Question:

Step 1: Determine the parameters $n$ and $p$ of the binomial distribution. For a binomial distribution, the mean is given by $np$ and the variance by $np(1-p)$. We are given: $$np = 8$$ $$np(1-p) = 4$$ Divide the variance by the mean: $$\frac{np(1-p)}{np} = \frac{4}{8}$$ $$1-p = \frac{1}{2}$$ $$p = \frac{1}{2}$$ Substitute $p = 1/2$ into the mean equation: $$n\left(\frac{1}{2}\right) = 8$$ $$n = 16$$ Thus, the random variable $X$ follows a binomial distribution with parameters $n=16$ and $p=1/2$. Step 2: Calculate the probabilities $P(X=0)$, $P(X=1)$, and $P(X=2)$. The probability mass function for a binomial distribution is $P(X=x) = \binom{n}{x} p^x (1-p)^{n-x}$. Given $n=16$ and $p=1/2$, we have $1-p=1/2$. So, $P(X=x) = \binom{16}{x} \left(\frac{1}{2}\right)^x \left(\frac{1}{2}\right)^{16-x} = \binom{16}{x} \left(\frac{1}{2}\right)^{16}$. $$P(X=0) = \binom{16}{0} \left(\frac{1}{2}\right)^{16} = 1 \cdot \frac{1}{2^{16}}$$ $$P(X=1) = \binom{16}{1} \left(\frac{1}{2}\right)^{16} = 16 \cdot \frac{1}{2^{16}}$$ $$P(X=2) = \binom{16}{2} \left(\frac{1}{2}\right)^{16} = \frac{16 \times 15}{2 \times 1} \cdot \frac{1}{2^{16}} = 120 \cdot \frac{1}{2^{16}}$$ Step 3: Sum the probabilities to find $P(X \leq 2)$ and determine $k$. $$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$$ $$P(X \leq 2) = \frac{1}{2^{16}} + \frac{16}{2^{16}} + \frac{120}{2^{16}}$$ $$P(X \leq 2) = \frac{1 + 16 + 120}{2^{16}}$$ $$P(X \leq 2) = \frac{137}{2^{16}}$$ We are given that $P(X \leq 2) = k/2^{16}$. By comparing the two expressions for $P(X \leq 2)$, we find: $$k = 137$$