Is the question wrong? Approximate binomial with normal distribution The results of the exam are a normal approximation to the binomial distribution. The maximum exam score is 100 points with µ = 60 and ơ = 7.
a) When X is a binomial random variable that can be approximated by the normal distribution, calculate the probability that a student scores between 80 and 90 points?
Is this question wrong? If X is a binomial random variable and X is the score, then X can't be a binomial random variable, because the score has more than two possible results. Therefore, this question is wrong. 
 A: No, the question is not "wrong" — the point you make reveals a confusion on your part about what a Binomial random variable is.

A Binomial random variable with parameters $n\geq 1$ and $p\in[0,1]$ takes values in $\{0,1,2,\dots, n\}$. It is the sum of $n$ independent Bernoulli random variables, each with parameter $p$.
In short: Bernoulli: two possible outcomes, $0$ and $1$. Binomial: $n+1$ possible outcomes (where $n$ is the parameter), $0,1,2,\dots,n$.
A: The wording is awkward, but I think there are two issues worth exploring.
First, if $X \stackrel{aprx}{\sim} \mathsf{Norm}(\mu = 60, \sigma = 7),$
then find $P(80 \le X \le 90) = P(79.5 < X < 90.5),$ which can be
evaluated by standardizing and using printed standard normal CDF tables 
or by statistical software as as very
nearly $0.0005.$ In R statistical software on gets:
diff(pnorm(c(79.95, 80.5), 60, 7))
## 0.0004833435

Note: There are  negligible probabilities that this normal distribution
takes 'prohibited' values above 100 or below 0, but it is customary to
ignore them.
Second, one can ask, "What binomial distribution might be approximated by 
$\mathsf{Norm}(\mu = 60, \sigma = 7)\,?$"
From $\mu = np = 60$ and $\sigma = np(1-p) = 50 \approx 7^2,$ one can
deduce that $n = 360$ and $p = 1/6.$ So perhaps the exam has
$n = 360$ questions with a probability $p \approx 1/6$ that any one of
them is answered correctly.
If $Y \sim \mathsf{Binom}(n = 360, p = 1/6),$ then the exact
value $P(80 \le Y \le 90) = P(Y \le 90) - P(Y \le 79) \approx 0.0013$
can be found using software:
diff(pbinom(c(79,80), 360, 1/6))
## 0.001253314

The absolute error of the various approximations is small, but the relative
error is large for the probability involved in this problem, which is far into the right tail. Perhaps the 'convenient approximation' $50 \approx 7^2$ (used to get round numbers)
should be improved a bit.
The figure below shows that $\mathsf{Norm}(\mu = 60, \sigma = 7)$ and
$\mathsf{Binom}(n = 360, p = 1/6)$ are a "pretty good" match.
 
