A problem with Markov Inequality

Question

From past experience, a professor knows that the test score of a student taking her final examination is a random variable with mean $75$. Give an upper bound for the probability that a student’s test score will exceed $85$.

$$P(X \ge 85) \le \frac{75}{85}=\frac{15}{17}$$ with Markov Inequality

Repeat the same problem when it is known that the variance of a student’s test score is equal to $25$.

I can calculate $$P(X \ge 85) =P\left(\frac{X-[X]}{\sigma(X)}>\frac{85-75}{ \sqrt{25}}\right)=1-\Phi(2)=1-0.9772=0.0228$$ but the final result in the book is $ \le 0.2$

Answer 1

Using Chebyshev's inequality for the last part, you get

$$Pr(|X-75|\ge 10)\le\frac{25}{10^2}=0.25.$$

There's no reason to assume that the distribution is normal.