# A die is rolled 10 times and the number of twos that come up is tallied. If this experiment is repeated find the standard deviation for number of two?

Dec 31, 2016

The standard deviation is $\sqrt{10 \cdot \frac{1}{6} \cdot \frac{5}{6}} = \frac{5 \sqrt{2}}{6}$.

#### Explanation:

This experiment involves repeating identical independent trials (the rolling of the die), with the same condition for "success" each time (rolling a "2"). So, it will have a binomial distribution—this means the probability of rolling $k$ 2's will be

$\left(\begin{matrix}n \\ k\end{matrix}\right) {p}^{k} {\left(1 - p\right)}^{n - k} , \text{ } k = 0 , 1 , 2 , \ldots , n$

Where:

• $n$ is the number of trials in the experiment ($10$), and
• $p$ is the probability of success in each trial ($\frac{1}{6}$).

Binomial distributions have a mean of $n p$, and a standard deviation of $\sqrt{n p \left(1 - p\right)}$. Thus, the standard deviation $\sigma$ of this distribution is

$\sigma = \sqrt{n p \left(1 - p\right)} = \sqrt{10 \cdot \frac{1}{6} \cdot \left(1 - \frac{1}{6}\right)}$
$\textcolor{w h i t e}{\sigma = \sqrt{n p \left(1 - p\right)}} = \sqrt{10 \cdot \frac{1}{6} \cdot \frac{5}{6}}$
$\textcolor{w h i t e}{\sigma = \sqrt{n p \left(1 - p\right)}} = \sqrt{\frac{50}{36}}$
$\textcolor{w h i t e}{\sigma = \sqrt{n p \left(1 - p\right)}} = \sqrt{\frac{{5}^{2} \cdot 2}{6} ^ 2}$
$\textcolor{w h i t e}{\sigma = \sqrt{n p \left(1 - p\right)}} = \frac{5 \sqrt{2}}{6}$.