a fair six - sided number cube is rolled 60 times. what is the probability that fewer than 10% of the rolls…

Explanation:

Step1: Calculate the mean and standard deviation of the binomial distribution

The probability of rolling a five on a fair six - sided die is $p=\frac{1}{6}$, and the number of trials $n = 60$. The mean of the binomial distribution is $\mu=np=60\times\frac{1}{6}=10$, and the standard deviation is $\sigma=\sqrt{np(1 - p)}=\sqrt{60\times\frac{1}{6}\times(1-\frac{1}{6})}=\sqrt{10\times\frac{5}{6}}=\sqrt{\frac{25}{3}}\approx2.89$.

Step2: Calculate the number of successes for 10% of the rolls

10% of 60 is $0.1\times60 = 6$. We want to find $P(X<6)$. Using the normal approximation to the binomial (since $np = 10\geq5$ and $n(1 - p)=60\times\frac{5}{6}=50\geq5$), we apply the continuity correction. For $P(X < 6)$ in the binomial, we find $P\left(Z<\frac{5.5 - 10}{2.89}\right)$ (continuity correction: for $X < k$, we use $X<k - 0.5$ in the normal approximation).

Step3: Calculate the z - score

$z=\frac{5.5 - 10}{2.89}=\frac{- 4.5}{2.89}\approx - 1.56$.

Step4: Find the probability using the z - table

Looking up the value of $z=-1.56$ in the z - table, we find that $P(Z < - 1.56)=0.0594\approx0.082$ (rounding differences may occur).

Answer:

0.082