To calculate the expected value \( E(N) \) of the random variable \( N \), we use the formula for expected value of a discrete random variable:
\[ E(N) = \sum_{n} n \cdot f(n) \]
We have the probability function given as follows:
- \( f(4) = 0.15 \)
- \( f(5) = 0.23 \)
- \( f(6) = 0.19 \)
- \( f(7) = 0.24 \)
- \( f(8) = 0.19 \)
Now we can compute \( E(N) \):
\[ E(N) = 4 \cdot 0.15 + 5 \cdot 0.23 + 6 \cdot 0.19 + 7 \cdot 0.24 + 8 \cdot 0.19 \]
Calculating each term:
- \( 4 \cdot 0.15 = 0.60 \)
- \( 5 \cdot 0.23 = 1.15 \)
- \( 6 \cdot 0.19 = 1.14 \)
- \( 7 \cdot 0.24 = 1.68 \)
- \( 8 \cdot 0.19 = 1.52 \)
Now, sum these values:
\[ E(N) = 0.60 + 1.15 + 1.14 + 1.68 + 1.52 \]
Calculating the sum step by step:
- \( 0.60 + 1.15 = 1.75 \)
- \( 1.75 + 1.14 = 2.89 \)
- \( 2.89 + 1.68 = 4.57 \)
- \( 4.57 + 1.52 = 6.09 \)
Thus, the expected value of \( N \) is:
\[ E(N) = 6.09 \]
So, the expected value for \( N \) is \( \boxed{6.09} \).