Find the mean, variance, and standard deviation of the binomial distribution with the given values of $n$ and $p$.
\[
n=121, p=0.52
\]
The mean, $\mu$ is 63 . (Round to the nearest tenth as needed)
The variance, $\sigma^{2}$, is $\square$. (Round to the nearest tenth as needed)
The standard deviation, $\sigma$, is
(Round to the nearest tenth as needed.)

Step 1 ：The mean of a binomial distribution is given by \(\mu = np\). Substituting the given values, we get \(\mu = 121 \times 0.52 = 62.92\), which rounds to 63.

Step 2 ：The variance of a binomial distribution is given by \(\sigma^2 = np(1-p)\). Substituting the given values, we get \(\sigma^2 = 121 \times 0.52 \times (1-0.52) = 29.9264\), which rounds to 29.9.

Step 3 ：The standard deviation of a binomial distribution is the square root of the variance. So, \(\sigma = \sqrt{\sigma^2} = \sqrt{29.9264} = 5.47\), which rounds to 5.5.

Step 4 ：Thus, the mean is \(\boxed{63}\), the variance is \(\boxed{29.9}\), and the standard deviation is \(\boxed{5.5}\).