum $S_{50}=\sum_{i=1}^{50} X_i$ can be standardized, and then follows a standard normal distribution:
$$\frac{S_{50}-50E(X)}{\sqrt{50Var(X)}} \simeq N(0,1)$$
From this point, I think the probability that the number of successes is less than 10 is:
$$P(S_{50})=P(\frac{S_{50}-50E(X)}{\sqrt{50Var(X)}}\leq\frac{10-50\cdot20}{\sqrt{50\cdot 12}})$$
Where, on the right side, 10 is the number of successes, $E(X)$ is $np=50\cdot 0.4=20$, and the variance is $npq=12$. These values are from the binomial distribution. Am I correct in calculating $E(X)$ and $Var(X)$ in this way? It seems that these values must be calculated from a binomial distribution. Any help is really appreciated.
You double counted $50E(X)$ and $50V(X)$. This because $S_{50}$ is a binomial, sum of 50 iid bernulli, thus in this version of CLT mean and variance are referred to the bernulli.
One way (not absolutely accurate is this)
$$\mathbb{P}(S<10)=\mathbb{P}\left[Z<\frac{10-50\cdot0.4}{\sqrt{50\cdot0.4\cdot0.6}}\right]=\Phi(-2.88)\approx0.0019$$
This result is not quite accurate. To realize that, the exact value using the binomial is
$$\mathbb{P}(S<10)\approx 0.0008$$
To improve our CLT approx, just use the Continuity Correction Factor obtaining
$$\mathbb{P}(S<10)=\mathbb{P}\left[Z<\frac{9.5-50\cdot0.4}{\sqrt{50\cdot0.4\cdot0.6}}\right]=\Phi(-3.031)\approx0.0012$$
To understand this reasoning I suggest you to calculate, for example, $P(S<20)$ so you will appreciate the best approximation I showed you. In fact, with the first approximation you get
$$\mathbb{P}(S<20)=\mathbb{P}(Z<0)=0.5$$
while the exact result with the binomial is $0.4465$
Now using the continuity correction factor you get
$$\mathbb{P}(S<20)=\mathbb{P}\left(Z<\frac{19.5-20}{3.464}\right)=\Phi(-0.14434)\approx0.4426$$
which is very close to the exact result