Non-parametric binomial control problem

I'm trying to solve the following problem:

It is known that for the last few years the percentage of students passing the course 'non-parametric statistics' is no more than 35%. This year 10 students were randomly chosen as a sample and their grades were the following: $$3.0\space\space 8.0\space\space 7.0\space\space 4.0\space\space 4.0\space\space 0.0\space\space 2.0\space\space 6.0\space\space 2.0\space\space 5.0\space\space$$ The required grade to pass the course is 5.0 . Applying percentage control (or maybe quantile test - I can't find the proper translation from my language) check whether this year's results coincide with those of the previous years(using a=0.1).

This is my attempt at a solution:

$$\text{Let X be the grade of a randomly chosen student.}\\ P(X\ge5.0)\le0.35 \\ Y_i = \begin{cases} 1, & X_i\ge5.0 \\ 0, & \text{otherwise} \end{cases}\\ Y_i\sim B(1,p)\space\space\space\space\space\space p=P(Y_i=1)=P(X_i\ge5.0)\\ H_0:p\le0.35\space\space\space\space H_1:p>0.35\\ \text{or } H_0:X_{0.35}\ge5.0 \space\space\space\space H_1:X_{0.35}<5.0\\ \phi(\underline x)=\begin{cases} 1, &T>t \\ 0, & \text{otherwise} \end{cases}\\ T\sim B(10,0.35)\\P_{\frac{1}{2}}(T\le t)=1-a=0.9\\ \text{Looking at a binomial distribution table we find that t=5 for a=0.0949. }\\P_{\frac{1}{2}}(T\le 5)=0.9051\\ \text{We can also see that out of the 10 students, 4 passed the course.}\\\text{ So }τ=4

You are on the right track. Because you have doubts about English statistical terminology, I will go through the problem with my explanation of what is going on.

I suppose you are testing $$H_0: p \le 0.35$$ against $$H_a: p > .35,$$ where $$p$$ is the proportion who pass.

Your data show $$T = 4$$ passing out of $$n=10$$ randomly chosen subjects. Under the null hypowthesis, $$T \sim \mathsf{Binom}(n=10, p = .35).$$

g = c(3.0, 8.0, 7.0, 4.0, 4.0, 0.0, 2.0, 6.0, 2.0, 5.0)
sum(g >= 5)
[1] 4


We would reject $$H_0$$ for large values of $$T.$$ A pass rate this year of $$\hat p = 4/10 = 0.4 > 0.35$$ may be grounds for some encouragement. But is this enough improvement to be statistically significant?

The P-value of the test is $$P(T \ge 4) = 1 - P(X \le 5) = 0.0949.$$ [Computation in R, where pbinom is a binomial CDF.]

1 - pbinom(5, 10, .35)
0.09493408


In order to reject $$H_0,$$ at level $$\alpha = 0.1,$$ the P-value has to be smaller than 0.1, and it is (just barely). So you have evidence (at the 10% level of significance) to claim that the pass rate has improved.

Notes: (1) This result is not significant at the 5% level, which is used more often in practice than the 10% level used here. (2) Depending on your course, you may be expected to solve some similar problems using normal approximations to binomial probabilities. But $$n$$ is not large enough here for a useful normal approximation.