Use $n=6$ and $p=0.65$ to complete parts (a) through (d) below. (a) Construct a binomial probability distribution with the given parameters. \begin{tabular}{cc} $\mathbf{x}$ & $\mathbf{P}(\mathbf{x})$ \\ \hline 0 & 0.0018 \\ \hline 1 & 0.0205 \\ \hline 2 & 0.0951 \\ \hline 3 & 0.2355 \\ \hline 4 & 0.3280 \\ \hline 5 & 0.2437 \\ \hline 6 & 0.0754 \\ \hline \end{tabular} (Round to four decimal places as needed.) (b) Compute the mean and standard deviation of the random variable using $\mu_{X}=\sum[x \cdot P(x)]$ and $\sigma_{X}=\sqrt{\sum\left[x^{2} \cdot P(x)\right]-\mu_{X}^{2}}$. $\mu_{X}=\square$ (Round to two decimal places as needed.) w an example Get more help . Clear all Check answer

Step 1 ：Construct a binomial probability distribution with the given parameters. The table is as follows: \[\begin{tabular}{cc} \(\mathbf{x}\) & \(\mathbf{P}(\mathbf{x})\) \\ \hline 0 & 0.0018 \\ \hline 1 & 0.0205 \\ \hline 2 & 0.0951 \\ \hline 3 & 0.2355 \\ \hline 4 & 0.3280 \\ \hline 5 & 0.2437 \\ \hline 6 & 0.0754 \\ \hline \end{tabular}\]

Step 2 ：Compute the mean of the random variable using the formula \(\mu_{X}=\sum[x \cdot P(x)]\). The mean is calculated as follows: \[\mu_{X} = 0\times0.0018 + 1\times0.0205 + 2\times0.0951 + 3\times0.2355 + 4\times0.3280 + 5\times0.2437 + 6\times0.0754 = 3.9001\]

Step 3 ：Compute the standard deviation of the random variable using the formula \(\sigma_{X}=\sqrt{\sum\left[x^{2} \cdot P(x)\right]-\mu_{X}^{2}}\). The standard deviation is calculated as follows: \[\sigma_{X} = \sqrt{(0^2\times0.0018 + 1^2\times0.0205 + 2^2\times0.0951 + 3^2\times0.2355 + 4^2\times0.3280 + 5^2\times0.2437 + 6^2\times0.0754) - (3.9001)^2} = 1.168126701175862\]

Step 4 ：Final Answer: The mean of the random variable is \(\boxed{3.90}\) and the standard deviation is \(\boxed{1.17}\).