Question

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

μ\mu

and

σ\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

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(a)\textbf{(a)}

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

μ=npμ=180×16μ=30σ=np(1p)σ=180×16×56σ=5\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*}

(b)\textbf{(b)}

Because 40=30+2×5=μ+2σ=P97.540=30+2\times 5=\mu+2\sigma=P_{97.5} the probability that a 66 will be rolled more than 4040 times is 10097.5=2.5%100-97.5=\boldsymbol{2.5\%}.

(c)\textbf{(c)}

Because 30=μ=P5030=\mu=P_{50} and 35=30+5=μ+σ=P8435=30+5=\mu+\sigma=P_{84} the probability that a 66 will be rolled between 3030 and 3535 times is 8450=34%84-50=\boldsymbol{34\%}.

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