## Related questions with answers

Suppose that an honest die is rolled n=180 times. Let the random variable X represent the number of times the number 6 is rolled. (a) Find the mean and standard deviation for the distribution of X. (Hint: Use the dishonest-coin principle with p=1/6 to find

$\mu$

and

$\sigma$

.) (b) Find the probability that a 6 will be rolled more than 40 times. (c) Find the probability that a 6 will be rolled somewhere between 30 and 35 times.

Solution

Verified$\textbf{(a)}$

Using a given hint find the mean and the standard deviation.

$\begin{align*} \mu&=np\\ \mu&=180\times \dfrac{1}{6}\\ \boldsymbol{\mu}&=\boldsymbol{30}\\ \sigma&=\sqrt{np(1-p)}\\ \sigma&=\sqrt{180\times \dfrac{1}{6} \times \dfrac{5}{6}}\\ \boldsymbol{\sigma}&=\boldsymbol{5} \end{align*}$

$\textbf{(b)}$

Because $40=30+2\times 5=\mu+2\sigma=P_{97.5}$ the probability that a $6$ will be rolled more than $40$ times is $100-97.5=\boldsymbol{2.5\%}$.

$\textbf{(c)}$

Because $30=\mu=P_{50}$ and $35=30+5=\mu+\sigma=P_{84}$ the probability that a $6$ will be rolled between $30$ and $35$ times is $84-50=\boldsymbol{34\%}$.

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