Problem Analysis #7

[THOUGHT] The random variable $N$ follows a binomial distribution with parameters $n = 20$ and $p = 1/2$. The fourth moment $\mathbb{E}[N^4]$ is computed using the formula for the fourth moment of a binomial distribution, which involves Stirling numbers of the second kind. The formula is:

where $S(4,1) = 1$, $S(4,2) = 7$, $S(4,3) = 6$, and $S(4,4) = 1$.

Substituting $n = 20$ and $p = 1/2$:

• First term: $1 \cdot 20 \cdot \frac{1}{2} = 10$
• Second term: $7 \cdot 20 \cdot 19 \cdot \left(\frac{1}{2}\right)^2 = 7 \cdot 380 \cdot 0.25 = 7 \cdot 95 = 665$
• Third term: $6 \cdot 20 \cdot 19 \cdot 18 \cdot \left(\frac{1}{2}\right)^3 = 6 \cdot 6840 \cdot 0.125 = 6 \cdot 855 = 5130$
• Fourth term: $1 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot \left(\frac{1}{2}\right)^4 = 20 \cdot 19 \cdot 18 \cdot 17 \cdot 0.0625$

Calculate $20 \cdot 19 = 380$, $380 \cdot 18 = 6840$, $6840 \cdot 17 = 116280$, and $116280 \cdot 0.0625 = 116280 / 16 = 7267.5$.

Summing the terms: $10 + 665 = 675$, $675 + 5130 = 5805$, $5805 + 7267.5 = 13072.5$.

Thus, $\mathbb{E}[N^4] = 13072.5$.

Now, $K = 1000 \cdot \mathbb{E}[N^4] = 1000 \cdot 13072.5 = 13072500$.

The exact integer value of $K$ is 13072500. $\boxed{13072500}$

[THOUGHT] We are to compute the fourth moment of a binomial random variable N ~ Binomial(20, 1/2), then multiply by 1000 to get K. The exact value of K is an integer. We use Python to compute the sum of k^4 * C(20,k) for k from 0 to 20, then multiply by 1000 and divide by 2^20 to get the integer K. [/THOUGHT] [PYTHON] import math n = 20 total = 0 for k in range(0, n+1): binom = math.comb(n, k) total += k4 * binom K = (1000 * total) // (220) print(K) [/PYTHON]

$\boxed{13072500}$