To solve this problem, let's define the events and use the information provided to find the probability of students who are good at calculus.

Let:
- \( A \): the event that a student is good in calculus.
- \( B \): the event that a student is good in statistics.

From the problem, we know:
1. \( P(A \cup B) = 0.7 \) (the proportion of students who are good in either calculus or statistics).
2. \( P(A \cup B') = 0.9 \) (the proportion of students who are good in calculus or not good in statistics). Here, \( B' \) is the complement of \( B \) (not good in statistics).

We also know the following relationship:
\[
P(A \cup B') = P(A) + P(B') - P(A \cap B')
\]

Since \( B' \) is the complement of \( B \):
\[
P(B') = 1 - P(B)
\]

Substituting this into our equation gives:
\[
P(A \cup (1 - B)) = P(A) + (1 - P(B)) - P(A \cap (1 - B))
\]
\[
0.9 = P(A) + 1 - P(B) - P(A \cap B')
\]

We can rewrite \( P(A \cap B') \) as:
\[
P(A \cap B') = P(A) - P(A \cap B) \quad \text{(because } A \cap B' \text{ is essentially } A \text{ without } B\text{)}
\]

Substituting \( P(A \cap B') \) into the equation yields:
\[
0.9 = P(A) + 1 - P(B) - (P(A) - P(A \cap B))
\]
\[
0.9 = P(A) + 1 - P(B) - P(A) + P(A \cap B)
\]
\[
0.9 = 1 - P(B) + P(A \cap B)
\]

Now let's also express the first event:
From \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \):
\[
0.7 = P(A) + P(B) - P(A \cap B)
\]

Now we can solve these two equations:
1. \( 0.9 = 1 - P(B) + P(A \cap B) \) simplifies to:
\[
P(B) = 1 + P(A \cap B) - 0.9 \Rightarrow P(B) = P(A \cap B) + 0.1
\]

2. From \( 0.7 = P(A) + P(B) - P(A \cap B) \):
Substitute \( P(B) \):
\[
0.7 = P(A) + (P(A \cap B) + 0.1) - P(A \cap B)
\]
\[
0.7 = P(A) + 0.1
\]
\[
P(A) = 0.7 - 0.1 = 0.6
\]

Therefore, the probability of a randomly selected student being good in calculus is \( \boxed{0.6} \).