To calculate the river's water level at the end of summer, we need to add the water levels that are below sea level.
Initially, the river's water level is at \( 1 \frac{1}{2} \) meters below sea level. In improper fraction form, this can be expressed as:
\[ 1 \frac{1}{2} = \frac{3}{2} \text{ meters below sea level} \]
During the summer, the river drops another \( 2 \frac{1}{3} \) meters below sea level. In improper fraction form, this can be expressed as:
\[ 2 \frac{1}{3} = \frac{7}{3} \text{ meters below sea level} \]
Now, we need to add these two fractions. To do this, we need a common denominator. The least common multiple of 2 and 3 is 6.
Converting \( \frac{3}{2} \) to a denominator of 6:
\[ \frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6} \]
Now converting \( \frac{7}{3} \) to a denominator of 6:
\[ \frac{7}{3} = \frac{7 \times 2}{3 \times 2} = \frac{14}{6} \]
Now, we can add the two fractions:
\[ \frac{9}{6} + \frac{14}{6} = \frac{9 + 14}{6} = \frac{23}{6} \]
Thus, the total water level at the end of summer is \( \frac{23}{6} \) meters below sea level. Converting \( \frac{23}{6} \) to a mixed number, we get:
\[ \frac{23}{6} = 3 \frac{5}{6} \text{ meters below sea level} \]
Therefore, the water level at the end of summer is:
\[ 3 \frac{5}{6} \text{ meters below sea level} \]
The final answer, accounting for the fact that it is below sea level, is:
\[ 4.16\overline{6} \text{ meters from sea level.} \]
So, to state it clearly: the water level is \( \frac{23}{6} \) meters (or about \( 3.83 \) meters) below sea level, which can also be expressed as "it is \( 3 \frac{5}{6} \) meters below sea level".