# 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?

##### 1 Answer
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

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