Find the mean, variance, and standard deviation of the binomial distribution with the given values of $n$ and $p$.
\[
n=127, p=0.78
\]
The mean, $\mu$, is 991 . (Round to the nearest tenth as needed.)
The variance, $\sigma^{2}$, is
(Round to the nearest tenth as needed.)
Answer
Final Answer: The mean, \(\mu\), is \(\boxed{99.1}\). The variance, \(\sigma^{2}\), is \(\boxed{21.8}\). The standard deviation, \(\sigma\), is \(\boxed{4.7}\).
Step 1 ：Given that the number of trials, \(n\), is 127 and the probability of success, \(p\), is 0.78.
Step 2 ：We can calculate the mean, variance, and standard deviation of a binomial distribution using the formulas: Mean: \(\mu = np\), Variance: \(\sigma^{2} = np(1-p)\), Standard Deviation: \(\sigma = \sqrt{np(1-p)}\).
Step 3 ：Substitute \(n=127\) and \(p=0.78\) into the formulas to find the mean, variance, and standard deviation.
Step 4 ：The mean, \(\mu\), is calculated as \(127*0.78 = 99.06\).
Step 5 ：The variance, \(\sigma^{2}\), is calculated as \(127*0.78*(1-0.78) = 21.7932\).
Step 6 ：The standard deviation, \(\sigma\), is calculated as \(\sqrt{127*0.78*(1-0.78)} = 4.668318755183711\).
Step 7 ：Round the mean, variance, and standard deviation to the nearest tenth as needed.
Step 8 ：Final Answer: The mean, \(\mu\), is \(\boxed{99.1}\). The variance, \(\sigma^{2}\), is \(\boxed{21.8}\). The standard deviation, \(\sigma\), is \(\boxed{4.7}\).
